Show that $\exists $ a $3\times 3$ invertible matrix $M\neq I_3$ with entries in $\mathbb Z_2$ such that $M^7=I_3$.  My try:  Consider the space $X=GL_3(\mathbb Z_2)$ .Now $|GL_3(\mathbb Z_2)|=168$.By Sylow's Theorem $X$ has a Sylow-$7$ subgroup of order $7$ say $H$ .Since $H$ is cyclic so $H$ has an element  of order $7$.Hence $\exists  A$ such that $A^7=I_3$.  But how to find the matrix $A$? I can't proceed further.  I found that this question has been asked before.I don't want  a hit and trial method for solving this question.Is there a definite way?
120852
Show that there exists a $3 × 3$ invertible matrix $M$ (which is not the identity matrix) with entries in the field $\mathbb{Z}/2\mathbb{Z}$ such that $M^7 = $Identity matrix.  All I could do was use hit and try method. I was checking different matrices which might satisfy this condition. Of course, it's a bad approach.
71010
Let $G$ be the group  of all $2 \times 2$ matrices of the form $\begin{pmatrix} a &amp; b \\ 0 &amp; d\end{pmatrix}$ where $ad \neq 0$ under matrix multiplication. Let  $N=\left\{A \in G \; \colon \; A = \begin{pmatrix}1 &amp; b \\ 0 &amp; 1\end{pmatrix} \right\}$ be  a subset  of the group $G$. Prove that  $N$ is  a normal subgroup of $G$ and prove that  $G/N$ is abelian group.   Here is my attempt!  To prove $N$ is normal I consider  the group homomorphism $f \colon G \to \mathbb R^*$ given by $f(B) = \det(B)$ for all $B$ in $G$. And I see that $f(N)$ is all the singleton $\{1\}$ since $\{1\}$ as a subgroup of $\mathbb R^*$  is normal, it follows that $N$ is also normal. Is this proof helpful  here? Then how to prove that $G/N$ is Abelian? I know $G/N$ is a collection of left cosets.   Thank you.
100707
Prove that any element of $GL_{2}(\mathbb{Z})$ of finite order has order $1,2,3,4,6$ using Field Theory. My idea is to reduce such a finite order matrix say $A$ with order $n$ to modulo a prime $p$. $\det A=\pm1$ so $A$ will land inside $G=\{M\in GL_{2}(\mathbb{F}_{p}):\det M=\pm 1\}$. $|G|=2p(p^{2}-1)$. So $n\mid 2p(p^{2}-1)$ for all prime $p$. I don't know how to proceed after this.
88691
True\False?  For every matrix $A\in M_{2}( \mathbb{C}) $ there's  $X\in M_{2}( \mathbb{C})$ such that $X^2=A$.  I know that every complexed matrix has a Jordan form matrix $J$ such that $P^{-1}CP=J$, But it's not diagonalizable for sure.  Thanks
63094
So i have this $3\times 3$ matrix  $$     \begin{pmatrix}     -2 &amp; 0 &amp; 0 \\     3 &amp; 1 &amp; -6 \\     0 &amp; 0 &amp; -2 \\     \end{pmatrix} $$  I want to check if the matrix is diagonizable. First thing i do is find the roots : namely $\lambda_1 = −2$ of multiplicity $2,$ and $\lambda_2 = 1$ of multiplicity $1.$  In order to ensure that the matrix is diagonalizable we need to find a basis of $\mathbb{R}^3$ whose elements are eigenvectors for $A.$ In particular, since we have the eigenvalue $\lambda_1=-2$ having multiplicity $2$ we need to check that the dimension of the eigenspace $E(\lambda_1)$ is equal to $2.$ We compute the eigenspaces of $A.$  $\DeclareMathOperator\Nul{Nul}$On the answer sheet is says that we need to find $\Nul(A+2Id)$... but why $2Id?$ Isn't is supposed to be $-2,$ so $\Nul(-2Id-A)$??   I do not understand this $\Nul(A+2Id).$ I always thought that we should compute $\Nul(-2Id-A),$ so what do i need to do ?   $\Nul(-2Id-A)$ gives me a different result from $\Nul(A+2Id).$  Does that mean that since $-2$ has a multiplicity of $2,$ we need to work with $2$ and not $-2$?  Somebody please help with a clear and simple explanation ! Thanks!
1817
There is an invertible $n×n$ matrix with each entry $0$ or $1$. Now we negate ($x\to 1-x$) the diagonal of the matrix, and it is still invertible. How many such matrices exist for a given $n$?  Can we derive a general formula? If $n=2$ there are $2$ such matrices, if $n=3$ there are $48$. But I have no idea whether a formula exists for higher $n$.
31401
I tried getting $n_{2}$ and $n_{7}$ , denote the number of Sylow-2-Subgroups and Sylow-7-Subgroups respectively.  I got two cases for $n_{2} = 1 , 7 $ and for  $n_{7} = 1 , 8$ , i noticed that if $n_{2} = 1$ and $n_{7}= 1$ ,then we are done since they are unique subgroups and hence will be normal in $G$.  Next how to proceed with the cases of $n_{2} = 7$ and $n_7 = 8$.  Please help?
47255
Consider the matrix $$ M=\left(\begin{matrix} 0&amp;0&amp;0&amp;1\\ 0&amp;0&amp;0&amp;0\\ 0&amp;0&amp;0&amp;0\\ 0&amp;0&amp;0&amp;0\\ \end{matrix}\right). $$ Is there a matrix $P\in{\Bbb C}^{4\times 4}$ such that $P^n=M$ for some $n&gt;1$?    One obvious fact is that if such $P$ exists, then $P$ must be nilpotent. However, I have no idea how to deal with this problem. Furthermore, what if $M$ is an arbitrary nilpotent matrix with index $k$?
16851
I'm asked to show that there are infinitely many units in the ring $\mathbb Z[\sqrt 3]$.   But I don't really see a good approach to this one, so far.   Some thoughts: The inverse of $a+\sqrt3 b$ should be given by $\pm(2 - \sqrt 3 b)$, since the norm   $$N: \mathbb Z[\sqrt{3}] \to \mathbb Z, \qquad N(x+\sqrt{3} y) = x^2 - 3y^2$$  is multiplicative. So if $(a+\sqrt{3}b)^{-1}=x+\sqrt 3 y$, then   $$1 = N(1) = N((a+\sqrt{3}b)(x + \sqrt{3}y)) = (a^2 - 3b^2)(x^2 - 3y^2)$$  Hence we must have $\pm 1 = (a^2 - 3b^2) = (a+\sqrt{3}b)(a-\sqrt{3}b)$.  Therefore I need to show that there are infinitely many $a,b$ such that $a^2 - 3b^2 = \pm 1$.   Here I don't know how to proceed.   Maybe rewriting as $a = \sqrt{3b^2 \pm 1}$, and now trying to prove that there are infinitely many $b$ for which $3b^2 \pm 1$ is a square?  A hint would be appreciated! =) Thanks.
85646
Let $G$ be a finite subgroup of $GL_n(\Bbb{Q})$. I want to show the existence of a matrix $A\in GL_n(\Bbb{Q})$ with the property that $AGA^{-1} \subseteq M_n(\Bbb{Z})$.
54848
For fixed prime power $q$ show that the general linear group $GL_n(\mathbb F_q)$ of invertible matrices with entries in the finite field $\mathbb F_q$ has an element of order $q^n-1$.   I tried to show this question with showing diagonal matrix but i can't find element directly  competible with order i think i am on wrong way please give me clue ?
77671
I am trying to solve this problem for homework but unable to get anything. The question is to find the number of invertible 2x2 matrices in mod p? Each entry can be from the set {0,1,2....(p-1)} where p is prime.
53648
Consider $$X=\Big\{A \in M_3(\Bbb R): \rho_A(x)=x^3-3x^2+2x-1\Big\}$$ where $\rho_A(x)$ is the characteristic polynomial of $A$ and $M_3(\Bbb R)$ is the space of all $3 \times 3$ matrices over $\Bbb R$. Is $X$ compact in $M_3(\Bbb R)$ ?   My try: I confused with only the $2x$ term in $\rho_A(x)$. Because , if the $2x$ term not appear , then $X$ becomes  $$X=\{A \in M_3(\Bbb R): \text{trace}(A)=3,\det A=1\} $$ which is unbounded, since $$(\forall n \in \Bbb N):\begin{pmatrix} 1&amp;0 &amp; n\\0&amp;1&amp;0\\0&amp;0&amp;1 \end{pmatrix} \in X$$ But here the problem is, the appearance of $2$.  I Know  $$2=A_{11}+A_{22}+A_{33}$$ so I think in this case the set becomes bounded and closed Any help?
97671
Let $A$ be a real square matrix en let $I$ be the identity matrix with the same size. Let $n&gt;0$ be the smallest natural number for which there exist  $a_0,a_1,...a_{n-1} \in {R}$ such that $A^n+a_{n-1}A^{n-1}+....+a_1A+a_0I=O$.  Show that A is invertible if and only if $a_0\neq0 $.  So i need to proof    $A$ invertible $\rightarrow$$a_0 \neq 0$ $a_0 \neq0 \rightarrow$ $A$ invertible    I don't really know where to start with 1, and with 2 I tried to prove it with a contradiction but I don't think it's right either.
69096
Let $G$ be a group of order 24, and we shall assume there there exist a non-normal 2-sylow group in $G$. i want to show that it's not simple.  first i have showed that there are exactly three 2-sylow group's.   and now i am trying to show that there is a non trivial homomorphism from $G$ to $S_3$. i know that $S_3\cong D_3$  and indeed there are three 2-sylow groups in $D_3$ , the reflection's. and there is one $3-$sylow group in $D_3$  so i wanted to creat a homomorphism that take the different p-sylow groups in $G$  and set them in right $p-$sylow group's in $D_3$  but for that i need to show that the intersection of every two $2-$sylow gouts in $G$  is trivial. and i can't manage to do that....
6343
I'm asked to show that $G=GL_2(\Bbb Z/2\Bbb Z)$ is isomorphic to $S_3$. I have few ideas but I don't manage to put them all together in order to obtain a satisying answer. I first tried using Cayley's theorem ($G$ is isomorphic to a subgroup of $S_6$), and I also noticed that $\operatorname{Card}(G)=\operatorname{Card}(S3)=6$ &amp; that they're both non-abelian group.  Is this enough to say that considering $S_3$ is a subgroup of $S_6$ with the same cardinality than $G$, it has to be isomorphic to it ?  Could anyone give me some elements to get a more rigorous proof or lead me to an other path to show this statement ?  Thanks in advance
78361
Let $A$ be an $n\times n$ invertible matrix. Let $a$ be a number in $\mathbb{F}$, let $\alpha$ be a row $n$-tuple of numbers from $\mathbb{F}$ and let $\beta$ be a column $n$-tuple of numbers from $\mathbb{F}$. Show that $$ |A|^{-1}\left|\begin{array}{cc}   a &amp; \alpha \\   \beta &amp; A \end{array}\right|=a-\alpha A^{-1}\beta $$  So far, I have the following: $$   \begin{pmatrix}     1 &amp; 0\\     -\beta a^{-1} &amp; I   \end{pmatrix}   \begin{pmatrix}     a &amp; \alpha\\     \beta &amp; A   \end{pmatrix}=\begin{pmatrix}     a &amp; \alpha\\     0 &amp; A-a^{-1}\beta\alpha   \end{pmatrix}\\   \left|\begin{array}{cc}   a &amp; \alpha \\   \beta &amp; A   \end{array}\right|=\left|\begin{array}{cc}     a &amp; \alpha\\     0 &amp; A-a^{-1}\beta\alpha   \end{array}\right|=|a||A-a^{-1}\beta\alpha|=|aA-\beta\alpha|\\   |A^{-1}||aA-\beta\alpha|=|a-A^{-1}\beta\alpha| $$  I don't know where to go from here...