Details For MMT-003 Solved Assignment

IGNOU MMT-003 Assignment Question Paper 2023

1. Which of the following statements are true? Give reasons for your answers. Marks will only be given for valid justification of your answers.

i) If $G$$G$GG$G$ is a finite abelian group and $p$$p$pp$p$ is a prime factor of $o(G)$$o(G)$o(G)o(G)$o(G)$, then the number of Sylow p-subgroups of $\mathrm{G}$$\mathrm{G}$Gmathrm{G}$\mathrm{G}$ is a prime.
ii) The minimum polynomial of $5^{1/3}$$5^{1/3}$5^(1//3)5^{1 / 3}$5^{1/3}$ over $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$ is $x^{1/3}$$x^{1/3}$x^(1//3)x^{1 / 3}$x^{1/3}$.
iii) ${\mathbb{Z}}_{\mathrm{m}\mathrm{n}}\simeq {\mathbb{Z}}_{\mathrm{m}}\times {\mathbb{Z}}_{\mathrm{h}}\mathrm{\forall }\mathrm{m},\mathrm{n}\in \mathbb{Z}$${\mathbb{Z}}_{\mathrm{m}\mathrm{n}}\simeq {\mathbb{Z}}_{\mathrm{m}}\times {\mathbb{Z}}_{\mathrm{h}}\mathrm{\forall }\mathrm{m},\mathrm{n}\in \mathbb{Z}$Z_(mn)≃Z_(m)xxZ_(h)AAm,ninZmathbb{Z}_{mathrm{mn}} simeq mathbb{Z}_{mathrm{m}} times mathbb{Z}_{mathrm{h}} forall mathrm{m}, mathrm{n} in mathbb{Z}${\mathbb{Z}}_{\mathrm{m}\mathrm{n}}\simeq {\mathbb{Z}}_{\mathrm{m}}\times {\mathbb{Z}}_{\mathrm{h}}\mathrm{\forall }\mathrm{m},\mathrm{n}\in \mathbb{Z}$
iv) If $G$$G$GG$G$ is a finite group and $m\mid o(G),m\in \mathbb{N}$$m\mid o(G),m\in \mathbb{N}$m∣o(G),m inNm mid o(G), m in mathbb{N}$m\mid o(G),m\in \mathbb{N}$, then $G$$G$GG$G$ has a subgroup of order $m$$m$mm$m$.
v) If ${\beta }_1$${\beta }_1$beta_(1)beta_{1}${\beta }_1$ and ${\beta }_2$${\beta }_2$beta_(2)beta_{2}${\beta }_2$ are two ${15}^{\text{th}}$${15}^{\text{th }}$15^(“th “)15^{text {th }}${15}^{\text{th }}$ roots of unity, then $\mathbb{Q}\left({\beta }_1\right)=\mathbb{Q}\left({\beta }_2\right)$$\mathbb{Q}\left({\beta }_1\right)=\mathbb{Q}\left({\beta }_2\right)$Q(beta_(1))=Q(beta_(2))mathbb{Q}left(beta_{1}right)=mathbb{Q}left(beta_{2}right)$\mathbb{Q}\left({\beta }_1\right)=\mathbb{Q}\left({\beta }_2\right)$.
vi) There exists an extension field of ${\mathbb{Z}}_3$${\mathbb{Z}}_3$Z_(3)mathbb{Z}_{3}${\mathbb{Z}}_3$ of order 25 .
vii) Every group of order 18 has a normal subgroup of order 2.
viii) If $\mathrm{I}$$\mathrm{I}$Imathrm{I}$\mathrm{I}$ and $\mathrm{J}$$\mathrm{J}$Jmathrm{J}$\mathrm{J}$ are ideals of a ring $\mathrm{R}$$\mathrm{R}$Rmathrm{R}$\mathrm{R}$, then $\mathrm{I}\mathrm{J}=\mathrm{I}\cap \mathrm{J}$$\mathrm{I}\mathrm{J}=\mathrm{I}\cap \mathrm{J}$IJ=InnJmathrm{IJ}=mathrm{I} cap mathrm{J}$\mathrm{I}\mathrm{J}=\mathrm{I}\cap \mathrm{J}$.
ix) If $\mathrm{f}:\mathrm{R}\to \mathrm{S}$$\mathrm{f}:\mathrm{R}\to \mathrm{S}$f:RrarrSmathrm{f}: mathrm{R} rightarrow mathrm{S}$\mathrm{f}:\mathrm{R}\to \mathrm{S}$ is a ring homomorphism and $\mathrm{I}$$\mathrm{I}$Imathrm{I}$\mathrm{I}$ is an ideal of $\mathrm{R}$$\mathrm{R}$Rmathrm{R}$\mathrm{R}$, then $\mathrm{f}(\mathrm{I})$$\mathrm{f}(\mathrm{I})$f(I)mathrm{f}(mathrm{I})$\mathrm{f}(\mathrm{I})$ is an ideal of S.
$\mathrm{x}$$\mathrm{x}$xmathrm{x}$\mathrm{x}$ Every prime ideal of an integral domain is a maximal ideal.

2. a) Let $G$$G$GG$G$ be a group and let $H\le G,K\le G,o(H)=o(K)=p$$H\le G,K\le G,o(H)=o(K)=p$H <= G,K <= G,o(H)=o(K)=pH leq G, K leq G, o(H)=o(K)=p$H\le G,K\le G,o(H)=o(K)=p$, a prime. Show that either $\mathrm{H}\cap \mathrm{K}=\{\mathrm{e}\}$$\mathrm{H}\cap \mathrm{K}=\{\mathrm{e}\}$HnnK={e}mathrm{H} cap mathrm{K}={mathrm{e}}$\mathrm{H}\cap \mathrm{K}=\{\mathrm{e}\}$ or $\mathrm{H}=\mathrm{K}$$\mathrm{H}=\mathrm{K}$H=Kmathrm{H}=mathrm{K}$\mathrm{H}=\mathrm{K}$. Is this result still true if $\mathrm{p}$$\mathrm{p}$pmathrm{p}$\mathrm{p}$ is not a prime? Justify your answer.

b) Let $\mathrm{G}$$\mathrm{G}$Gmathrm{G}$\mathrm{G}$ be the group of all rigid motions of a plane and $\mathrm{S}$$\mathrm{S}$Smathrm{S}$\mathrm{S}$ be the set of all rectangles in the plane. Show that $G$$G$GG$G$ acts on S . Also obtain the orbit and stabiliser of a square under this action.
c) Let $G$$G$GG$G$ be a finite group and $H$$H$HH$H$ be a normal subgroup of $G$$G$GG$G$. Prove that $H=\cup C_x$$H=\cup C_x$H=uuC_(x)H=cup C_{x}$H=\cup C_x$ where the $C_x$$C_x$C_(x)C_{x}$C_x$ are all the distinct conjugacy classes of $G$$G$GG$G$ such that $H\cap C_x\ne \varnothing$$H\cap C_x\ne \varnothing$H nnC_(x)!=O/H cap C_{x} neq varnothing$H\cap C_x\ne \varnothing$.

3. a) Find the number of Sylow 5-subgroups, Sylow 7-subgroups and Sylow 2-subgroups $A_5$$A_5$A_(5)A_{5}$A_5$ has.

b) Let $G_1$$G_1$G_(1)G_{1}$G_1$ and $G_2$$G_2$G_(2)G_{2}$G_2$ be finite groups such that $p$$p$pp$p$ divides $\left|G_1\right|$$\left|G_1\right|$|G_(1)|left|G_{1}right|$\left|G_1\right|$ and $\left|G_2\right|$$\left|G_2\right|$|G_(2)|left|G_{2}right|$\left|G_2\right|$. Prove that the Sylow p-subgroups of $G_1\times G_2$$G_1\times G_2$G_(1)xxG_(2)G_{1} times G_{2}$G_1\times G_2$ are precisely of the form $P_1\times P_2$$P_1\times P_2$P_(1)xxP_(2)P_{1} times P_{2}$P_1\times P_2$, where $P_1$$P_1$P_(1)P_{1}$P_1$ and $P_2$$P_2$P_(2)P_{2}$P_2$ are Sylow p-subgroups of $G_1$$G_1$G_(1)G_{1}$G_1$ and $G_2$$G_2$G_(2)G_{2}$G_2$, respectively.

4. a) Write $\left[\begin{array}{ccc}1& -1& 0\\ 2& 0& 3\\ 5& 5& 1\end{array}\right]$$\left[\begin{array}{ccc}1& -1& 0\\ 2& 0& 3\\ 5& 5& 1\end{array}\right]$[[1,-1,0],[2,0,3],[5,5,1]]left[begin{array}{ccc}1 & -1 & 0 \ 2 & 0 & 3 \ 5 & 5 & 1end{array}right]$\left[\begin{array}{ccc}1& -1& 0\\ 2& 0& 3\\ 5& 5& 1\end{array}\right]$ as a product of $\mathrm{O}(3)$$\mathrm{O}(3)$O(3)mathrm{O}(3)$\mathrm{O}(3)$ and an element of $B_3(\mathbb{R})$$B_3(\mathbb{R})$B_(3)(R)B_{3}(mathbb{R})$B_3(\mathbb{R})$.

b) Show that $\mathrm{S}\mathrm{U}(2)$$\mathrm{S}\mathrm{U}(2)$SU(2)mathrm{SU}(2)$\mathrm{S}\mathrm{U}(2)$ and ${\mathrm{S}}^3$${\mathrm{S}}^3$S^(3)mathrm{S}^{3}${\mathrm{S}}^3$ are structurally the same.

5. a) Find all the possible abelian groups, up to isomorphism, of order 900 .

b) Construct the free group on the set $\{\alpha ,\beta ,\gamma \}$$\{\alpha ,\beta ,\gamma \}${alpha,beta,gamma}{alpha, beta, gamma}$\{\alpha ,\beta ,\gamma \}$. Further, check if it is a free abelian group or not.

6. a) Check whether or not the ring $\mathrm{R}={\mathbb{Z}}_3[\mathrm{x}]/<{\mathrm{x}}^6-1>$$\mathrm{R}={\mathbb{Z}}_3[\mathrm{x}]/<{\mathrm{x}}^6-1>$R=Z_(3)[x]// < x^(6)-1 >mathrm{R}=mathbb{Z}_{3}[mathrm{x}] /<mathrm{x}^{6}-1>$\mathrm{R}={\mathbb{Z}}_3[\mathrm{x}]/<{\mathrm{x}}^6-1>$
   i) is finite;
   ii) has zero divisors;
   iii) has nilpotent elements.

b) Give two distinct rings whose quotient field is $\{a+ib\mid a,b\in \mathbb{R}\}$$\{a+ib\mid a,b\in \mathbb{R}\}${a+ib∣a,b inR}{a+i b mid a, b in mathbb{R}}$\{a+ib\mid a,b\in \mathbb{R}\}$. Justify your answer.
c) Evaluate the following Legendre Symbols:
i) $\left(\frac{139}{431}\right)$$\left(\frac{139}{431}\right)$((139)/(431))left(frac{139}{431}right)$\left(\frac{139}{431}\right)$
ii) $\left(\frac{149}{439}\right)$$\left(\frac{149}{439}\right)$((149)/(439))left(frac{149}{439}right)$\left(\frac{149}{439}\right)$
d) Check whether 9782957210008 is a valid ISBN number.

7. a) Check whether or not $\mathbb{Z}[\sqrt{-7}]$$\mathbb{Z}[\sqrt{-7}]$Z[sqrt(-7)]mathbb{Z}[sqrt{-7}]$\mathbb{Z}[\sqrt{-7}]$ is a Euclidean domain.

b) Use the division algorithm to find the inverse of $\overline{18}$$\overline{18}$bar(18)overline{18}$\overline{18}$ in ${\mathbb{Z}}_{35}$${\mathbb{Z}}_{35}$Z_(35)mathbb{Z}_{35}${\mathbb{Z}}_{35}$.

8. i) Let $G$$G$GG$G$ be a group of automorphisms of a field $K$$K$KK$K$. Is the fixed field $K^G$$K^G$K^(G)K^{G}$K^G$ a subfield of $\mathrm{K}$$\mathrm{K}$Kmathrm{K}$\mathrm{K}$ ? Why, or why not?

ii) Find $K^G$$K^G$K^(G)K^{G}$K^G$, where $K=\mathbb{Q}(i,\sqrt{3}),G=G(K/\mathbb{Q})$$K=\mathbb{Q}(i,\sqrt{3}),G=G(K/\mathbb{Q})$K=Q(i,sqrt3),G=G(K//Q)K=mathbb{Q}(i, sqrt{3}), G=G(K / mathbb{Q})$K=\mathbb{Q}(i,\sqrt{3}),G=G(K/\mathbb{Q})$.


[stray-random categories=trigonometry]

MMT-003 Sample Solution 2023

MMT-003 Solved Assignment 2023

Question:-01

1. Which of the following statements are true? Give reasons for your answers. Marks will only be given for valid justification of your answers.
   i) If $\mathrm{G}$$\mathrm{G}$Gmathrm{G}$\mathrm{G}$ is a finite abelian group and $\mathrm{p}$$\mathrm{p}$pmathrm{p}$\mathrm{p}$ is a prime factor of $\mathrm{o}(\mathrm{G})$$\mathrm{o}(\mathrm{G})$o(G)mathrm{o}(mathrm{G})$\mathrm{o}(\mathrm{G})$, then the number of Sylow p-subgroups of $\mathrm{G}$$\mathrm{G}$Gmathrm{G}$\mathrm{G}$ is a prime.

Answer:
The statement “If $G$$G$GG$G$ is a finite abelian group and $p$$p$pp$p$ is a prime factor of $o(G)$$o(G)$o(G)o(G)$o(G)$, then the number of Sylow $p$$p$pp$p$-subgroups of $G$$G$GG$G$ is a prime” is false.

Justification:

In a finite abelian group $G$$G$GG$G$, every Sylow $p$$p$pp$p$-subgroup is normal. Therefore, all Sylow $p$$p$pp$p$-subgroups are conjugate to each other, and they are all isomorphic. In an abelian group, the number of Sylow $p$$p$pp$p$-subgroups is given by $n_p$$n_p$n_(p)n_p$n_p$, and according to Sylow’s Third Theorem, $n_p$$n_p$n_(p)n_p$n_p$ divides $|G|$$|G|$|G||G|$|G|$ and $n_p\equiv 1\mathrm{mod}p$$n_p\equiv 1\mathrm{mod}p$n_(p)-=1modpn_p equiv 1 mod p$n_p\equiv 1\mathrm{mod}p$.
However, there is no requirement that $n_p$$n_p$n_(p)n_p$n_p$ must be a prime number.

Counterexample:

Consider the abelian group $G={\mathbb{Z}}_4\times {\mathbb{Z}}_2$$G={\mathbb{Z}}_4\times {\mathbb{Z}}_2$G=Z_(4)xxZ_(2)G = mathbb{Z}_4 times mathbb{Z}_2$G={\mathbb{Z}}_4\times {\mathbb{Z}}_2$, where ${\mathbb{Z}}_n$${\mathbb{Z}}_n$Z_(n)mathbb{Z}_n${\mathbb{Z}}_n$ denotes the integers modulo $n$$n$nn$n$. The order of $G$$G$GG$G$ is $o(G)=4\times 2=8$$o(G)=4\times 2=8$o(G)=4xx2=8o(G) = 4 times 2 = 8$o(G)=4\times 2=8$.
The prime factors of $o(G)$$o(G)$o(G)o(G)$o(G)$ are $2$$2$22$2$ and $2$$2$22$2$ (i.e., $8=2^3$$8=2^3$8=2^(3)8 = 2^3$8=2^3$).
The Sylow $2$$2$22$2$-subgroups of $G$$G$GG$G$ are generated by $(2,0)$$(2,0)$(2,0)(2, 0)$(2,0)$, $(0,1)$$(0,1)$(0,1)(0, 1)$(0,1)$, and $(2,1)$$(2,1)$(2,1)(2, 1)$(2,1)$. So, there are $3$$3$33$3$ Sylow $2$$2$22$2$-subgroups, and $3$$3$33$3$ is not a prime divisor of $8$$8$88$8$.
Thus, the statement is false, as demonstrated by this counterexample.

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ii) The minimum polynomial of $5^{1/3}$$5^{1/3}$5^(1//3)5^{1 / 3}$5^{1/3}$ over $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$ is $x^{1/3}$$x^{1/3}$x^(1//3)x^{1 / 3}$x^{1/3}$.
Answer:
The statement “The minimum polynomial of $5^{1/3}$$5^{1/3}$5^(1//3)5^{1/3}$5^{1/3}$ over $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$ is $x^{1/3}$$x^{1/3}$x^(1//3)x^{1/3}$x^{1/3}$” is false.

Justification:

The minimum polynomial of an algebraic number $\alpha$$\alpha$alphaalpha$\alpha$ over a field $F$$F$FF$F$ is the unique monic polynomial $f(x)$$f(x)$f(x)f(x)$f(x)$ of least degree such that $f(\alpha )=0$$f(\alpha )=0$f(alpha)=0f(alpha) = 0$f(\alpha )=0$ and $f(x)$$f(x)$f(x)f(x)$f(x)$ has coefficients in $F$$F$FF$F$.
For $5^{1/3}$$5^{1/3}$5^(1//3)5^{1/3}$5^{1/3}$, the minimum polynomial over $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$ is $x^3-5$$x^3-5$x^(3)-5x^3 – 5$x^3-5$, not $x^{1/3}$$x^{1/3}$x^(1//3)x^{1/3}$x^{1/3}$.

1. $x^3-5$$x^3-5$x^(3)-5x^3 – 5$x^3-5$ is a polynomial with coefficients in $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$.
2. $(5^{1/3}{)}^3-5=5-5=0$$(5^{1/3}{)}^3-5=5-5=0$(5^(1//3))^(3)-5=5-5=0(5^{1/3})^3 – 5 = 5 – 5 = 0$(5^{1/3}{)}^3-5=5-5=0$, so $5^{1/3}$$5^{1/3}$5^(1//3)5^{1/3}$5^{1/3}$ is a root of $x^3-5$$x^3-5$x^(3)-5x^3 – 5$x^3-5$.
3. $x^3-5$$x^3-5$x^(3)-5x^3 – 5$x^3-5$ is irreducible over $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$, which means it has the least degree among all polynomials that have $5^{1/3}$$5^{1/3}$5^(1//3)5^{1/3}$5^{1/3}$ as a root and have coefficients in $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$.

Moreover, $x^{1/3}$$x^{1/3}$x^(1//3)x^{1/3}$x^{1/3}$ is not even a polynomial over $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$ because its exponent $1/3$$1/3$1//31/3$1/3$ is not a non-negative integer.
Thus, the statement is false.

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iii) ${\mathbb{Z}}_{\mathrm{m}\mathrm{n}}\simeq {\mathbb{Z}}_{\mathrm{m}}\times {\mathbb{Z}}_{\mathrm{n}}\mathrm{\forall }\mathrm{m},\mathrm{n}\in \mathbb{Z}$${\mathbb{Z}}_{\mathrm{m}\mathrm{n}}\simeq {\mathbb{Z}}_{\mathrm{m}}\times {\mathbb{Z}}_{\mathrm{n}}\mathrm{\forall }\mathrm{m},\mathrm{n}\in \mathbb{Z}$Z_(mn)≃Z_(m)xxZ_(n)AAm,ninZmathbb{Z}_{mathrm{mn}} simeq mathbb{Z}_{mathrm{m}} times mathbb{Z}_{mathrm{n}} forall mathrm{m}, mathrm{n} in mathbb{Z}${\mathbb{Z}}_{\mathrm{m}\mathrm{n}}\simeq {\mathbb{Z}}_{\mathrm{m}}\times {\mathbb{Z}}_{\mathrm{n}}\mathrm{\forall }\mathrm{m},\mathrm{n}\in \mathbb{Z}$
Answer:
The statement “${\mathbb{Z}}_{mn}\simeq {\mathbb{Z}}_m\times {\mathbb{Z}}_n\mathrm{\forall }m,n\in \mathbb{Z}$${\mathbb{Z}}_{mn}\simeq {\mathbb{Z}}_m\times {\mathbb{Z}}_n\mathrm{\forall }m,n\in \mathbb{Z}$Z_(mn)≃Z_(m)xxZ_(n)AA m,n inZmathbb{Z}_{mn} simeq mathbb{Z}_{m} times mathbb{Z}_{n} forall m, n in mathbb{Z}${\mathbb{Z}}_{mn}\simeq {\mathbb{Z}}_m\times {\mathbb{Z}}_n\mathrm{\forall }m,n\in \mathbb{Z}$” is false.

Justification:

The isomorphism ${\mathbb{Z}}_{mn}\simeq {\mathbb{Z}}_m\times {\mathbb{Z}}_n$${\mathbb{Z}}_{mn}\simeq {\mathbb{Z}}_m\times {\mathbb{Z}}_n$Z_(mn)≃Z_(m)xxZ_(n)mathbb{Z}_{mn} simeq mathbb{Z}_{m} times mathbb{Z}_{n}${\mathbb{Z}}_{mn}\simeq {\mathbb{Z}}_m\times {\mathbb{Z}}_n$ holds if and only if $m$$m$mm$m$ and $n$$n$nn$n$ are coprime (i.e., $gcd(m,n)=1$$gcd(m,n)=1$gcd(m,n)=1gcd(m, n) = 1$gcd(m,n)=1$).

Counterexample:

Let $m=4$$m=4$m=4m = 4$m=4$ and $n=2$$n=2$n=2n = 2$n=2$. Then ${\mathbb{Z}}_{4\times 2}={\mathbb{Z}}_8$${\mathbb{Z}}_{4\times 2}={\mathbb{Z}}_8$Z_(4xx2)=Z_(8)mathbb{Z}_{4 times 2} = mathbb{Z}_8${\mathbb{Z}}_{4\times 2}={\mathbb{Z}}_8$.
However, ${\mathbb{Z}}_4\times {\mathbb{Z}}_2$${\mathbb{Z}}_4\times {\mathbb{Z}}_2$Z_(4)xxZ_(2)mathbb{Z}_4 times mathbb{Z}_2${\mathbb{Z}}_4\times {\mathbb{Z}}_2$ is not isomorphic to ${\mathbb{Z}}_8$${\mathbb{Z}}_8$Z_(8)mathbb{Z}_8${\mathbb{Z}}_8$.

1. ${\mathbb{Z}}_8$${\mathbb{Z}}_8$Z_(8)mathbb{Z}_8${\mathbb{Z}}_8$ has elements $[0],[1],[2],[3],[4],[5],[6],[7]$$[0],[1],[2],[3],[4],[5],[6],[7]$[0],[1],[2],[3],[4],[5],[6],[7][0], [1], [2], [3], [4], [5], [6], [7]$[0],[1],[2],[3],[4],[5],[6],[7]$ and is cyclic, generated by $[1]$$[1]$[1][1]$[1]$.
2. ${\mathbb{Z}}_4\times {\mathbb{Z}}_2$${\mathbb{Z}}_4\times {\mathbb{Z}}_2$Z_(4)xxZ_(2)mathbb{Z}_4 times mathbb{Z}_2${\mathbb{Z}}_4\times {\mathbb{Z}}_2$ has elements $(0,0),(0,1),(1,0),(1,1),(2,0),(2,1),(3,0),(3,1)$$(0,0),(0,1),(1,0),(1,1),(2,0),(2,1),(3,0),(3,1)$(0,0),(0,1),(1,0),(1,1),(2,0),(2,1),(3,0),(3,1)(0,0), (0,1), (1,0), (1,1), (2,0), (2,1), (3,0), (3,1)$(0,0),(0,1),(1,0),(1,1),(2,0),(2,1),(3,0),(3,1)$ and is not cyclic.

Since ${\mathbb{Z}}_8$${\mathbb{Z}}_8$Z_(8)mathbb{Z}_8${\mathbb{Z}}_8$ is cyclic and ${\mathbb{Z}}_4\times {\mathbb{Z}}_2$${\mathbb{Z}}_4\times {\mathbb{Z}}_2$Z_(4)xxZ_(2)mathbb{Z}_4 times mathbb{Z}_2${\mathbb{Z}}_4\times {\mathbb{Z}}_2$ is not, they cannot be isomorphic.
Thus, the statement is false, as demonstrated by this counterexample.

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iv) If $\mathrm{G}$$\mathrm{G}$Gmathrm{G}$\mathrm{G}$ is a finite group and $\mathrm{m}\mid \mathrm{o}(\mathrm{G}),\mathrm{m}\in \mathbb{N}$$\mathrm{m}\mid \mathrm{o}(\mathrm{G}),\mathrm{m}\in \mathbb{N}$m∣o(G),minNmathrm{m} mid mathrm{o}(mathrm{G}), mathrm{m} in mathbb{N}$\mathrm{m}\mid \mathrm{o}(\mathrm{G}),\mathrm{m}\in \mathbb{N}$, then $\mathrm{G}$$\mathrm{G}$Gmathrm{G}$\mathrm{G}$ has a subgroup of order $\mathrm{m}$$\mathrm{m}$mmathrm{m}$\mathrm{m}$.
Answer:
The statement “If $G$$G$GG$G$ is a finite group and $m\mid o(G),m\in \mathbb{N}$$m\mid o(G),m\in \mathbb{N}$m∣o(G),m inNm mid o(G), m in mathbb{N}$m\mid o(G),m\in \mathbb{N}$, then $G$$G$GG$G$ has a subgroup of order $m$$m$mm$m$” is false.

Justification:

The statement would be true if $G$$G$GG$G$ were a cyclic group, as every divisor of the order of a cyclic group corresponds to a unique subgroup. However, the statement is not generally true for all finite groups.

Counterexample:

Consider the symmetric group $S_3$$S_3$S_(3)S_3$S_3$ with $o(S_3)=6$$o(S_3)=6$o(S_(3))=6o(S_3) = 6$o(S_3)=6$. The divisors of 6 are $1,2,3,6$$1,2,3,6$1,2,3,61, 2, 3, 6$1,2,3,6$.
While $S_3$$S_3$S_(3)S_3$S_3$ does have subgroups of orders 1, 2, 3, and 6, it does not have a subgroup of order 4, even though 4 is a natural number ($m\in \mathbb{N}$$m\in \mathbb{N}$m inNm in mathbb{N}$m\in \mathbb{N}$).
Thus, the statement is false, as demonstrated by this counterexample.

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v) If ${\beta }_1$${\beta }_1$beta_(1)beta_1${\beta }_1$ and ${\beta }_2$${\beta }_2$beta_(2)beta_2${\beta }_2$ are two ${15}^{\text{th}}$${15}^{\text{th }}$15^(“th “)15^{text {th }}${15}^{\text{th }}$ roots of unity, then $\mathbb{Q}\left({\beta }_1\right)=\mathbb{Q}\left({\beta }_2\right)$$\mathbb{Q}\left({\beta }_1\right)=\mathbb{Q}\left({\beta }_2\right)$Q(beta_(1))=Q(beta_(2))mathbb{Q}left(beta_1right)=mathbb{Q}left(beta_2right)$\mathbb{Q}\left({\beta }_1\right)=\mathbb{Q}\left({\beta }_2\right)$.
Answer:
The statement “If ${\beta }_1$${\beta }_1$beta_(1)beta_1${\beta }_1$ and ${\beta }_2$${\beta }_2$beta_(2)beta_2${\beta }_2$ are two ${15}^{\text{th}}$${15}^{\text{th}}$15^(“th”)15^{text{th}}${15}^{\text{th}}$ roots of unity, then $\mathbb{Q}({\beta }_1)=\mathbb{Q}({\beta }_2)$$\mathbb{Q}({\beta }_1)=\mathbb{Q}({\beta }_2)$Q(beta_(1))=Q(beta_(2))mathbb{Q}(beta_1) = mathbb{Q}(beta_2)$\mathbb{Q}({\beta }_1)=\mathbb{Q}({\beta }_2)$” is true.

Justification:

The ${15}^{\text{th}}$${15}^{\text{th}}$15^(“th”)15^{text{th}}${15}^{\text{th}}$ roots of unity are the complex numbers of the form
$${\beta }_k=\cos \left(\frac{2\pi k}{15}\right)+i\sin \left(\frac{2\pi k}{15}\right),$$$${\beta }_k=\cos \left(\frac{2\pi k}{15}\right)+i\sin \left(\frac{2\pi k}{15}\right),$$beta _(k)=cos((2pi k)/(15))+i sin((2pi k)/(15)),beta_k = cosleft(frac{2pi k}{15}right) + i sinleft(frac{2pi k}{15}right),$${\beta }_k=\cos \left(\frac{2\pi k}{15}\right)+i\sin \left(\frac{2\pi k}{15}\right),$$
where $i$$i$ii$i$ is the imaginary unit and $k=0,1,2,\dots ,14$$k=0,1,2,\dots ,14$k=0,1,2,dots,14k = 0, 1, 2, ldots, 14$k=0,1,2,\dots ,14$.
These roots of unity are all solutions to the equation $x^{15}-1=0$$x^{15}-1=0$x^(15)-1=0x^{15} – 1 = 0$x^{15}-1=0$, and they generate a cyclotomic field $\mathbb{Q}({\zeta }_{15})$$\mathbb{Q}({\zeta }_{15})$Q(zeta_(15))mathbb{Q}(zeta_{15})$\mathbb{Q}({\zeta }_{15})$, where ${\zeta }_{15}=e^{2\pi i/15}$${\zeta }_{15}=e^{2\pi i/15}$zeta_(15)=e^(2pi i//15)zeta_{15} = e^{2pi i / 15}${\zeta }_{15}=e^{2\pi i/15}$.
Now, any ${15}^{\text{th}}$${15}^{\text{th}}$15^(“th”)15^{text{th}}${15}^{\text{th}}$ root of unity can be expressed as a power of ${\zeta }_{15}$${\zeta }_{15}$zeta_(15)zeta_{15}${\zeta }_{15}$. Specifically, ${\beta }_1={\zeta }_{15}^{k_1}$${\beta }_1={\zeta }_{15}^{k_1}$beta_(1)=zeta_(15)^(k_(1))beta_1 = zeta_{15}^{k_1}${\beta }_1={\zeta }_{15}^{k_1}$ and ${\beta }_2={\zeta }_{15}^{k_2}$${\beta }_2={\zeta }_{15}^{k_2}$beta_(2)=zeta_(15)^(k_(2))beta_2 = zeta_{15}^{k_2}${\beta }_2={\zeta }_{15}^{k_2}$ for some $k_1,k_2\in \{0,1,2,\dots ,14\}$$k_1,k_2\in \{0,1,2,\dots ,14\}$k_(1),k_(2)in{0,1,2,dots,14}k_1, k_2 in {0, 1, 2, ldots, 14}$k_1,k_2\in \{0,1,2,\dots ,14\}$.
Since both ${\beta }_1$${\beta }_1$beta_(1)beta_1${\beta }_1$ and ${\beta }_2$${\beta }_2$beta_(2)beta_2${\beta }_2$ can be expressed using powers of ${\zeta }_{15}$${\zeta }_{15}$zeta_(15)zeta_{15}${\zeta }_{15}$, it follows that $\mathbb{Q}({\beta }_1)$$\mathbb{Q}({\beta }_1)$Q(beta_(1))mathbb{Q}(beta_1)$\mathbb{Q}({\beta }_1)$ and $\mathbb{Q}({\beta }_2)$$\mathbb{Q}({\beta }_2)$Q(beta_(2))mathbb{Q}(beta_2)$\mathbb{Q}({\beta }_2)$ are both subfields of $\mathbb{Q}({\zeta }_{15})$$\mathbb{Q}({\zeta }_{15})$Q(zeta_(15))mathbb{Q}(zeta_{15})$\mathbb{Q}({\zeta }_{15})$.
Moreover, ${\beta }_1$${\beta }_1$beta_(1)beta_1${\beta }_1$ and ${\beta }_2$${\beta }_2$beta_(2)beta_2${\beta }_2$ generate the same field as ${\zeta }_{15}$${\zeta }_{15}$zeta_(15)zeta_{15}${\zeta }_{15}$ because they are powers of ${\zeta }_{15}$${\zeta }_{15}$zeta_(15)zeta_{15}${\zeta }_{15}$. Therefore, $\mathbb{Q}({\beta }_1)=\mathbb{Q}({\zeta }_{15})$$\mathbb{Q}({\beta }_1)=\mathbb{Q}({\zeta }_{15})$Q(beta_(1))=Q(zeta_(15))mathbb{Q}(beta_1) = mathbb{Q}(zeta_{15})$\mathbb{Q}({\beta }_1)=\mathbb{Q}({\zeta }_{15})$ and $\mathbb{Q}({\beta }_2)=\mathbb{Q}({\zeta }_{15})$$\mathbb{Q}({\beta }_2)=\mathbb{Q}({\zeta }_{15})$Q(beta_(2))=Q(zeta_(15))mathbb{Q}(beta_2) = mathbb{Q}(zeta_{15})$\mathbb{Q}({\beta }_2)=\mathbb{Q}({\zeta }_{15})$, which implies $\mathbb{Q}({\beta }_1)=\mathbb{Q}({\beta }_2)$$\mathbb{Q}({\beta }_1)=\mathbb{Q}({\beta }_2)$Q(beta_(1))=Q(beta_(2))mathbb{Q}(beta_1) = mathbb{Q}(beta_2)$\mathbb{Q}({\beta }_1)=\mathbb{Q}({\beta }_2)$.
Thus, the statement is true.

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vi) There exists an extension field of ${\mathbb{Z}}_3$${\mathbb{Z}}_3$Z_(3)mathbb{Z}_3${\mathbb{Z}}_3$ of order 25 .
Answer:
The statement “There exists an extension field of ${\mathbb{Z}}_3$${\mathbb{Z}}_3$Z_(3)mathbb{Z}_3${\mathbb{Z}}_3$ of order 25″ is false.

Justification:

An extension field $K$$K$KK$K$ of a finite field $F$$F$FF$F$ of order $p^n$$p^n$p^(n)p^n$p^n$ (where $p$$p$pp$p$ is a prime and $n$$n$nn$n$ is a positive integer) must have order $p^m$$p^m$p^(m)p^m$p^m$ for some integer $m>n$$m>n$m > nm > n$m>n$. This is because the extension field $K$$K$KK$K$ can be thought of as a vector space over $F$$F$FF$F$, and the size of $K$$K$KK$K$ must be $p^m$$p^m$p^(m)p^m$p^m$ to be compatible with the vector space structure.
In the given statement, ${\mathbb{Z}}_3$${\mathbb{Z}}_3$Z_(3)mathbb{Z}_3${\mathbb{Z}}_3$ is a field of order $3$$3$33$3$, and we are asked about an extension field of order $25$$25$2525$25$. The number $25=5^2$$25=5^2$25=5^(2)25 = 5^2$25=5^2$ is not of the form $3^m$$3^m$3^(m)3^m$3^m$ for any integer $m$$m$mm$m$.
Therefore, there cannot exist an extension field of ${\mathbb{Z}}_3$${\mathbb{Z}}_3$Z_(3)mathbb{Z}_3${\mathbb{Z}}_3$ with order $25$$25$2525$25$.
Thus, the statement is false.

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vii) Every group of order 18 has a normal subgroup of order 2 .
Answer:
The statement “Every group of order 18 has a normal subgroup of order 2” is false.

Justification:

A group of order 18 has order $2\times 9$$2\times 9$2xx92 times 9$2\times 9$, which is divisible by the primes 2 and 3. According to the Sylow theorems, for a group $G$$G$GG$G$ of order $18=2\times 9$$18=2\times 9$18=2xx918 = 2 times 9$18=2\times 9$, the number $n_2$$n_2$n_(2)n_2$n_2$ of Sylow 2-subgroups must divide 9 and be congruent to 1 modulo 2. The possible values for $n_2$$n_2$n_(2)n_2$n_2$ are therefore 1, 3, or 9.

Counterexample:

Consider the dihedral group $D_9$$D_9$D_(9)D_9$D_9$, which is the group of symmetries of a regular 9-gon. The order of $D_9$$D_9$D_(9)D_9$D_9$ is 18. In $D_9$$D_9$D_(9)D_9$D_9$, the number of Sylow 2-subgroups is 9, and none of them are normal in $D_9$$D_9$D_(9)D_9$D_9$.
Thus, $D_9$$D_9$D_(9)D_9$D_9$ is a group of order 18 that does not have a normal subgroup of order 2, making the statement false.

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viii) If $\mathrm{I}$$\mathrm{I}$Imathrm{I}$\mathrm{I}$ and $\mathrm{J}$$\mathrm{J}$Jmathrm{J}$\mathrm{J}$ are ideals of a ring $\mathrm{R}$$\mathrm{R}$Rmathrm{R}$\mathrm{R}$, then $\mathrm{I}\mathrm{J}=\mathrm{I}\cap \mathrm{J}$$\mathrm{I}\mathrm{J}=\mathrm{I}\cap \mathrm{J}$IJ=InnJmathrm{IJ}=mathrm{I} cap mathrm{J}$\mathrm{I}\mathrm{J}=\mathrm{I}\cap \mathrm{J}$.
Answer:
The statement “If $I$$I$II$I$ and $J$$J$JJ$J$ are ideals of a ring $R$$R$RR$R$, then $IJ=I\cap J$$IJ=I\cap J$IJ=I nn JIJ = I cap J$IJ=I\cap J$” is false.

Justification:

1. Product of Ideals $IJ$$IJ$IJIJ$IJ$: The product $IJ$$IJ$IJIJ$IJ$ is defined as the set $\{\sum _{i=1}^n{a}_i{b}_i\mid a_i\in I,b_i\in J,n\in \mathbb{N}\}$$\{\sum _{i=1}^n a_i{b}_i\mid a_i\in I,b_i\in J,n\in \mathbb{N}\}${sum_(i=1)^(n)a_(i)b_(i)∣a_(i)in I,b_(i)in J,n inN}{ sum_{i=1}^{n} a_i b_i mid a_i in I, b_i in J, n in mathbb{N} }$\{\sum _{i=1}^n{a}_i{b}_i\mid a_i\in I,b_i\in J,n\in \mathbb{N}\}$. It is also an ideal of $R$$R$RR$R$.
2. Intersection of Ideals $I\cap J$$I\cap J$I nn JI cap J$I\cap J$: The intersection $I\cap J$$I\cap J$I nn JI cap J$I\cap J$ is the set of all elements that are both in $I$$I$II$I$ and $J$$J$JJ$J$. It is also an ideal of $R$$R$RR$R$.

Counterexample:

Consider the ring $R=\mathbb{Z}$$R=\mathbb{Z}$R=ZR = mathbb{Z}$R=\mathbb{Z}$ of integers. Let $I=2\mathbb{Z}$$I=2\mathbb{Z}$I=2ZI = 2mathbb{Z}$I=2\mathbb{Z}$ and $J=3\mathbb{Z}$$J=3\mathbb{Z}$J=3ZJ = 3mathbb{Z}$J=3\mathbb{Z}$.

1. $IJ=\{2\times 3,2\times 6,2\times 9,\dots ,4\times 3,4\times 6,\dots \}=6\mathbb{Z}$$IJ=\{2\times 3,2\times 6,2\times 9,\dots ,4\times 3,4\times 6,\dots \}=6\mathbb{Z}$IJ={2xx3,2xx6,2xx9,dots,4xx3,4xx6,dots}=6ZIJ = { 2 times 3, 2 times 6, 2 times 9, ldots, 4 times 3, 4 times 6, ldots } = 6mathbb{Z}$IJ=\{2\times 3,2\times 6,2\times 9,\dots ,4\times 3,4\times 6,\dots \}=6\mathbb{Z}$
2. $I\cap J=\{n\in \mathbb{Z}\mid n\text{is divisible by both 2 and 3}\}=6\mathbb{Z}$$I\cap J=\{n\in \mathbb{Z}\mid n\text{ is divisible by both 2 and 3}\}=6\mathbb{Z}$I nn J={n inZ∣n” is divisible by both 2 and 3″}=6ZI cap J = { n in mathbb{Z} mid n text{ is divisible by both 2 and 3} } = 6mathbb{Z}$I\cap J=\{n\in \mathbb{Z}\mid n\text{ is divisible by both 2 and 3}\}=6\mathbb{Z}$

In this example, $IJ=I\cap J$$IJ=I\cap J$IJ=I nn JIJ = I cap J$IJ=I\cap J$, but this is not generally true for all ideals $I$$I$II$I$ and $J$$J$JJ$J$ in all rings $R$$R$RR$R$.
For instance, consider the ring $R=\mathbb{Z}[x]$$R=\mathbb{Z}[x]$R=Z[x]R = mathbb{Z}[x]$R=\mathbb{Z}[x]$ of all polynomials with integer coefficients. Let $I=(2,x)$$I=(2,x)$I=(2,x)I = (2, x)$I=(2,x)$ and $J=(3,x)$$J=(3,x)$J=(3,x)J = (3, x)$J=(3,x)$.

1. $IJ=(6,2x,3x,x^2)$$IJ=(6,2x,3x,x^2)$IJ=(6,2x,3x,x^(2))IJ = (6, 2x, 3x, x^2)$IJ=(6,2x,3x,x^2)$
2. $I\cap J=(6,x)$$I\cap J=(6,x)$I nn J=(6,x)I cap J = (6, x)$I\cap J=(6,x)$

Here, $IJ\ne I\cap J$$IJ\ne I\cap J$IJ!=I nn JIJ neq I cap J$IJ\ne I\cap J$, disproving the statement.
Thus, the statement is false, as demonstrated by this counterexample.

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ix) If $\mathrm{f}:\mathrm{R}\to \mathrm{S}$$\mathrm{f}:\mathrm{R}\to \mathrm{S}$f:RrarrSmathrm{f}: mathrm{R} rightarrow mathrm{S}$\mathrm{f}:\mathrm{R}\to \mathrm{S}$ is a ring homomorphism and $\mathrm{I}$$\mathrm{I}$Imathrm{I}$\mathrm{I}$ is an ideal of $\mathrm{R}$$\mathrm{R}$Rmathrm{R}$\mathrm{R}$, then $\mathrm{f}(\mathrm{I})$f(I)mathrm{f}(mathrm{I})$\mathrm{f}(\mathrm{I})$ is an ideal of $\mathrm{S}$$\mathrm{S}$Smathrm{S}$\mathrm{S}$.
Answer:
The statement “If $f:R\to S$$f:R\to S$f:R rarr Sf: R rightarrow S$f:R\to S$ is a ring homomorphism and $I$$I$II$I$ is an ideal of $R$$R$RR$R$, then $f(I)$$f(I)$f(I)f(I)$f(I)$ is an ideal of $S$$S$SS$S$” is false.

Justification:

An ideal $I$$I$II$I$ of a ring $R$$R$RR$R$ is a subset of $R$$R$RR$R$ that is closed under addition and multiplication by any element in $R$$R$RR$R$. A ring homomorphism $f:R\to S$$f:R\to S$f:R rarr Sf: R rightarrow S$f:R\to S$ preserves addition and multiplication but does not necessarily map ideals to ideals.

Counterexample:

Consider the ring $R=\mathbb{Z}$$R=\mathbb{Z}$R=ZR = mathbb{Z}$R=\mathbb{Z}$ and the ring $S=\mathbb{Z}/2\mathbb{Z}$$S=\mathbb{Z}/2\mathbb{Z}$S=Z//2ZS = mathbb{Z}/2mathbb{Z}$S=\mathbb{Z}/2\mathbb{Z}$ (the integers modulo 2). Let $f:R\to S$$f:R\to S$f:R rarr Sf: R rightarrow S$f:R\to S$ be the natural projection map defined by $f(x)=x\mathrm{mod}2$$f(x)=x\mathrm{mod}2$f(x)=xmod2f(x) = x mod 2$f(x)=x\mathrm{mod}2$.
Let $I=2\mathbb{Z}$$I=2\mathbb{Z}$I=2ZI = 2mathbb{Z}$I=2\mathbb{Z}$ be the ideal of even integers in $R$$R$RR$R$.
The image $f(I)$$f(I)$f(I)f(I)$f(I)$ consists of the set $\{0\}$$\{0\}${0}{0}$\{0\}$ in $S$$S$SS$S$, since all even integers are mapped to 0 in $\mathbb{Z}/2\mathbb{Z}$$\mathbb{Z}/2\mathbb{Z}$Z//2Zmathbb{Z}/2mathbb{Z}$\mathbb{Z}/2\mathbb{Z}$.
However, $\{0\}$$\{0\}${0}{0}$\{0\}$ is not an ideal in $S$$S$SS$S$ because it is not closed under multiplication by any element in $S$$S$SS$S$. Specifically, $0\times 1=0$$0\times 1=0$0xx1=00 times 1 = 0$0\times 1=0$ but $1$$1$11$1$ is not in $\{0\}$$\{0\}${0}{0}$\{0\}$.
Thus, $f(I)$$f(I)$f(I)f(I)$f(I)$ is not an ideal in $S$$S$SS$S$, making the statement false.

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x) Every prime ideal of an integral domain is a maximal ideal.
Answer:
The statement “Every prime ideal of an integral domain is a maximal ideal” is false.

Justification:

1. Prime Ideal: An ideal $P$$P$PP$P$ in a commutative ring $R$$R$RR$R$ is called a prime ideal if for any $a,b\in R$$a,b\in R$a,b in Ra, b in R$a,b\in R$, $ab\in P$$ab\in P$ab in Pab in P$ab\in P$ implies $a\in P$$a\in P$a in Pa in P$a\in P$ or $b\in P$$b\in P$b in Pb in P$b\in P$.
2. Maximal Ideal: An ideal $M$$M$MM$M$ in a commutative ring $R$$R$RR$R$ is called a maximal ideal if $M\ne R$$M\ne R$M!=RM neq R$M\ne R$ and there is no ideal $N$$N$NN$N$ such that $M⊊N⊊R$$M⊊N⊊R$M⊊N⊊RM subsetneq N subsetneq R$M⊊N⊊R$.

Counterexample:

Consider the integral domain $R=\mathbb{Z}[x]$$R=\mathbb{Z}[x]$R=Z[x]R = mathbb{Z}[x]$R=\mathbb{Z}[x]$, the ring of all polynomials with integer coefficients.
Let $P=(x)$$P=(x)$P=(x)P = (x)$P=(x)$, the ideal generated by $x$$x$xx$x$. This is a prime ideal because $\mathbb{Z}[x]/(x)\cong \mathbb{Z}$$\mathbb{Z}[x]/(x)\cong \mathbb{Z}$Z[x]//(x)~=Zmathbb{Z}[x]/(x) cong mathbb{Z}$\mathbb{Z}[x]/(x)\cong \mathbb{Z}$, which is an integral domain. However, $P$$P$PP$P$ is not maximal because it is properly contained in other ideals, such as $(x,2)$$(x,2)$(x,2)(x, 2)$(x,2)$, which in turn is properly contained in $R$$R$RR$R$.
Thus, $P$$P$PP$P$ is a prime ideal that is not maximal, disproving the statement.
Therefore, the statement is false, as demonstrated by this counterexample.

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Question:-02

2. a) Let $\mathrm{G}$$\mathrm{G}$Gmathrm{G}$\mathrm{G}$ be a group and let $\mathrm{H}\le \mathrm{G},\mathrm{K}\le \mathrm{G},\mathrm{o}(\mathrm{H})=\mathrm{o}(\mathrm{K})=\mathrm{p}$$\mathrm{H}\le \mathrm{G},\mathrm{K}\le \mathrm{G},\mathrm{o}(\mathrm{H})=\mathrm{o}(\mathrm{K})=\mathrm{p}$H <= G,K <= G,o(H)=o(K)=pmathrm{H} leq mathrm{G}, mathrm{K} leq mathrm{G}, mathrm{o}(mathrm{H})=mathrm{o}(mathrm{K})=mathrm{p}$\mathrm{H}\le \mathrm{G},\mathrm{K}\le \mathrm{G},\mathrm{o}(\mathrm{H})=\mathrm{o}(\mathrm{K})=\mathrm{p}$, a prime. Show that either $\mathrm{H}\cap \mathrm{K}=\{\mathrm{e}\}$$\mathrm{H}\cap \mathrm{K}=\{\mathrm{e}\}$HnnK={e}mathrm{H} cap mathrm{K}={mathrm{e}}$\mathrm{H}\cap \mathrm{K}=\{\mathrm{e}\}$ or $\mathrm{H}=\mathrm{K}$$\mathrm{H}=\mathrm{K}$H=Kmathrm{H}=mathrm{K}$\mathrm{H}=\mathrm{K}$. Is this result still true if $\mathrm{p}$$\mathrm{p}$pmathrm{p}$\mathrm{p}$ is not a prime? Justify your answer.

Answer:

Part 1: $o(H)=o(K)=p$$o(H)=o(K)=p$o(H)=o(K)=po(H) = o(K) = p$o(H)=o(K)=p$, a prime

Let $G$$G$GG$G$ be a group and $H,K\le G$$H,K\le G$H,K <= GH, K leq G$H,K\le G$ with $o(H)=o(K)=p$$o(H)=o(K)=p$o(H)=o(K)=po(H) = o(K) = p$o(H)=o(K)=p$, where $p$$p$pp$p$ is a prime number.

1. Case 1: $H\cap K\ne \{e\}$$H\cap K\ne \{e\}$H nn K!={e}H cap K neq { e }$H\cap K\ne \{e\}$
   • If $H\cap K$$H\cap K$H nn KH cap K$H\cap K$ contains an element other than the identity $e$$e$ee$e$, then $H\cap K$$H\cap K$H nn KH cap K$H\cap K$ must be a subgroup of both $H$$H$HH$H$ and $K$$K$KK$K$ (since the intersection of two subgroups is always a subgroup).
   • The order of $H\cap K$$H\cap K$H nn KH cap K$H\cap K$ must divide the order of $H$$H$HH$H$ and $K$$K$KK$K$ by Lagrange’s theorem. Since $o(H)=o(K)=p$$o(H)=o(K)=p$o(H)=o(K)=po(H) = o(K) = p$o(H)=o(K)=p$ and $p$$p$pp$p$ is a prime, the only divisors are 1 and $p$$p$pp$p$.
   • Since $H\cap K$$H\cap K$H nn KH cap K$H\cap K$ contains an element other than $e$$e$ee$e$, its order must be $p$$p$pp$p$, which means $H\cap K=H=K$$H\cap K=H=K$H nn K=H=KH cap K = H = K$H\cap K=H=K$.
2. Case 2: $H\cap K=\{e\}$$H\cap K=\{e\}$H nn K={e}H cap K = { e }$H\cap K=\{e\}$
   • If $H\cap K$$H\cap K$H nn KH cap K$H\cap K$ contains only the identity $e$$e$ee$e$, then $H\cap K=\{e\}$$H\cap K=\{e\}$H nn K={e}H cap K = { e }$H\cap K=\{e\}$.

So, either $H\cap K=\{e\}$$H\cap K=\{e\}$H nn K={e}H cap K = { e }$H\cap K=\{e\}$ or $H=K$$H=K$H=KH = K$H=K$.

Part 2: $o(H)=o(K)$$o(H)=o(K)$o(H)=o(K)o(H) = o(K)$o(H)=o(K)$ not necessarily a prime

If $p$$p$pp$p$ is not a prime, the result may not hold. For example, consider $G={\mathbb{Z}}_{12}$$G={\mathbb{Z}}_{12}$G=Z_(12)G = mathbb{Z}_{12}$G={\mathbb{Z}}_{12}$, $H=\{0,4,8\}$$H=\{0,4,8\}$H={0,4,8}H = { 0, 4, 8 }$H=\{0,4,8\}$, and $K=\{0,6,12\}$$K=\{0,6,12\}$K={0,6,12}K = { 0, 6, 12 }$K=\{0,6,12\}$. Here, $o(H)=o(K)=3$$o(H)=o(K)=3$o(H)=o(K)=3o(H) = o(K) = 3$o(H)=o(K)=3$, which is not a prime. We have $H\cap K=\{0\}$$H\cap K=\{0\}$H nn K={0}H cap K = { 0 }$H\cap K=\{0\}$, but $H\ne K$$H\ne K$H!=KH neq K$H\ne K$.
Thus, the result is not necessarily true if $p$$p$pp$p$ is not a prime.

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b) Let $\mathrm{G}$$\mathrm{G}$Gmathrm{G}$\mathrm{G}$ be the group of all rigid motions of a plane and $\mathrm{S}$$\mathrm{S}$Smathrm{S}$\mathrm{S}$ be the set of all rectangles in the plane. Show that $\mathrm{G}$$\mathrm{G}$Gmathrm{G}$\mathrm{G}$ acts on $\mathrm{S}$$\mathrm{S}$Smathrm{S}$\mathrm{S}$. Also obtain the orbit and stabiliser of a square under this action.
Answer:
To show that $G$$G$GG$G$ acts on $S$$S$SS$S$, we need to satisfy two conditions:

1. For each $g\in G$$g\in G$g in Gg in G$g\in G$ and $s\in S$$s\in S$s in Ss in S$s\in S$, $g\cdot s$$g\cdot s$g*sg cdot s$g\cdot s$ is also in $S$$S$SS$S$.
2. For each $g_1,g_2\in G$$g_1,g_2\in G$g_(1),g_(2)in Gg_1, g_2 in G$g_1,g_2\in G$ and $s\in S$$s\in S$s in Ss in S$s\in S$, $(g_1\cdot g_2)\cdot s=g_1\cdot (g_2\cdot s)$$(g_1\cdot g_2)\cdot s=g_1\cdot (g_2\cdot s)$(g_(1)*g_(2))*s=g_(1)*(g_(2)*s)(g_1 cdot g_2) cdot s = g_1 cdot (g_2 cdot s)$(g_1\cdot g_2)\cdot s=g_1\cdot (g_2\cdot s)$.

Condition 1: Closure

A rigid motion in the plane consists of rotations, translations, and reflections. When we apply any of these to a rectangle, we still get a rectangle. Therefore, $G$$G$GG$G$ is closed under the action on $S$$S$SS$S$.

Condition 2: Associativity

Let $g_1,g_2$$g_1,g_2$g_(1),g_(2)g_1, g_2$g_1,g_2$ be any two rigid motions and $s$$s$ss$s$ be any rectangle. Then $(g_1\cdot g_2)\cdot s$$(g_1\cdot g_2)\cdot s$(g_(1)*g_(2))*s(g_1 cdot g_2) cdot s$(g_1\cdot g_2)\cdot s$ is the rectangle obtained by first applying $g_2$$g_2$g_(2)g_2$g_2$ to $s$$s$ss$s$ and then applying $g_1$$g_1$g_(1)g_1$g_1$ to the result. This is the same as first applying $g_2$$g_2$g_(2)g_2$g_2$ to $s$$s$ss$s$ to get $g_2\cdot s$$g_2\cdot s$g_(2)*sg_2 cdot s$g_2\cdot s$ and then applying $g_1$$g_1$g_(1)g_1$g_1$ to $g_2\cdot s$$g_2\cdot s$g_(2)*sg_2 cdot s$g_2\cdot s$ to get $g_1\cdot (g_2\cdot s)$$g_1\cdot (g_2\cdot s)$g_(1)*(g_(2)*s)g_1 cdot (g_2 cdot s)$g_1\cdot (g_2\cdot s)$. Therefore, $(g_1\cdot g_2)\cdot s=g_1\cdot (g_2\cdot s)$$(g_1\cdot g_2)\cdot s=g_1\cdot (g_2\cdot s)$(g_(1)*g_(2))*s=g_(1)*(g_(2)*s)(g_1 cdot g_2) cdot s = g_1 cdot (g_2 cdot s)$(g_1\cdot g_2)\cdot s=g_1\cdot (g_2\cdot s)$.
So both conditions are satisfied, and $G$$G$GG$G$ acts on $S$$S$SS$S$.

Orbit and Stabilizer for a Square

Orbit

The orbit of a square under this action is the set of all rectangles that can be obtained by applying a rigid motion to the square. Since a square is a special case of a rectangle, any rigid motion (rotation, translation, reflection) will still produce a square. Therefore, the orbit of a square under this action is the set of all squares in the plane.

Stabilizer

The stabilizer of a square is the set of all rigid motions that leave the square invariant. This includes:

1. The identity motion (doing nothing).
2. 90-degree rotations about the center of the square.
3. 180-degree rotations about the center of the square.
4. 270-degree rotations about the center of the square.
5. Reflections about the axes of symmetry of the square.

So, the stabilizer consists of these 8 rigid motions.
Thus, we have shown that $G$$G$GG$G$ acts on $S$$S$SS$S$ and have obtained the orbit and stabilizer of a square under this action.

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c) Let $\mathrm{G}$$\mathrm{G}$Gmathrm{G}$\mathrm{G}$ be a finite group and $\mathrm{H}$$\mathrm{H}$Hmathrm{H}$\mathrm{H}$ be a normal subgroup of $\mathrm{G}$$\mathrm{G}$Gmathrm{G}$\mathrm{G}$. Prove that $\mathrm{H}=\cup {\mathrm{C}}_{\mathrm{x}}$$\mathrm{H}=\cup {\mathrm{C}}_{\mathrm{x}}$H=uuC_(x)mathrm{H}=cup mathrm{C}_{mathrm{x}}$\mathrm{H}=\cup {\mathrm{C}}_{\mathrm{x}}$ where the ${\mathrm{C}}_{\mathrm{x}}$${\mathrm{C}}_{\mathrm{x}}$C_(x)mathrm{C}_{mathrm{x}}${\mathrm{C}}_{\mathrm{x}}$ are all the distinct conjugacy classes of $\mathrm{G}$$\mathrm{G}$Gmathrm{G}$\mathrm{G}$ such that $\mathrm{H}\cap {\mathrm{C}}_{\mathrm{x}}\ne \varnothing$$\mathrm{H}\cap {\mathrm{C}}_{\mathrm{x}}\ne \varnothing$HnnC_(x)!=O/mathrm{H} cap mathrm{C}_{mathrm{x}} neq varnothing$\mathrm{H}\cap {\mathrm{C}}_{\mathrm{x}}\ne \varnothing$.
Answer:
To prove that $H=\cup C_x$$H=\cup C_x$H=uuC_(x)H = cup C_x$H=\cup C_x$ where $C_x$$C_x$C_(x)C_x$C_x$ are all the distinct conjugacy classes of $G$$G$GG$G$ such that $H\cap C_x\ne \varnothing$$H\cap C_x\ne \varnothing$H nnC_(x)!=O/H cap C_x neq varnothing$H\cap C_x\ne \varnothing$, we need to show two things:

1. $H\subseteq \cup C_x$$H\subseteq \cup C_x$H sube uuC_(x)H subseteq cup C_x$H\subseteq \cup C_x$
2. $\cup C_x\subseteq H$$\cup C_x\subseteq H$uuC_(x)sube Hcup C_x subseteq H$\cup C_x\subseteq H$

1. $H\subseteq \cup C_x$$H\subseteq \cup C_x$H sube uuC_(x)H subseteq cup C_x$H\subseteq \cup C_x$

Take any element $h\in H$$h\in H$h in Hh in H$h\in H$. Since $H$$H$HH$H$ is a normal subgroup of $G$$G$GG$G$, it is invariant under conjugation by elements of $G$$G$GG$G$. That is, for any $g\in G$$g\in G$g in Gg in G$g\in G$, $g^{-1}hg\in H$$g^{-1}hg\in H$g^(-1)hg in Hg^{-1}hg in H$g^{-1}hg\in H$.
The conjugacy class $C_h$$C_h$C_(h)C_h$C_h$ of $h$$h$hh$h$ in $G$$G$GG$G$ is defined as $C_h=\{g^{-1}hg\mid g\in G\}$$C_h=\{g^{-1}hg\mid g\in G\}$C_(h)={g^(-1)hg∣g in G}C_h = { g^{-1}hg mid g in G }$C_h=\{g^{-1}hg\mid g\in G\}$. Since $H$$H$HH$H$ is normal, $C_h\subseteq H$$C_h\subseteq H$C_(h)sube HC_h subseteq H$C_h\subseteq H$.
Therefore, $h\in C_h$$h\in C_h$h inC_(h)h in C_h$h\in C_h$ and $C_h\cap H\ne \varnothing$$C_h\cap H\ne \varnothing$C_(h)nn H!=O/C_h cap H neq varnothing$C_h\cap H\ne \varnothing$. So, $h\in \cup C_x$$h\in \cup C_x$h in uuC_(x)h in cup C_x$h\in \cup C_x$.
This shows that $H\subseteq \cup C_x$$H\subseteq \cup C_x$H sube uuC_(x)H subseteq cup C_x$H\subseteq \cup C_x$.

2. $\cup C_x\subseteq H$$\cup C_x\subseteq H$uuC_(x)sube Hcup C_x subseteq H$\cup C_x\subseteq H$

Take any element $x$$x$xx$x$ such that $C_x\cap H\ne \varnothing$$C_x\cap H\ne \varnothing$C_(x)nn H!=O/C_x cap H neq varnothing$C_x\cap H\ne \varnothing$. Let $h\in C_x\cap H$$h\in C_x\cap H$h inC_(x)nn Hh in C_x cap H$h\in C_x\cap H$.
The conjugacy class $C_x$$C_x$C_(x)C_x$C_x$ is defined as $C_x=\{g^{-1}xg\mid g\in G\}$$C_x=\{g^{-1}xg\mid g\in G\}$C_(x)={g^(-1)xg∣g in G}C_x = { g^{-1}xg mid g in G }$C_x=\{g^{-1}xg\mid g\in G\}$.
Since $h\in C_x$$h\in C_x$h inC_(x)h in C_x$h\in C_x$, there exists some $g\in G$$g\in G$g in Gg in G$g\in G$ such that $h=g^{-1}xg$$h=g^{-1}xg$h=g^(-1)xgh = g^{-1}xg$h=g^{-1}xg$.
Since $h\in H$$h\in H$h in Hh in H$h\in H$ and $H$$H$HH$H$ is a normal subgroup, $g^{-1}hg\in H$$g^{-1}hg\in H$g^(-1)hg in Hg^{-1}hg in H$g^{-1}hg\in H$. But $g^{-1}hg=x$$g^{-1}hg=x$g^(-1)hg=xg^{-1}hg = x$g^{-1}hg=x$, so $x\in H$$x\in H$x in Hx in H$x\in H$.
This shows that $C_x\subseteq H$$C_x\subseteq H$C_(x)sube HC_x subseteq H$C_x\subseteq H$ for every $x$$x$xx$x$ such that $C_x\cap H\ne \varnothing$$C_x\cap H\ne \varnothing$C_(x)nn H!=O/C_x cap H neq varnothing$C_x\cap H\ne \varnothing$.
Therefore, $\cup C_x\subseteq H$$\cup C_x\subseteq H$uuC_(x)sube Hcup C_x subseteq H$\cup C_x\subseteq H$.

Conclusion

Since $H\subseteq \cup C_x$$H\subseteq \cup C_x$H sube uuC_(x)H subseteq cup C_x$H\subseteq \cup C_x$ and $\cup C_x\subseteq H$$\cup C_x\subseteq H$uuC_(x)sube Hcup C_x subseteq H$\cup C_x\subseteq H$, we have $H=\cup C_x$$H=\cup C_x$H=uuC_(x)H = cup C_x$H=\cup C_x$ where $C_x$$C_x$C_(x)C_x$C_x$ are all the distinct conjugacy classes of $G$$G$GG$G$ such that $H\cap C_x\ne \varnothing$$H\cap C_x\ne \varnothing$H nnC_(x)!=O/H cap C_x neq varnothing$H\cap C_x\ne \varnothing$.
This completes the proof.

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