# IGNOU MMT-003 Solved Assignment 2023 | M.Sc. MACS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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## IGNOU MMT-003 Assignment Question Paper 2023

mmt-003-2023-ext-aa2d2d76-613f-4e79-9c79-c97616952506
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\left(G\right)$$o\left(G\right)$o(G)o(G)$o\left(G\right)$, then the number of Sylow p-subgroups of $\mathrm{G}$$\mathrm{G}$G\mathrm{G}$\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}$Q\mathbb{Q}$\mathbb{Q}$ is ${x}^{1/3}$${x}^{1/3}$x^(1//3)x^{1 / 3}${x}^{1/3}$.
iii) ${\mathbb{Z}}_{\mathrm{m}\mathrm{n}}\simeq {\mathbb{Z}}_{\mathrm{m}}×{\mathbb{Z}}_{\mathrm{h}}\mathrm{\forall }\mathrm{m},\mathrm{n}\in \mathbb{Z}$${\mathbb{Z}}_{\mathrm{m}\mathrm{n}}\simeq {\mathbb{Z}}_{\mathrm{m}}×{\mathbb{Z}}_{\mathrm{h}}\mathrm{\forall }\mathrm{m},\mathrm{n}\in \mathbb{Z}$Z_(mn)≃Z_(m)xxZ_(h)AAm,ninZ\mathbb{Z}_{\mathrm{mn}} \simeq \mathbb{Z}_{\mathrm{m}} \times \mathbb{Z}_{\mathrm{h}} \forall \mathrm{m}, \mathrm{n} \in \mathbb{Z}${\mathbb{Z}}_{\mathrm{m}\mathrm{n}}\simeq {\mathbb{Z}}_{\mathrm{m}}×{\mathbb{Z}}_{\mathrm{h}}\mathrm{\forall }\mathrm{m},\mathrm{n}\in \mathbb{Z}$
iv) If $G$$G$GG$G$ is a finite group and $m\mid o\left(G\right),m\in \mathbb{N}$$m\mid o\left(G\right),m\in \mathbb{N}$m∣o(G),m inNm \mid o(G), m \in \mathbb{N}$m\mid o\left(G\right),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^(“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}$$\mathrm{I}$I\mathrm{I}$\mathrm{I}$ and $\mathrm{J}$$\mathrm{J}$J\mathrm{J}$\mathrm{J}$ are ideals of a ring $\mathrm{R}$$\mathrm{R}$R\mathrm{R}$\mathrm{R}$, then $\mathrm{I}\mathrm{J}=\mathrm{I}\cap \mathrm{J}$$\mathrm{I}\mathrm{J}=\mathrm{I}\cap \mathrm{J}$IJ=InnJ\mathrm{IJ}=\mathrm{I} \cap \mathrm{J}$\mathrm{I}\mathrm{J}=\mathrm{I}\cap \mathrm{J}$.
ix) If $\mathrm{f}:\mathrm{R}\to \mathrm{S}$$\mathrm{f}:\mathrm{R}\to \mathrm{S}$f:RrarrS\mathrm{f}: \mathrm{R} \rightarrow \mathrm{S}$\mathrm{f}:\mathrm{R}\to \mathrm{S}$ is a ring homomorphism and $\mathrm{I}$$\mathrm{I}$I\mathrm{I}$\mathrm{I}$ is an ideal of $\mathrm{R}$$\mathrm{R}$R\mathrm{R}$\mathrm{R}$, then $\mathrm{f}\left(\mathrm{I}\right)$$\mathrm{f}\left(\mathrm{I}\right)$f(I)\mathrm{f}(\mathrm{I})$\mathrm{f}\left(\mathrm{I}\right)$ is an ideal of S.
$\mathrm{x}$$\mathrm{x}$x\mathrm{x}$\mathrm{x}$ Every prime ideal of an integral domain is a maximal ideal.
1. a) Let $G$$G$GG$G$ be a group and let $H\le G,K\le G,o\left(H\right)=o\left(K\right)=p$$H\le G,K\le G,o\left(H\right)=o\left(K\right)=p$H <= G,K <= G,o(H)=o(K)=pH \leq G, K \leq G, o(H)=o(K)=p$H\le G,K\le G,o\left(H\right)=o\left(K\right)=p$, a prime. Show that either $\mathrm{H}\cap \mathrm{K}=\left\{\mathrm{e}\right\}$$\mathrm{H}\cap \mathrm{K}=\left\{\mathrm{e}\right\}$HnnK={e}\mathrm{H} \cap \mathrm{K}=\{\mathrm{e}\}$\mathrm{H}\cap \mathrm{K}=\left\{\mathrm{e}\right\}$ or $\mathrm{H}=\mathrm{K}$$\mathrm{H}=\mathrm{K}$H=K\mathrm{H}=\mathrm{K}$\mathrm{H}=\mathrm{K}$. Is this result still true if $\mathrm{p}$$\mathrm{p}$p\mathrm{p}$\mathrm{p}$ is not a prime? Justify your answer.
b) Let $\mathrm{G}$$\mathrm{G}$G\mathrm{G}$\mathrm{G}$ be the group of all rigid motions of a plane and $\mathrm{S}$$\mathrm{S}$S\mathrm{S}$\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$.
1. 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 $|{G}_{1}|$$\left|{G}_{1}\right|$|G_(1)|\left|G_{1}\right|$|{G}_{1}|$ and $|{G}_{2}|$$\left|{G}_{2}\right|$|G_(2)|\left|G_{2}\right|$|{G}_{2}|$. Prove that the Sylow p-subgroups of ${G}_{1}×{G}_{2}$${G}_{1}×{G}_{2}$G_(1)xxG_(2)G_{1} \times G_{2}${G}_{1}×{G}_{2}$ are precisely of the form ${P}_{1}×{P}_{2}$${P}_{1}×{P}_{2}$P_(1)xxP_(2)P_{1} \times P_{2}${P}_{1}×{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.
1. a) Write $\left[\begin{array}{ccc}1& -1& 0\\ 2& 0& 3\\ 5& 5& 1\end{array}\right]$$\left[\begin{array}{ccc}1& -1& 0\\ 2& 0& 3\\ 5& 5& 1\end{array}\right]$[[1,-1,0],[2,0,3],[5,5,1]]\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 0 & 3 \\ 5 & 5 & 1\end{array}\right]$\left[\begin{array}{ccc}1& -1& 0\\ 2& 0& 3\\ 5& 5& 1\end{array}\right]$ as a product of $\mathrm{O}\left(3\right)$$\mathrm{O}\left(3\right)$O(3)\mathrm{O}(3)$\mathrm{O}\left(3\right)$ and an element of ${B}_{3}\left(\mathbb{R}\right)$${B}_{3}\left(\mathbb{R}\right)$B_(3)(R)B_{3}(\mathbb{R})${B}_{3}\left(\mathbb{R}\right)$.
b) Show that $\mathrm{S}\mathrm{U}\left(2\right)$$\mathrm{S}\mathrm{U}\left(2\right)$SU(2)\mathrm{SU}(2)$\mathrm{S}\mathrm{U}\left(2\right)$ and ${\mathrm{S}}^{3}$${\mathrm{S}}^{3}$S^(3)\mathrm{S}^{3}${\mathrm{S}}^{3}$ are structurally the same.
1. a) Find all the possible abelian groups, up to isomorphism, of order 900 .
b) Construct the free group on the set $\left\{\alpha ,\beta ,\gamma \right\}$$\left\{\alpha ,\beta ,\gamma \right\}${alpha,beta,gamma}\{\alpha, \beta, \gamma\}$\left\{\alpha ,\beta ,\gamma \right\}$. Further, check if it is a free abelian group or not.
1. a) Check whether or not the ring $\mathrm{R}={\mathbb{Z}}_{3}\left[\mathrm{x}\right]/<{\mathrm{x}}^{6}-1>$$\mathrm{R}={\mathbb{Z}}_{3}\left[\mathrm{x}\right]/<{\mathrm{x}}^{6}-1>$R=Z_(3)[x]// < x^(6)-1 >\mathrm{R}=\mathbb{Z}_{3}[\mathrm{x}] /<\mathrm{x}^{6}-1>$\mathrm{R}={\mathbb{Z}}_{3}\left[\mathrm{x}\right]/<{\mathrm{x}}^{6}-1>$
i) is finite;
ii) has zero divisors;
iii) has nilpotent elements.
b) Give two distinct rings whose quotient field is $\left\{a+ib\mid a,b\in \mathbb{R}\right\}$$\left\{a+ib\mid a,b\in \mathbb{R}\right\}${a+ib∣a,b inR}\{a+i b \mid a, b \in \mathbb{R}\}$\left\{a+ib\mid a,b\in \mathbb{R}\right\}$. 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.
1. a) Check whether or not $\mathbb{Z}\left[\sqrt{-7}\right]$$\mathbb{Z}\left[\sqrt{-7}\right]$Z[sqrt(-7)]\mathbb{Z}[\sqrt{-7}]$\mathbb{Z}\left[\sqrt{-7}\right]$ is a Euclidean domain.
b) Use the division algorithm to find the inverse of $\overline{18}$$\overline{18}$bar(18)\overline{18}$\overline{18}$ in ${\mathbb{Z}}_{35}$${\mathbb{Z}}_{35}$Z_(35)\mathbb{Z}_{35}${\mathbb{Z}}_{35}$.
1. i) Let $G$$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}$$\mathrm{K}$K\mathrm{K}$\mathrm{K}$ ? Why, or why not?
ii) Find ${K}^{G}$${K}^{G}$K^(G)K^{G}${K}^{G}$, where $K=\mathbb{Q}\left(i,\sqrt{3}\right),G=G\left(K/\mathbb{Q}\right)$$K=\mathbb{Q}\left(i,\sqrt{3}\right),G=G\left(K/\mathbb{Q}\right)$K=Q(i,sqrt3),G=G(K//Q)K=\mathbb{Q}(i, \sqrt{3}), G=G(K / \mathbb{Q})$K=\mathbb{Q}\left(i,\sqrt{3}\right),G=G\left(K/\mathbb{Q}\right)$.
$$sin^2\left(\frac{\theta }{2}\right)=\frac{1-cos\:\theta }{2}$$

## 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}$$\mathrm{G}$G\mathrm{G}$\mathrm{G}$ is a finite abelian group and $\mathrm{p}$$\mathrm{p}$p\mathrm{p}$\mathrm{p}$ is a prime factor of $\mathrm{o}\left(\mathrm{G}\right)$$\mathrm{o}\left(\mathrm{G}\right)$o(G)\mathrm{o}(\mathrm{G})$\mathrm{o}\left(\mathrm{G}\right)$, then the number of Sylow p-subgroups of $\mathrm{G}$$\mathrm{G}$G\mathrm{G}$\mathrm{G}$ is a prime.
The statement “If $G$$G$GG$G$ is a finite abelian group and $p$$p$pp$p$ is a prime factor of $o\left(G\right)$$o\left(G\right)$o(G)o(G)$o\left(G\right)$, then the number of Sylow $p$$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|$$|G|$|G||G|$|G|$ and ${n}_{p}\equiv 1\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p$${n}_{p}\equiv 1\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p$n_(p)-=1modpn_p \equiv 1 \mod p${n}_{p}\equiv 1\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}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}×{\mathbb{Z}}_{2}$$G={\mathbb{Z}}_{4}×{\mathbb{Z}}_{2}$G=Z_(4)xxZ_(2)G = \mathbb{Z}_4 \times \mathbb{Z}_2$G={\mathbb{Z}}_{4}×{\mathbb{Z}}_{2}$, where ${\mathbb{Z}}_{n}$${\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\left(G\right)=4×2=8$$o\left(G\right)=4×2=8$o(G)=4xx2=8o(G) = 4 \times 2 = 8$o\left(G\right)=4×2=8$.
The prime factors of $o\left(G\right)$$o\left(G\right)$o(G)o(G)$o\left(G\right)$ 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$$G$GG$G$ are generated by $\left(2,0\right)$$\left(2,0\right)$(2,0)(2, 0)$\left(2,0\right)$, $\left(0,1\right)$$\left(0,1\right)$(0,1)(0, 1)$\left(0,1\right)$, and $\left(2,1\right)$$\left(2,1\right)$(2,1)(2, 1)$\left(2,1\right)$. 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}$Q\mathbb{Q}$\mathbb{Q}$ is ${x}^{1/3}$${x}^{1/3}$x^(1//3)x^{1 / 3}${x}^{1/3}$.
The statement “The minimum polynomial of ${5}^{1/3}$${5}^{1/3}$5^(1//3)5^{1/3}${5}^{1/3}$ over $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$ is ${x}^{1/3}$${x}^{1/3}$x^(1//3)x^{1/3}${x}^{1/3}$” is false.

### Justification:

The minimum polynomial of an algebraic number $\alpha$$\alpha$alpha\alpha$\alpha$ over a field $F$$F$FF$F$ is the unique monic polynomial $f\left(x\right)$$f\left(x\right)$f(x)f(x)$f\left(x\right)$ of least degree such that $f\left(\alpha \right)=0$$f\left(\alpha \right)=0$f(alpha)=0f(\alpha) = 0$f\left(\alpha \right)=0$ and $f\left(x\right)$$f\left(x\right)$f(x)f(x)$f\left(x\right)$ has coefficients in $F$$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}$Q\mathbb{Q}$\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}$Q\mathbb{Q}$\mathbb{Q}$.
2. $\left({5}^{1/3}{\right)}^{3}-5=5-5=0$$\left({5}^{1/3}{\right)}^{3}-5=5-5=0$(5^(1//3))^(3)-5=5-5=0(5^{1/3})^3 – 5 = 5 – 5 = 0$\left({5}^{1/3}{\right)}^{3}-5=5-5=0$, so ${5}^{1/3}$${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}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$, which means it has the least degree among all polynomials that have ${5}^{1/3}$${5}^{1/3}$5^(1//3)5^{1/3}${5}^{1/3}$ as a root and have coefficients in $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$.
Moreover, ${x}^{1/3}$${x}^{1/3}$x^(1//3)x^{1/3}${x}^{1/3}$ is not even a polynomial over $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$ because its exponent $1/3$$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}}×{\mathbb{Z}}_{\mathrm{n}}\mathrm{\forall }\mathrm{m},\mathrm{n}\in \mathbb{Z}$${\mathbb{Z}}_{\mathrm{m}\mathrm{n}}\simeq {\mathbb{Z}}_{\mathrm{m}}×{\mathbb{Z}}_{\mathrm{n}}\mathrm{\forall }\mathrm{m},\mathrm{n}\in \mathbb{Z}$Z_(mn)≃Z_(m)xxZ_(n)AAm,ninZ\mathbb{Z}_{\mathrm{mn}} \simeq \mathbb{Z}_{\mathrm{m}} \times \mathbb{Z}_{\mathrm{n}} \forall \mathrm{m}, \mathrm{n} \in \mathbb{Z}${\mathbb{Z}}_{\mathrm{m}\mathrm{n}}\simeq {\mathbb{Z}}_{\mathrm{m}}×{\mathbb{Z}}_{\mathrm{n}}\mathrm{\forall }\mathrm{m},\mathrm{n}\in \mathbb{Z}$
The statement “${\mathbb{Z}}_{mn}\simeq {\mathbb{Z}}_{m}×{\mathbb{Z}}_{n}\mathrm{\forall }m,n\in \mathbb{Z}$${\mathbb{Z}}_{mn}\simeq {\mathbb{Z}}_{m}×{\mathbb{Z}}_{n}\mathrm{\forall }m,n\in \mathbb{Z}$Z_(mn)≃Z_(m)xxZ_(n)AA m,n inZ\mathbb{Z}_{mn} \simeq \mathbb{Z}_{m} \times \mathbb{Z}_{n} \forall m, n \in \mathbb{Z}${\mathbb{Z}}_{mn}\simeq {\mathbb{Z}}_{m}×{\mathbb{Z}}_{n}\mathrm{\forall }m,n\in \mathbb{Z}$” is false.

### Justification:

The isomorphism ${\mathbb{Z}}_{mn}\simeq {\mathbb{Z}}_{m}×{\mathbb{Z}}_{n}$${\mathbb{Z}}_{mn}\simeq {\mathbb{Z}}_{m}×{\mathbb{Z}}_{n}$Z_(mn)≃Z_(m)xxZ_(n)\mathbb{Z}_{mn} \simeq \mathbb{Z}_{m} \times \mathbb{Z}_{n}${\mathbb{Z}}_{mn}\simeq {\mathbb{Z}}_{m}×{\mathbb{Z}}_{n}$ holds if and only if $m$$m$mm$m$ and $n$$n$nn$n$ are coprime (i.e., $gcd\left(m,n\right)=1$$gcd\left(m,n\right)=1$gcd(m,n)=1\gcd(m, n) = 1$gcd\left(m,n\right)=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}}_{4×2}={\mathbb{Z}}_{8}$${\mathbb{Z}}_{4×2}={\mathbb{Z}}_{8}$Z_(4xx2)=Z_(8)\mathbb{Z}_{4 \times 2} = \mathbb{Z}_8${\mathbb{Z}}_{4×2}={\mathbb{Z}}_{8}$.
However, ${\mathbb{Z}}_{4}×{\mathbb{Z}}_{2}$${\mathbb{Z}}_{4}×{\mathbb{Z}}_{2}$Z_(4)xxZ_(2)\mathbb{Z}_4 \times \mathbb{Z}_2${\mathbb{Z}}_{4}×{\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 $\left[0\right],\left[1\right],\left[2\right],\left[3\right],\left[4\right],\left[5\right],\left[6\right],\left[7\right]$$\left[0\right],\left[1\right],\left[2\right],\left[3\right],\left[4\right],\left[5\right],\left[6\right],\left[7\right]$[0],[1],[2],[3],[4],[5],[6],[7][0], [1], [2], [3], [4], [5], [6], [7]$\left[0\right],\left[1\right],\left[2\right],\left[3\right],\left[4\right],\left[5\right],\left[6\right],\left[7\right]$ and is cyclic, generated by $\left[1\right]$$\left[1\right]$[1][1]$\left[1\right]$.
2. ${\mathbb{Z}}_{4}×{\mathbb{Z}}_{2}$${\mathbb{Z}}_{4}×{\mathbb{Z}}_{2}$Z_(4)xxZ_(2)\mathbb{Z}_4 \times \mathbb{Z}_2${\mathbb{Z}}_{4}×{\mathbb{Z}}_{2}$ has elements $\left(0,0\right),\left(0,1\right),\left(1,0\right),\left(1,1\right),\left(2,0\right),\left(2,1\right),\left(3,0\right),\left(3,1\right)$$\left(0,0\right),\left(0,1\right),\left(1,0\right),\left(1,1\right),\left(2,0\right),\left(2,1\right),\left(3,0\right),\left(3,1\right)$(0,0),(0,1),(1,0),(1,1),(2,0),(2,1),(3,0),(3,1)(0,0), (0,1), (1,0), (1,1), (2,0), (2,1), (3,0), (3,1)$\left(0,0\right),\left(0,1\right),\left(1,0\right),\left(1,1\right),\left(2,0\right),\left(2,1\right),\left(3,0\right),\left(3,1\right)$ and is not cyclic.
Since ${\mathbb{Z}}_{8}$${\mathbb{Z}}_{8}$Z_(8)\mathbb{Z}_8${\mathbb{Z}}_{8}$ is cyclic and ${\mathbb{Z}}_{4}×{\mathbb{Z}}_{2}$${\mathbb{Z}}_{4}×{\mathbb{Z}}_{2}$Z_(4)xxZ_(2)\mathbb{Z}_4 \times \mathbb{Z}_2${\mathbb{Z}}_{4}×{\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}$$\mathrm{G}$G\mathrm{G}$\mathrm{G}$ is a finite group and $\mathrm{m}\mid \mathrm{o}\left(\mathrm{G}\right),\mathrm{m}\in \mathbb{N}$$\mathrm{m}\mid \mathrm{o}\left(\mathrm{G}\right),\mathrm{m}\in \mathbb{N}$m∣o(G),minN\mathrm{m} \mid \mathrm{o}(\mathrm{G}), \mathrm{m} \in \mathbb{N}$\mathrm{m}\mid \mathrm{o}\left(\mathrm{G}\right),\mathrm{m}\in \mathbb{N}$, then $\mathrm{G}$$\mathrm{G}$G\mathrm{G}$\mathrm{G}$ has a subgroup of order $\mathrm{m}$$\mathrm{m}$m\mathrm{m}$\mathrm{m}$.
The statement “If $G$$G$GG$G$ is a finite group and $m\mid o\left(G\right),m\in \mathbb{N}$$m\mid o\left(G\right),m\in \mathbb{N}$m∣o(G),m inNm \mid o(G), m \in \mathbb{N}$m\mid o\left(G\right),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}$${S}_{3}$S_(3)S_3${S}_{3}$ with $o\left({S}_{3}\right)=6$$o\left({S}_{3}\right)=6$o(S_(3))=6o(S_3) = 6$o\left({S}_{3}\right)=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^(“th “)15^{\text {th }} roots of unity, then $\mathbb{Q}\left({\beta }_{1}\right)=\mathbb{Q}\left({\beta }_{2}\right)$$\mathbb{Q}\left({\beta }_{1}\right)=\mathbb{Q}\left({\beta }_{2}\right)$Q(beta_(1))=Q(beta_(2))\mathbb{Q}\left(\beta_1\right)=\mathbb{Q}\left(\beta_2\right)$\mathbb{Q}\left({\beta }_{1}\right)=\mathbb{Q}\left({\beta }_{2}\right)$.
The statement “If ${\beta }_{1}$${\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}\left({\beta }_{1}\right)=\mathbb{Q}\left({\beta }_{2}\right)$$\mathbb{Q}\left({\beta }_{1}\right)=\mathbb{Q}\left({\beta }_{2}\right)$Q(beta_(1))=Q(beta_(2))\mathbb{Q}(\beta_1) = \mathbb{Q}(\beta_2)$\mathbb{Q}\left({\beta }_{1}\right)=\mathbb{Q}\left({\beta }_{2}\right)$” is true.

### Justification:

The ${15}^{\text{th}}$${15}^{\text{th}}$15^(“th”)15^{\text{th}}${15}^{\text{th}}$ roots of unity are the complex numbers of the form
${\beta }_{k}=\mathrm{cos}\left(\frac{2\pi k}{15}\right)+i\mathrm{sin}\left(\frac{2\pi k}{15}\right),$${\beta }_{k}=\mathrm{cos}\left(\frac{2\pi k}{15}\right)+i\mathrm{sin}\left(\frac{2\pi k}{15}\right),$beta _(k)=cos((2pi k)/(15))+i sin((2pi k)/(15)),\beta_k = \cos\left(\frac{2\pi k}{15}\right) + i \sin\left(\frac{2\pi k}{15}\right),${\beta }_{k}=\mathrm{cos}\left(\frac{2\pi k}{15}\right)+i\mathrm{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$${x}^{15}-1=0$x^(15)-1=0x^{15} – 1 = 0${x}^{15}-1=0$, and they generate a cyclotomic field $\mathbb{Q}\left({\zeta }_{15}\right)$$\mathbb{Q}\left({\zeta }_{15}\right)$Q(zeta_(15))\mathbb{Q}(\zeta_{15})$\mathbb{Q}\left({\zeta }_{15}\right)$, where ${\zeta }_{15}={e}^{2\pi i/15}$${\zeta }_{15}={e}^{2\pi i/15}$zeta_(15)=e^(2pi i//15)\zeta_{15} = e^{2\pi i / 15}${\zeta }_{15}={e}^{2\pi i/15}$.
Now, any ${15}^{\text{th}}$${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}}$${\beta }_{2}={\zeta }_{15}^{{k}_{2}}$beta_(2)=zeta_(15)^(k_(2))\beta_2 = \zeta_{15}^{k_2}${\beta }_{2}={\zeta }_{15}^{{k}_{2}}$ for some ${k}_{1},{k}_{2}\in \left\{0,1,2,\dots ,14\right\}$${k}_{1},{k}_{2}\in \left\{0,1,2,\dots ,14\right\}$k_(1),k_(2)in{0,1,2,dots,14}k_1, k_2 \in \{0, 1, 2, \ldots, 14\}${k}_{1},{k}_{2}\in \left\{0,1,2,\dots ,14\right\}$.
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}$${\zeta }_{15}$zeta_(15)\zeta_{15}${\zeta }_{15}$, it follows that $\mathbb{Q}\left({\beta }_{1}\right)$$\mathbb{Q}\left({\beta }_{1}\right)$Q(beta_(1))\mathbb{Q}(\beta_1)$\mathbb{Q}\left({\beta }_{1}\right)$ and $\mathbb{Q}\left({\beta }_{2}\right)$$\mathbb{Q}\left({\beta }_{2}\right)$Q(beta_(2))\mathbb{Q}(\beta_2)$\mathbb{Q}\left({\beta }_{2}\right)$ are both subfields of $\mathbb{Q}\left({\zeta }_{15}\right)$$\mathbb{Q}\left({\zeta }_{15}\right)$Q(zeta_(15))\mathbb{Q}(\zeta_{15})$\mathbb{Q}\left({\zeta }_{15}\right)$.
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}\left({\beta }_{1}\right)=\mathbb{Q}\left({\zeta }_{15}\right)$$\mathbb{Q}\left({\beta }_{1}\right)=\mathbb{Q}\left({\zeta }_{15}\right)$Q(beta_(1))=Q(zeta_(15))\mathbb{Q}(\beta_1) = \mathbb{Q}(\zeta_{15})$\mathbb{Q}\left({\beta }_{1}\right)=\mathbb{Q}\left({\zeta }_{15}\right)$ and $\mathbb{Q}\left({\beta }_{2}\right)=\mathbb{Q}\left({\zeta }_{15}\right)$$\mathbb{Q}\left({\beta }_{2}\right)=\mathbb{Q}\left({\zeta }_{15}\right)$Q(beta_(2))=Q(zeta_(15))\mathbb{Q}(\beta_2) = \mathbb{Q}(\zeta_{15})$\mathbb{Q}\left({\beta }_{2}\right)=\mathbb{Q}\left({\zeta }_{15}\right)$, which implies $\mathbb{Q}\left({\beta }_{1}\right)=\mathbb{Q}\left({\beta }_{2}\right)$$\mathbb{Q}\left({\beta }_{1}\right)=\mathbb{Q}\left({\beta }_{2}\right)$Q(beta_(1))=Q(beta_(2))\mathbb{Q}(\beta_1) = \mathbb{Q}(\beta_2)$\mathbb{Q}\left({\beta }_{1}\right)=\mathbb{Q}\left({\beta }_{2}\right)$.
Thus, the statement is true.

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