Details For MMT-003 Solved Assignment

IGNOU MMT-003 Assignment Question Paper 2024

mmt-003-assignment-question-paper-1f02de37-022a-430d-b961-44d586763fd8

1. Which of the following statements are true and which are false? Give reasons for your answer.
   (a) If a finite group $G$$G$GG$G$ acts on a finite set $S$$S$SS$S$, then $G_{s_1}=G_{s_2}$$G_{s_1}=G_{s_2}$G_(s_(1))=G_(s_(2))G_{s_1}=G_{s_2}$G_{s_1}=G_{s_2}$ for all $s_1,s_2\in X$$s_1,s_2\in X$s_(1),s_(2)in Xs_1, s_2 in X$s_1,s_2\in X$.
   (b) There are exactly 8 elements of order 3 in $S_4$$S_4$S_(4)S_4$S_4$.
   (c) If $F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$F=Q(root(5)(2),root(3)(5))F=mathbb{Q}(sqrt[5]{2}, sqrt[3]{5})$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$, then $[F:\mathbb{Q}]=8$$[F:\mathbb{Q}]=8$[F:Q]=8[F: mathbb{Q}]=8$[F:\mathbb{Q}]=8$.
   (d) ${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$F_(7)(sqrt3)=F_(7)(sqrt5)mathbf{F}_7(sqrt{3})=mathbf{F}_7(sqrt{5})${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$.
   (e) For any $\alpha \in {\mathbb{F}}_{2^5}^{\ast },\alpha \ne 1,{\mathbb{F}}_{2^5}={\mathbb{F}}_2[\alpha ]$$\alpha \in {\mathbb{F}}_{2^5}^{\ast },\alpha \ne 1,{\mathbb{F}}_{2^5}={\mathbb{F}}_2[\alpha ]$alpha inF_(2^(5))^(**),alpha!=1,F_(2^(5))=F_(2)[alpha]alpha in mathbb{F}_{2^5}^*, alpha neq 1, mathbb{F}_{2^5}=mathbb{F}_2[alpha]$\alpha \in {\mathbb{F}}_{2^5}^{\ast },\alpha \ne 1,{\mathbb{F}}_{2^5}={\mathbb{F}}_2[\alpha ]$.
2. (a) Consider the natural action of $G{L}_2(\mathbb{R})$$G{L}_2(\mathbb{R})$GL_(2)(R)G L_2(mathbb{R})$G{L}_2(\mathbb{R})$ on ${\mathbf{M}}_2(\mathbb{R})$${\mathbf{M}}_2(\mathbb{R})$M_(2)(R)mathbf{M}_2(mathbb{R})${\mathbf{M}}_2(\mathbb{R})$, the set of $2\times 2$$2\times 2$2xx22 times 2$2\times 2$ real matrices, by left multiplication.
   (i) Under this action, if $\det (\mathbf{x})\ne \mathbf{0}$$\det (\mathbf{x})\ne \mathbf{0}$det(x)!=0operatorname{det}(mathbf{x}) neq mathbf{0}$\det (\mathbf{x})\ne \mathbf{0}$, show that the stabiliser of $\mathbf{x}\in M_2(\mathbb{R})$$\mathbf{x}\in M_2(\mathbb{R})$xinM_(2)(R)mathbf{x} in M_2(mathbb{R})$\mathbf{x}\in M_2(\mathbb{R})$ is $\{\mathbf{I}\}$$\{\mathbf{I}\}${I}{mathbf{I}}$\{\mathbf{I}\}$, where $\mathbf{I}$$\mathbf{I}$Imathbf{I}$\mathbf{I}$ is the $2\times 2$$2\times 2$2xx22 times 2$2\times 2$ identity matrix.
   (ii) Suppose that $\det (\mathbf{x})=\mathbf{0}$$\det (\mathbf{x})=\mathbf{0}$det(x)=0operatorname{det}(mathbf{x})=mathbf{0}$\det (\mathbf{x})=\mathbf{0}$ in the remaining parts of this exercise. We will show that the stabiliser of $\mathbf{x}$$\mathbf{x}$xmathbf{x}$\mathbf{x}$ is infinite. If $\mathbf{x}=\mathbf{0}$$\mathbf{x}=\mathbf{0}$x=0mathbf{x}=mathbf{0}$\mathbf{x}=\mathbf{0}$, the stabiliser of $\mathbf{x}$$\mathbf{x}$xmathbf{x}$\mathbf{x}$ is $G{L}_2(\mathbb{R})$$G{L}_2(\mathbb{R})$GL_(2)(R)G L_2(mathbb{R})$G{L}_2(\mathbb{R})$. So suppose $\mathbf{x}\ne \mathbf{0}$$\mathbf{x}\ne \mathbf{0}$x!=0mathbf{x} neq mathbf{0}$\mathbf{x}\ne \mathbf{0}$. Let us write $\mathbf{x}=\left[\begin{array}{ll}a& c\\ b& d\end{array}\right]$$\mathbf{x}=\left[\begin{array}{l}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}c\\ b\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}d\end{array}\right]$x=[[a,c],[b,d]]mathbf{x}=left[begin{array}{ll}a & c \ b & dend{array}right]$\mathbf{x}=\left[\begin{array}{ll}a& c\\ b& d\end{array}\right]$. Then, $\left[\begin{array}{l}a\\ b\end{array}\right]=\lambda \left[\begin{array}{l}c\\ d\end{array}\right]$$\left[\begin{array}{l}a\\ b\end{array}\right]=\lambda \left[\begin{array}{l}c\\ d\end{array}\right]$[[a],[b]]=lambda[[c],[d]]left[begin{array}{l}a \ bend{array}right]=lambdaleft[begin{array}{l}c \ dend{array}right]$\left[\begin{array}{l}a\\ b\end{array}\right]=\lambda \left[\begin{array}{l}c\\ d\end{array}\right]$ for non-zero $\lambda \in \mathbb{R}$$\lambda \in \mathbb{R}$lambda inRlambda in mathbb{R}$\lambda \in \mathbb{R}$. Why ?
   (iii) Let $\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ b^{\mathrm{\prime }}\end{array}\right]$$\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ b^{\mathrm{\prime }}\end{array}\right]$[[a^(‘)],[b^(‘)]]left[begin{array}{l}a^{prime} \ b^{prime}end{array}right]$\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ b^{\mathrm{\prime }}\end{array}\right]$ be a vector that is not a scalar multiple of $\left[\begin{array}{l}a\\ b\end{array}\right]$$\left[\begin{array}{l}a\\ b\end{array}\right]$[[a],[b]]left[begin{array}{l}a \ bend{array}right]$\left[\begin{array}{l}a\\ b\end{array}\right]$. Show that there is a matrix $\mathbf{b}=$$\mathbf{b}=$b=mathbf{b}=$\mathbf{b}=$ $\left[\begin{array}{ll}u& v\\ w& z\end{array}\right]$$\left[\begin{array}{l}u\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}v\\ w\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}z\end{array}\right]$[[u,v],[w,z]]left[begin{array}{ll}u & v \ w & zend{array}right]$\left[\begin{array}{ll}u& v\\ w& z\end{array}\right]$ such that $\mathbf{b}\left[\begin{array}{l}a\\ b\end{array}\right]=\mathbf{0}$$\mathbf{b}\left[\begin{array}{l}a\\ b\end{array}\right]=\mathbf{0}$b[[a],[b]]=0mathbf{b}left[begin{array}{l}a \ bend{array}right]=mathbf{0}$\mathbf{b}\left[\begin{array}{l}a\\ b\end{array}\right]=\mathbf{0}$ and $\mathbf{b}\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ b^{\mathrm{\prime }}\end{array}\right]=\alpha \left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ b^{\mathrm{\prime }}\end{array}\right]$$\mathbf{b}\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ b^{\mathrm{\prime }}\end{array}\right]=\alpha \left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ b^{\mathrm{\prime }}\end{array}\right]$b[[a^(‘)],[b^(‘)]]=alpha[[a^(‘)],[b^(‘)]]mathbf{b}left[begin{array}{l}a^{prime} \ b^{prime}end{array}right]=alphaleft[begin{array}{l}a^{prime} \ b^{prime}end{array}right]$\mathbf{b}\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ b^{\mathrm{\prime }}\end{array}\right]=\alpha \left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ b^{\mathrm{\prime }}\end{array}\right]$.(Hint: Set up two sets of simultaneous equations in two unknowns and argue why they have a solution.)
   (iv) Check that $\mathbf{I}-\mathbf{b}$$\mathbf{I}-\mathbf{b}$I-bmathbf{I}-mathbf{b}$\mathbf{I}-\mathbf{b}$ is in the stabiliser of $\mathbf{x}$$\mathbf{x}$xmathbf{x}$\mathbf{x}$. Also, show that there are infinitely many choices of $\alpha$$\alpha$alphaalpha$\alpha$ for which $\mathbf{I}-\mathbf{b}$$\mathbf{I}-\mathbf{b}$I-bmathbf{I}-mathbf{b}$\mathbf{I}-\mathbf{b}$ is invertible.
   (b) Let $H$$H$HH$H$ be a finite group and, for some prime $p$$p$pp$p$, let $P$$P$PP$P$ be a p-Sylow subgroup of $H$$H$HH$H$ which is normal in $H$$H$HH$H$. Suppose $H$$H$HH$H$ is normal in $K$$K$KK$K$, where $K$$K$KK$K$ is a finite group. Then, show that $P$$P$PP$P$ is normal in $K$$K$KK$K$.
   (c) Find the elementary divisors and invariant factors of ${\mathbb{Z}}_8\times {\mathbb{Z}}_{12}\times {\mathbb{Z}}_{15}$${\mathbb{Z}}_8\times {\mathbb{Z}}_{12}\times {\mathbb{Z}}_{15}$Z_(8)xxZ_(12)xxZ_(15)mathbb{Z}_8 times mathbb{Z}_{12} times mathbb{Z}_{15}${\mathbb{Z}}_8\times {\mathbb{Z}}_{12}\times {\mathbb{Z}}_{15}$.
3. Describe the set of primes $p$$p$pp$p$ for which $x^2-11$$x^2-11$x^(2)-11x^2-11$x^2-11$ splits into linear factors over ${\mathbb{Z}}_p$${\mathbb{Z}}_p$Z_(p)mathbb{Z}_p${\mathbb{Z}}_p$.
4. (a) Determine, up to isomorphism, all the finite groups with exactly 2 conjugacy classes.
   (b) Is there a finite group with class equation $1+1+2+2+2+2+2+2$$1+1+2+2+2+2+2+2$1+1+2+2+2+2+2+21+1+2+2+2+2+2+2$1+1+2+2+2+2+2+2$ ?
   (c) Compute the following:
   a) $\left(\frac{173}{211}\right)$$\left(\frac{173}{211}\right)$((173)/(211))left(frac{173}{211}right)$\left(\frac{173}{211}\right)$
   b) $\left(\frac{167}{239}\right)$$\left(\frac{167}{239}\right)$((167)/(239))left(frac{167}{239}right)$\left(\frac{167}{239}\right)$.
5. (a) Let $F(\alpha )$$F(\alpha )$F(alpha)F(alpha)$F(\alpha )$ be a finite extension $F$$F$FF$F$ of odd degree(greater than 1). Show that $F\left({\alpha }^2\right)=F(\alpha )$$F\left({\alpha }^2\right)=F(\alpha )$F(alpha^(2))=F(alpha)Fleft(alpha^2right)=F(alpha)$F\left({\alpha }^2\right)=F(\alpha )$.
   (b) Let $F\subset K$$F\subset K$F sub KF subset K$F\subset K$ and let $\alpha ,\beta \in K$$\alpha ,\beta \in K$alpha,beta in Kalpha, beta in K$\alpha ,\beta \in K$ be algebraic over $\mathrm{F}$$\mathrm{F}$Fmathrm{F}$\mathrm{F}$ of degree $\mathrm{m}$$\mathrm{m}$mmathrm{m}$\mathrm{m}$ and $\mathrm{n}$$\mathrm{n}$nmathrm{n}$\mathrm{n}$, respectively. Show that $[F(\alpha ,\beta ):F]\le mn$$[F(\alpha ,\beta ):F]\le mn$[F(alpha,beta):F] <= mn[F(alpha, beta): F] leq m n$[F(\alpha ,\beta ):F]\le mn$. What can you say about $[F(\alpha ,\beta ):F]$$[F(\alpha ,\beta ):F]$[F(alpha,beta):F][F(alpha, beta): F]$[F(\alpha ,\beta ):F]$ if $\mathrm{m}$$\mathrm{m}$mmathrm{m}$\mathrm{m}$ and $\mathrm{n}$$\mathrm{n}$nmathrm{n}$\mathrm{n}$ are coprime?
   (c) Find $[\mathbb{Q}(\sqrt[3]{2},\omega ):\mathbb{Q}]$$[\mathbb{Q}(\sqrt[3]{2},\omega ):\mathbb{Q}]$[Q(root(3)(2),omega):Q][mathbb{Q}(sqrt[3]{2}, omega): mathbb{Q}]$[\mathbb{Q}(\sqrt[3]{2},\omega ):\mathbb{Q}]$ where ${\omega }^3=1,\omega \ne 1$${\omega }^3=1,\omega \ne 1$omega^(3)=1,omega!=1omega^3=1, omega neq 1${\omega }^3=1,\omega \ne 1$.
6. (a) If $\mathrm{char}(F)\ne 2$$\mathrm{char}(F)\ne 2$char(F)!=2operatorname{char}(F) neq 2$\mathrm{char}(F)\ne 2$, show that a polynomial $a{x}^2+bx+c$$a{x}^2+bx+c$ax^(2)+bx+ca x^2+b x+c$a{x}^2+bx+c$ is irreducible iff $b^2-4ac\notin {\mathbb{F}}^{\ast 2}$$b^2-4ac\notin {\mathbb{F}}^{\ast 2}$b^(2)-4ac!inF^(**2)b^2-4 a c notin mathbb{F}^{* 2}$b^2-4ac\notin {\mathbb{F}}^{\ast 2}$ where ${\mathbb{F}}^{\ast 2}$${\mathbb{F}}^{\ast 2}$F^(**2)mathbb{F}^{* 2}${\mathbb{F}}^{\ast 2}$ is the group of squares in ${\mathbb{F}}^{\ast }$${\mathbb{F}}^{\ast }$F^(**)mathbb{F}^*${\mathbb{F}}^{\ast }$.
   (b) By looking at the factorisation of $x^9-x\in {\mathbb{F}}_3[x]$$x^9-x\in {\mathbb{F}}_3[x]$x^(9)-x inF_(3)[x]x^9-x in mathbb{F}_3[x]$x^9-x\in {\mathbb{F}}_3[x]$ guess the number of irreducible polynomials of degree 2 over ${\mathbb{F}}_3$${\mathbb{F}}_3$F_(3)mathbb{F}_3${\mathbb{F}}_3$. Find all the irreducible polynomials of degree 2 over ${\mathbb{F}}_3$${\mathbb{F}}_3$F_(3)mathbb{F}_3${\mathbb{F}}_3$.
   (c) If $\mathbb{F}$$\mathbb{F}$Fmathbb{F}$\mathbb{F}$ is a finite field show that there is always an irreducible polynomial of the form $x^3-x+a$$x^3-x+a$x^(3)-x+ax^3-x+a$x^3-x+a$ where $a\in F$$a\in F$a in Fa in F$a\in F$.(Hint: Show that $x\mapsto x^3-x$$x\mapsto x^3-x$x|->x^(3)-xx mapsto x^3-x$x\mapsto x^3-x$ is not a surjective map.)
7. (a) Suppose that $M=\left[\begin{array}{ll}A& B\\ C& D\end{array}\right]$$M=\left[\begin{array}{l}A\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}B\\ C\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}D\end{array}\right]$M=[[A,B],[C,D]]M=left[begin{array}{ll}A & B \ C & Dend{array}right]$M=\left[\begin{array}{ll}A& B\\ C& D\end{array}\right]$ is $2n\times 2n$$2n\times 2n$2n xx2n2 n times 2 n$2n\times 2n$ matrix where $A,B,C$$A,B,C$A,B,CA, B, C$A,B,C$ and $D$$D$DD$D$ are $n\times n$$n\times n$n xx nn times n$n\times n$ matrices. Show that $M$$M$MM$M$ is symplectic if and only if the following conditions are satisfied:

$$\begin{array}{rl}& A^{t}D-C^{t}B=\mathbf{I}\\ & A^{t}C-C^{t}A=\mathbf{0}\\ & B^{t}D-D^{t}B=\mathbf{0}\end{array}$$$$\begin{array}{r}{A}^{t}D-C^{t}B=\mathbf{I}\\ A^{t}C-C^{t}A=\mathbf{0}\\ B^{t}D-D^{t}B=\mathbf{0}\end{array}$${:[A^(t)D-C^(t)B=I],[A^(t)C-C^(t)A=0],[B^(t)D-D^(t)B=0]:}begin{aligned}
& A^t D-C^t B=mathbf{I} \
& A^t C-C^t A=mathbf{0} \
& B^t D-D^t B=mathbf{0}
end{aligned}$$\begin{array}{rl}& A^{t}D-C^{t}B=\mathbf{I}\\ & A^{t}C-C^{t}A=\mathbf{0}\\ & B^{t}D-D^{t}B=\mathbf{0}\end{array}$$
(Hint: Use block matrix multiplication.)
Also, check that the matrix $\left[\begin{array}{cc}\mathbf{0}& -A\\ A& \mathbf{0}\end{array}\right]$$\left[\begin{array}{cc}\mathbf{0}& -A\\ A& \mathbf{0}\end{array}\right]$[[0,-A],[A,0]]left[begin{array}{cc}mathbf{0} & -A \ A & mathbf{0}end{array}right]$\left[\begin{array}{cc}\mathbf{0}& -A\\ A& \mathbf{0}\end{array}\right]$, where $A$$A$AA$A$ is a $n\times n$$n\times n$n xx nn times n$n\times n$ orthogonal matrix, is a symplectic matrix.
(b) The aim of this exercise is to show that $S{P}_2(\mathbb{R})$$S{P}_2(\mathbb{R})$SP_(2)(R)S P_2(mathbb{R})$S{P}_2(\mathbb{R})$ acts transitively on ${\mathbb{R}}^2\mathrm{\setminus }\{0\}$${\mathbb{R}}^2\mathrm{\setminus }\{0\}$R^(2)\{0}mathbb{R}^2 backslash{0}${\mathbb{R}}^2\mathrm{\setminus }\{0\}$.
(c) Show that
(i) Show that a matrix $\left[\begin{array}{ll}a& b\\ c& d\end{array}\right]\in G{L}_2(\mathbb{R})$$\left[\begin{array}{l}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b\\ c\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}d\end{array}\right]\in G{L}_2(\mathbb{R})$[[a,b],[c,d]]in GL_(2)(R)left[begin{array}{ll}a & b \ c & dend{array}right] in G L_2(mathbb{R})$\left[\begin{array}{ll}a& b\\ c& d\end{array}\right]\in G{L}_2(\mathbb{R})$ is symplectic if and only if $ad-bc=1$$ad-bc=1$ad-bc=1a d-b c=1$ad-bc=1$.
(ii) Show that, to prove that ${\mathrm{S}\mathrm{P}}_2(\mathbb{R})$${\mathrm{S}\mathrm{P}}_2(\mathbb{R})$SP_(2)(R)mathrm{SP}_2(mathbb{R})${\mathrm{S}\mathrm{P}}_2(\mathbb{R})$ acts transitively on $G{L}_2(\mathbb{R})$$G{L}_2(\mathbb{R})$GL_(2)(R)G L_2(mathbb{R})$G{L}_2(\mathbb{R})$, it is enough to show that, for any vector $\left[\begin{array}{l}a\\ b\end{array}\right]\ne \mathbf{0}\in {\mathbb{R}}^2$$\left[\begin{array}{l}a\\ b\end{array}\right]\ne \mathbf{0}\in {\mathbb{R}}^2$[[a],[b]]!=0inR^(2)left[begin{array}{l}a \ bend{array}right] neq mathbf{0} in mathbb{R}^2$\left[\begin{array}{l}a\\ b\end{array}\right]\ne \mathbf{0}\in {\mathbb{R}}^2$, there is a $2\times 2$$2\times 2$2xx22 times 2$2\times 2$ symplectic matrix with $\left[\begin{array}{l}a\\ b\end{array}\right]$$\left[\begin{array}{l}a\\ b\end{array}\right]$[[a],[b]]left[begin{array}{l}a \ bend{array}right]$\left[\begin{array}{l}a\\ b\end{array}\right]$ as the first column. (Hint: For any matrix $A$$A$AA$A$, what is $A\left[\begin{array}{l}1\\ 0\end{array}\right]$$A\left[\begin{array}{l}1\\ 0\end{array}\right]$A[[1],[0]]Aleft[begin{array}{l}1 \ 0end{array}right]$A\left[\begin{array}{l}1\\ 0\end{array}\right]$ ?)
(iii) Complete the proof by showing that, given any non-zero vector $\left[\begin{array}{l}a\\ b\end{array}\right]$$\left[\begin{array}{l}a\\ b\end{array}\right]$[[a],[b]]left[begin{array}{l}a \ bend{array}right]$\left[\begin{array}{l}a\\ b\end{array}\right]$, there is always a non-zero vector $\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ b^{\mathrm{\prime }}\end{array}\right]$$\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ b^{\mathrm{\prime }}\end{array}\right]$[[a^(‘)],[b^(‘)]]left[begin{array}{l}a^{prime} \ b^{prime}end{array}right]$\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ b^{\mathrm{\prime }}\end{array}\right]$ such that $\left[\begin{array}{ll}a& a^{\mathrm{\prime }}\\ b& b^{\mathrm{\prime }}\end{array}\right]$$\left[\begin{array}{l}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{a}^{\mathrm{\prime }}\\ b\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{b}^{\mathrm{\prime }}\end{array}\right]$[[a,a^(‘)],[b,b^(‘)]]left[begin{array}{ll}a & a^{prime} \ b & b^{prime}end{array}right]$\left[\begin{array}{ll}a& a^{\mathrm{\prime }}\\ b& b^{\mathrm{\prime }}\end{array}\right]$ is symplectic.
8. In this exercise, we ask you to find the Sylow $p$$p$pp$p$-subgroups of the dihedral group
$$D_n=⟨x,y:x^n,y^2,yxyx⟩,n\in \mathbb{N},n\ge 2.$$$$D_n=⟨x,y:x^n,y^2,yxyx⟩,n\in \mathbb{N},n\ge 2.$$D_(n)=(:x,y:x^(n),y^(2),yxyx:),n inN,n >= 2.D_n=leftlangle x, y: x^n, y^2, y x y xrightrangle, n in mathbb{N}, n geq 2 .$$D_n=⟨x,y:x^n,y^2,yxyx⟩,n\in \mathbb{N},n\ge 2.$$
(a) Let $p$$p$pp$p$ be an odd prime that divides $n,n=p^{r}l,p\nmid l$$n,n=p^{r}l,p\nmid l$n,n=p^(r)l,p∤ln, n=p^r l, p nmid l$n,n=p^{r}l,p\nmid l$. Suppose $C=⟨x^l⟩$$C=⟨x^l⟩$C=(:x^(l):)C=leftlangle x^lrightrangle$C=⟨x^l⟩$. Show that $C$$C$CC$C$ is the unique Sylow $p$$p$pp$p$-subgroup of $D_n$$D_n$D_(n)D_n$D_n$.
(b) Prove the relation
$$y^i{x}^j{y}^k{x}^l=\{\begin{array}{ll}{y}^i{x}^{j+l}& \text{if}\mathrm{k}\text{is even}\\ y^{i+k}{x}^{l-j}& \text{if}\mathrm{k}\text{is odd.}\end{array}$$$$y^i{x}^j{y}^k{x}^l=\left\{\begin{array}{l}{y}^i{x}^{j+l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}\mathrm{k}\text{is even}\\ y^{i+k}{x}^{l-j}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}\mathrm{k}\text{is odd.}\end{array}\right.$$y^(i)x^(j)y^(k)x^(l)={[y^(i)x^(j+l),” if “k” is even “],[y^(i+k)x^(l-j),” if “k” is odd. “]:}y^i x^j y^k x^l= begin{cases}y^i x^{j+l} & text { if } mathrm{k} text { is even } \ y^{i+k} x^{l-j} & text { if } mathrm{k} text { is odd. }end{cases}$$y^i{x}^j{y}^k{x}^l=\{\begin{array}{ll}{y}^i{x}^{j+l}& \text{if}\mathrm{k}\text{is even}\\ y^{i+k}{x}^{l-j}& \text{if}\mathrm{k}\text{is odd.}\end{array}$$
Further, find all the elements of order 2 in $D_n$$D_n$D_(n)D_n$D_n$.
(c) Find all the Sylow 2-subgroups of $D_n$$D_n$D_(n)D_n$D_n$ when $n$$n$nn$n$ is odd. Describe them in terms of $x$$x$xx$x$ and $y$$y$yy$y$.
(d) Suppose $n$$n$nn$n$ is even, $n=2^{k}m$$n=2^{k}m$n=2^(k)mn=2^k m$n=2^{k}m$, where $2\nmid m,k\ge 2$$2\nmid m,k\ge 2$2∤m,k >= 22 nmid m, k geq 2$2\nmid m,k\ge 2$. Let $N=⟨x^m⟩$$N=⟨x^m⟩$N=(:x^(m):)N=leftlangle x^mrightrangle$N=⟨x^m⟩$ and $H=⟨y⟩$$H=⟨y⟩$H=(:y:)H=langle yrangle$H=⟨y⟩$. Show that $HN$$HN$HNH N$HN$ is a subgroup of $D_n$$D_n$D_(n)D_n$D_n$. What is its order?
(e) Suppose $n$$n$nn$n$ is as in the previous part. Find all the Sylow 2-supgroups of $D_n$$D_n$D_(n)D_n$D_n$. Describe them in terms of $x$$x$xx$x$ and $y$$y$yy$y$.
9. (a) Let $G=⟨a,b\mid a^2,b^3,ab{a}^{-1}{b}^{-1}⟩$$G=⟨a,b\mid a^2,b^3,ab{a}^{-1}{b}^{-1}⟩$G=(:a,b∣a^(2),b^(3),aba^(-1)b^(-1):)G=leftlangle a, b mid a^2, b^3, a b a^{-1} b^{-1}rightrangle$G=⟨a,b\mid a^2,b^3,ab{a}^{-1}{b}^{-1}⟩$. Show that $G$$G$GG$G$ is the cyclic group of order six.
(b) Solve the following set of congruences:
$$\begin{array}{rlrl}x& \equiv 2& & (mod17)\\ 3x& \equiv 4& (mod19)\\ x& \equiv 7& & (mod23)\end{array}$$$$\begin{array}{r}x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\equiv 2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(mod17)\\ 3x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\equiv 4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(mod19)\\ x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\equiv 7\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(mod23)\end{array}$${:[x,-=2,,(mod 17)],[3x,-=4,(mod 19)],[x,-=7,,(mod 23)]:}begin{array}{rlrl}
x & equiv 2 & & (bmod 17) \
3 x & equiv 4 & (bmod 19) \
x & equiv 7 & & (bmod 23)
end{array}$$\begin{array}{rlrl}x& \equiv 2& & (mod17)\\ 3x& \equiv 4& (mod19)\\ x& \equiv 7& & (mod23)\end{array}$$
(c) Show that $\mathbb{Q}(\sqrt{-19})$$\mathbb{Q}(\sqrt{-19})$Q(sqrt(-19))mathbb{Q}(sqrt{-19})$\mathbb{Q}(\sqrt{-19})$ is not a UFD by giving two different factorisations of 20 .


[stray-random categories=trigonometry]

MMT-003 Sample Solution 2024

mmt-003-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

MMT-003 Solved Assignment 2024

1. Which of the following statements are true and which are false? Give reasons for your answer.
   (a) If a finite group $G$$G$GG$G$ acts on a finite set $S$$S$SS$S$, then $G_{s_1}=G_{s_2}$$G_{s_1}=G_{s_2}$G_(s_(1))=G_(s_(2))G_{s_1}=G_{s_2}$G_{s_1}=G_{s_2}$ for all $s_1,s_2\in X$$s_1,s_2\in X$s_(1),s_(2)in Xs_1, s_2 in X$s_1,s_2\in X$.

Answer:
The statement "If a finite group $G$$G$GG$G$ acts on a finite set $S$$S$SS$S$, then $G_{s_1}=G_{s_2}$$G_{s_1}=G_{s_2}$G_(s_(1))=G_(s_(2))G_{s_1}=G_{s_2}$G_{s_1}=G_{s_2}$ for all $s_1,s_2\in S$$s_1,s_2\in S$s_(1),s_(2)in Ss_1, s_2 in S$s_1,s_2\in S$" is false. Here, $G_{s_1}$$G_{s_1}$G_(s_(1))G_{s_1}$G_{s_1}$ and $G_{s_2}$$G_{s_2}$G_(s_(2))G_{s_2}$G_{s_2}$ denote the stabilizer subgroups of $s_1$$s_1$s_(1)s_1$s_1$ and $s_2$$s_2$s_(2)s_2$s_2$ in $S$$S$SS$S$, respectively. The stabilizer subgroup $G_s$$G_s$G_(s)G_s$G_s$ of an element $s\in S$$s\in S$s in Ss in S$s\in S$ is defined as the set of all elements in $G$$G$GG$G$ that fix $s$$s$ss$s$, i.e., $G_s=\{g\in G\mid g\cdot s=s\}$$G_s=\{g\in G\mid g\cdot s=s\}$G_(s)={g in G∣g*s=s}G_s = {g in G mid g cdot s = s}$G_s=\{g\in G\mid g\cdot s=s\}$.
Justification/Proof:
The statement claims that for any two elements $s_1$$s_1$s_(1)s_1$s_1$ and $s_2$$s_2$s_(2)s_2$s_2$ in the set $S$$S$SS$S$, their stabilizer subgroups in $G$$G$GG$G$ are equal. This would mean that every element of $G$$G$GG$G$ that fixes $s_1$$s_1$s_(1)s_1$s_1$ also fixes $s_2$$s_2$s_(2)s_2$s_2$ and vice versa. However, this is not necessarily true for all actions of $G$$G$GG$G$ on $S$$S$SS$S$. The equality of stabilizers for all elements in $S$$S$SS$S$ would imply a very rigid action, essentially saying that either the entire group fixes every element of $S$$S$SS$S$ or that the action is trivial (where $G$$G$GG$G$ acts on $S$$S$SS$S$ in such a way that every element of $G$$G$GG$G$ fixes every element of $S$$S$SS$S$).
Counterexample:
Consider the group $G=\{e,g\}$$G=\{e,g\}$G={e,g}G = {e, g}$G=\{e,g\}$ where $e$$e$ee$e$ is the identity element and $g$$g$gg$g$ is an element of order 2 (meaning $g^2=e$$g^2=e$g^(2)=eg^2 = e$g^2=e$), acting on the set $S=\{s_1,s_2\}$$S=\{s_1,s_2\}$S={s_(1),s_(2)}S = {s_1, s_2}$S=\{s_1,s_2\}$ with two elements. Define the action of $G$$G$GG$G$ on $S$$S$SS$S$ as follows:

• $e$$e$ee$e$ fixes both $s_1$$s_1$s_(1)s_1$s_1$ and $s_2$$s_2$s_(2)s_2$s_2$ (as it must, being the identity).
• $g$$g$gg$g$ swaps $s_1$$s_1$s_(1)s_1$s_1$ and $s_2$$s_2$s_(2)s_2$s_2$, i.e., $g\cdot s_1=s_2$$g\cdot s_1=s_2$g*s_(1)=s_(2)g cdot s_1 = s_2$g\cdot s_1=s_2$ and $g\cdot s_2=s_1$$g\cdot s_2=s_1$g*s_(2)=s_(1)g cdot s_2 = s_1$g\cdot s_2=s_1$.

For this action, we have:

• $G_{s_1}=\{e\}$$G_{s_1}=\{e\}$G_(s_(1))={e}G_{s_1} = {e}$G_{s_1}=\{e\}$ because only the identity element fixes $s_1$$s_1$s_(1)s_1$s_1$.
• $G_{s_2}=\{e\}$$G_{s_2}=\{e\}$G_(s_(2))={e}G_{s_2} = {e}$G_{s_2}=\{e\}$ because only the identity element fixes $s_2$$s_2$s_(2)s_2$s_2$.

In this specific example, $G_{s_1}$$G_{s_1}$G_(s_(1))G_{s_1}$G_{s_1}$ and $G_{s_2}$$G_{s_2}$G_(s_(2))G_{s_2}$G_{s_2}$ happen to be equal because the non-identity element of $G$$G$GG$G$ does not fix either of the elements of $S$$S$SS$S$ but rather swaps them. However, the action of $g$$g$gg$g$ clearly differentiates $s_1$$s_1$s_(1)s_1$s_1$ and $s_2$$s_2$s_(2)s_2$s_2$, showing that the nature of the action can lead to different behaviors for different elements of $S$$S$SS$S$, and it’s easy to construct examples where $G_{s_1}\ne G_{s_2}$$G_{s_1}\ne G_{s_2}$G_(s_(1))!=G_(s_(2))G_{s_1} neq G_{s_2}$G_{s_1}\ne G_{s_2}$ by adjusting the action.
For a more direct counterexample to the statement, consider a group $G$$G$GG$G$ acting on a set $S=\{s_1,s_2,s_3\}$$S=\{s_1,s_2,s_3\}$S={s_(1),s_(2),s_(3)}S = {s_1, s_2, s_3}$S=\{s_1,s_2,s_3\}$ where $G$$G$GG$G$ fixes $s_1$$s_1$s_(1)s_1$s_1$ and $s_2$$s_2$s_(2)s_2$s_2$ but not $s_3$$s_3$s_(3)s_3$s_3$, or any scenario where the action is not uniform across all elements of $S$$S$SS$S$. This shows that the original statement is false.
(b) There are exactly 8 elements of order 3 in $S_4$$S_4$S_(4)S_4$S_4$.
Answer:
To determine the truth of the statement "There are exactly 8 elements of order 3 in $S_4$$S_4$S_(4)S_4$S_4$," we need to understand the structure of the symmetric group $S_4$$S_4$S_(4)S_4$S_4$ and the concept of an element’s order.
Background:

• The symmetric group $S_4$$S_4$S_(4)S_4$S_4$ consists of all permutations of four elements, with a total of $4!=24$$4!=24$4!=244! = 24$4!=24$ elements.
• The order of an element in a group is the smallest positive integer $n$$n$nn$n$ such that the element raised to the $n$$n$nn$n$th power (under the group operation, which is composition in this case) equals the identity element. For permutations, this means repeating the permutation $n$$n$nn$n$ times returns the set to its original order.

Analysis:
Elements of order 3 in $S_4$$S_4$S_(4)S_4$S_4$ are those permutations that cycle three elements while leaving the fourth unchanged. These can be represented as 3-cycles, such as $(123)$$(123)$(123)(123)$(123)$, which means that 1 goes to 2, 2 goes to 3, and 3 goes back to 1, with the fourth element (in this case, 4) remaining fixed.
To count the elements of order 3, we look for all possible 3-cycles within $S_4$$S_4$S_(4)S_4$S_4$. A 3-cycle can be chosen by selecting 3 elements out of 4 to participate in the cycle, and the order in which these three elements are arranged matters (since $(123)$$(123)$(123)(123)$(123)$ is different from $(132)$$(132)$(132)(132)$(132)$, for example).
Counting 3-Cycles:
The number of ways to choose 3 elements out of 4 is given by the combination formula $(\genfrac{}{}{0ex}{}{4}{3})$$(\genfrac{}{}{0ex}{}{4}{3})$((4)/(3))binom{4}{3}$(\genfrac{}{}{0ex}{}{4}{3})$. For each selection of 3 elements, there are 2 ways to arrange them into a cycle (since for any three distinct elements $a$$a$aa$a$, $b$$b$bb$b$, and $c$$c$cc$c$, there are exactly two 3-cycles: $(abc)$$(abc)$(abc)(abc)$(abc)$ and $(acb)$$(acb)$(acb)(acb)$(acb)$).
Thus, the total number of 3-cycles (elements of order 3) in $S_4$$S_4$S_(4)S_4$S_4$ is:
$$(\genfrac{}{}{0ex}{}{4}{3})\times 2$$$$(\genfrac{}{}{0ex}{}{4}{3})\times 2$$((4)/(3))xx2binom{4}{3} times 2$$(\genfrac{}{}{0ex}{}{4}{3})\times 2$$
Let’s calculate this value:
$$(\genfrac{}{}{0ex}{}{4}{3})\times 2=4\times 2=8$$$$(\genfrac{}{}{0ex}{}{4}{3})\times 2=4\times 2=8$$((4)/(3))xx2=4xx2=8binom{4}{3} times 2 = 4 times 2 = 8$$(\genfrac{}{}{0ex}{}{4}{3})\times 2=4\times 2=8$$
After calculating, we find that there are indeed exactly 8 elements of order 3 in $S_4$$S_4$S_(4)S_4$S_4$, which means the statement is true. Each of these elements is a permutation that cycles three of the four elements back to their original positions after three applications, which is the definition of having order 3.
(c) If $F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$F=Q(root(5)(2),root(3)(5))F=mathbb{Q}(sqrt[5]{2}, sqrt[3]{5})$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$, then $[F:\mathbb{Q}]=8$$[F:\mathbb{Q}]=8$[F:Q]=8[F: mathbb{Q}]=8$[F:\mathbb{Q}]=8$.
Answer:
To evaluate the statement "If $F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$F=Q(root(5)(2),root(3)(5))F=mathbb{Q}(sqrt[5]{2}, sqrt[3]{5})$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$, then $[F:\mathbb{Q}]=8$$[F:\mathbb{Q}]=8$[F:Q]=8[F: mathbb{Q}]=8$[F:\mathbb{Q}]=8$," we need to understand the concept of field extensions and the degree of an extension.
Background:

• $F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$F=Q(root(5)(2),root(3)(5))F=mathbb{Q}(sqrt[5]{2}, sqrt[3]{5})$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$ is a field extension of $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$ that includes both $\sqrt[5]{2}$$\sqrt[5]{2}$root(5)(2)sqrt[5]{2}$\sqrt[5]{2}$ and $\sqrt[3]{5}$$\sqrt[3]{5}$root(3)(5)sqrt[3]{5}$\sqrt[3]{5}$.
• The degree of the extension $[F:\mathbb{Q}]$$[F:\mathbb{Q}]$[F:Q][F: mathbb{Q}]$[F:\mathbb{Q}]$ is the dimension of $F$$F$FF$F$ as a vector space over $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$.
• The degree of a composite extension can be determined by the multiplicativity of degrees in tower extensions: if $K\subseteq L\subseteq M$$K\subseteq L\subseteq M$K sube L sube MK subseteq L subseteq M$K\subseteq L\subseteq M$ are fields, then $[M:K]=[M:L][L:K]$$[M:K]=[M:L][L:K]$[M:K]=[M:L][L:K][M:K] = [M:L][L:K]$[M:K]=[M:L][L:K]$.

Analysis:

1. Consider the extension $\mathbb{Q}(\sqrt[5]{2})$$\mathbb{Q}(\sqrt[5]{2})$Q(root(5)(2))mathbb{Q}(sqrt[5]{2})$\mathbb{Q}(\sqrt[5]{2})$ over $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$. The minimal polynomial of $\sqrt[5]{2}$$\sqrt[5]{2}$root(5)(2)sqrt[5]{2}$\sqrt[5]{2}$ over $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$ is $x^5-2$$x^5-2$x^(5)-2x^5 – 2$x^5-2$, which is irreducible over $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$ by Eisenstein’s criterion (with prime $p=2$$p=2$p=2p=2$p=2$). Thus, $[\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]=5$$[\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]=5$[Q(root(5)(2)):Q]=5[mathbb{Q}(sqrt[5]{2}):mathbb{Q}] = 5$[\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]=5$.
2. Consider the extension $\mathbb{Q}(\sqrt[3]{5})$$\mathbb{Q}(\sqrt[3]{5})$Q(root(3)(5))mathbb{Q}(sqrt[3]{5})$\mathbb{Q}(\sqrt[3]{5})$ over $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$. The minimal polynomial of $\sqrt[3]{5}$$\sqrt[3]{5}$root(3)(5)sqrt[3]{5}$\sqrt[3]{5}$ over $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$ is $x^3-5$$x^3-5$x^(3)-5x^3 – 5$x^3-5$, which is also irreducible over $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$ by Eisenstein’s criterion (with prime $p=5$$p=5$p=5p=5$p=5$). Thus, $[\mathbb{Q}(\sqrt[3]{5}):\mathbb{Q}]=3$$[\mathbb{Q}(\sqrt[3]{5}):\mathbb{Q}]=3$[Q(root(3)(5)):Q]=3[mathbb{Q}(sqrt[3]{5}):mathbb{Q}] = 3$[\mathbb{Q}(\sqrt[3]{5}):\mathbb{Q}]=3$.
3. To find $[F:\mathbb{Q}]$$[F:\mathbb{Q}]$[F:Q][F: mathbb{Q}]$[F:\mathbb{Q}]$ where $F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$F=Q(root(5)(2),root(3)(5))F=mathbb{Q}(sqrt[5]{2}, sqrt[3]{5})$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$, we need to consider how $\sqrt[5]{2}$$\sqrt[5]{2}$root(5)(2)sqrt[5]{2}$\sqrt[5]{2}$ and $\sqrt[3]{5}$$\sqrt[3]{5}$root(3)(5)sqrt[3]{5}$\sqrt[3]{5}$ interact within the extension. If these roots were to generate independent extensions (i.e., the extensions do not intersect in a way that reduces the overall degree), then we might expect the degree of $F$$F$FF$F$ over $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$ to be the product of the degrees of the individual extensions. However, the actual degree of the combined extension depends on the interaction between these two extensions.

Given that the degrees of the individual extensions are 5 and 3, respectively, and assuming no prior knowledge about a direct interaction that would reduce the combined degree, one might naively expect the degree of the combined extension to be $5\times 3=15$$5\times 3=15$5xx3=155 times 3 = 15$5\times 3=15$. This is because the extensions are generated by roots of polynomials that are irreducible over $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$ and do not immediately suggest a common field that would reduce the degree of the extension.
Conclusion:
The statement "If $F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$F=Q(root(5)(2),root(3)(5))F=mathbb{Q}(sqrt[5]{2}, sqrt[3]{5})$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$, then $[F:\mathbb{Q}]=8$$[F:\mathbb{Q}]=8$[F:Q]=8[F: mathbb{Q}]=8$[F:\mathbb{Q}]=8$" is false. Based on the analysis above, without specific interaction that would reduce the degree, the expected degree of the field extension $F$$F$FF$F$ over $\mathbb{Q}$$\mathbb{Q}$Qmathbb{Q}$\mathbb{Q}$ would naively be calculated as $15$$15$1515$15$, not $8$$8$88$8$, assuming independent extensions generated by $\sqrt[5]{2}$$\sqrt[5]{2}$root(5)(2)sqrt[5]{2}$\sqrt[5]{2}$ and $\sqrt[3]{5}$$\sqrt[3]{5}$root(3)(5)sqrt[3]{5}$\sqrt[3]{5}$. However, the actual degree calculation requires detailed knowledge of how these extensions interact, and in general, the degree of such a combined extension is not simply the product of the degrees if there’s a nontrivial intersection. But in this specific case, without evidence of such an intersection that would lower the degree to $8$$8$88$8$, the initial analysis suggests the statement is incorrect.
(d) ${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$F_(7)(sqrt3)=F_(7)(sqrt5)mathbf{F}_7(sqrt{3})=mathbf{F}_7(sqrt{5})${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$.
Answer:
To evaluate the statement "${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$F_(7)(sqrt3)=F_(7)(sqrt5)mathbf{F}_7(sqrt{3})=mathbf{F}_7(sqrt{5})${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$," we need to understand the structure of finite fields, specifically the field extensions of ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)mathbf{F}_7${\mathbf{F}}_7$, the finite field with 7 elements.
Background:

• ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)mathbf{F}_7${\mathbf{F}}_7$ is the finite field with 7 elements, typically represented as $\{0,1,2,3,4,5,6\}$$\{0,1,2,3,4,5,6\}${0,1,2,3,4,5,6}{0, 1, 2, 3, 4, 5, 6}$\{0,1,2,3,4,5,6\}$ with arithmetic operations performed modulo 7.
• ${\mathbf{F}}_7(\sqrt{3})$${\mathbf{F}}_7(\sqrt{3})$F_(7)(sqrt3)mathbf{F}_7(sqrt{3})${\mathbf{F}}_7(\sqrt{3})$ denotes the smallest field extension of ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)mathbf{F}_7${\mathbf{F}}_7$ containing $\sqrt{3}$$\sqrt{3}$sqrt3sqrt{3}$\sqrt{3}$, and similarly, ${\mathbf{F}}_7(\sqrt{5})$${\mathbf{F}}_7(\sqrt{5})$F_(7)(sqrt5)mathbf{F}_7(sqrt{5})${\mathbf{F}}_7(\sqrt{5})$ denotes the smallest field extension of ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)mathbf{F}_7${\mathbf{F}}_7$ containing $\sqrt{5}$$\sqrt{5}$sqrt5sqrt{5}$\sqrt{5}$.
• In the context of finite fields, an element $a$$a$aa$a$ is said to have a square root in ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)mathbf{F}_7${\mathbf{F}}_7$ if there exists an element $b\in {\mathbf{F}}_7$$b\in {\mathbf{F}}_7$b inF_(7)b in mathbf{F}_7$b\in {\mathbf{F}}_7$ such that $b^2\equiv a\mathrm{mod}7$$b^2\equiv a\mathrm{mod}7$b^(2)-=amod7b^2 equiv a mod 7$b^2\equiv a\mathrm{mod}7$.