Let $H$ be the subgroup of the group $G$ of all $2 \times 2$ non-singular matrices whose entries are integers modulo a given prime $p$ consisting of those and only those matrices in $G$ whose determinant is $1$.   What is the order of $H$? And how to find it?   I've already managed to find the order of $G$. It is $p^4 - p^3 - p^2 + p$.   Last but not least, I would also like to be able to compute the order of $G$ and that of $H$ in the general case corresponding to the  $n \times n$ matrices for an arbitrary integer $n &gt; 2$.
121384
Let $p$ be a prime number and let $\mathbb{Z}_p$ denote the field of integers modulo $p$. Find the number of $2 × 2$ invertible matrices with entries from this field.  an I get some help how to solve the problem. thanks.
44474
Let $G$ is a group and $a, b \in G$. If $a$ has order $6$, then the order of $bab^{-1}$ is...  How to find this answer? Sorry for my bad question, but I need this for my study.
841
I don't know about the Sylow Theorems.  But I have been wondering about a proof of the fact that a group or order $pq$ where $p$ and $q$ are distinct primes must be cyclic. I can't quite work out the details, but here is the general idea. I would like help with filling in details. I assume that it is already known that $G$ has subgroup(s) of order $p$ and subgroup(s) of order $q$.  If $G$ is a group of order $pq$ ($p\neq q$), then I know that $G$ has a subgroup $H$ of order $p$ and a subgroup $K$ of order $q$. Then $H\simeq \mathbb{Z}_p$ and $K\simeq \mathbb{Z}_q$.   But then $H\oplus K \simeq \mathbb{Z}_{pq}$, so I would think that $H\oplus K \simeq G$. I guess one could do an internal direct product instead of an external direct product, but I don't know that $H$ and $K$ are normal subgroups.  I am asking for help completing this argument.  Edit: I see from the comments below that I might need to assume that the smaller prime does not divide the larger prime minus $1$. Or maybe it is enough to assume that the primes are greater than or equal to $3$ (Still distinct).
80359
Let $p,q$ be primes and let $G$ be a group of order $p^2q^2$, what's the best way to show $G$ is non-simple?      I know it suffices to show that one of the Sylow-p or Sylow-q subgroup of $G$ is normal, but the counting elements argument doesn't work here since different Sylow subgroups may have non-trivial intersection.
113406
Let $p$ be prime and consider the set of all matrices $$\begin{bmatrix} 1 &amp; a &amp; b \\ 0 &amp; 1 &amp; c \\ 0 &amp; 0 &amp; 1 \end{bmatrix}$$ where $a,b,c \in \mathbb{Z_p}$. This set forms a non-abelian group of order $p^3$.  Any ideas how to prove using group actions?
8324
Question:      Let $p$ be any prime and $n \geq 3$. Show that there exists a non-abelian group of order $p^n$.    Attempt:   Take $n = 3$. Writing $\mathbb Z_p \times \mathbb Z_p = \{e, \alpha_1, \ldots, \alpha_{p^2 - 1}\}$ and considering $$\begin{align}\tau_{jk} : \mathbb Z_p \times \mathbb Z_p&amp;\to \mathrm {Aut} (\mathbb Z_p)\,\\e &amp;\mapsto id\\\alpha_j &amp;\mapsto id \\\alpha_i &amp;\mapsto \rho_k  \end{align}$$  $\forall i \neq j$ where $\rho_k : \mathbb Z_p \to \mathbb Z_p$ is such that $\rho_k (1) = k$, with $k \in \{2, \ldots, p-1\}$. Then we take the group $(\mathbb Z_p \times \mathbb Z_p) \ltimes_{\tau_{jk}} \mathbb Z_p$.   Then $$(\alpha_i , 2) \ltimes_{\tau_{jk}} (\alpha_i, 1) = (2\alpha_i, 1  + \tau_{jk} (\alpha_i)(2)) = (2\alpha_i , 1  +2 k)$$  and $$(\alpha_i , 1) \ltimes_{\tau_{jk}} (\alpha_i, 2) = (2\alpha_i, 1  + \tau_{jk} (\alpha_i)(2)) = (2\alpha_i , 1  +k)$$  taking $k = 1$ for example we have that the group $(\mathbb Z_p \times \mathbb Z_p) \ltimes_{\tau_{jk}} \mathbb Z_p$ is not abelian.    I was wondering if this is a good aproach. If not, is there an easier way to reach the result?  If this is the "best" way then would the higher cases ($n &gt; 3$) be more of a notation play game?
111518
Proof Verification: A group of order $595$ has a normal Sylow 17-subgroup.   $|G|=595=5.7.17$  The divisors of $595$ are $1,5,7,17,35,85,119,595$.  $17|n_{17}-1\implies n_{17}=1,35\\7|n_7-1\implies n_7=1,85,119\\5|n_5-1\implies n_5=1$   If possible let $n_{17}=35.$ Then there is at least $35.17-34=561$ element of order $17.$ Then $n_7=1$ for otherwise the number of elements of order $17,7$ will exceed $595!$  Let $H_5,H_7$ be the normal Sylow 5 and Sylow $7$ subgroups of $G.$ Then $H_5H_7\le G$ and since $H_5\cap H_7=(e),G=H_5\times H_7\simeq H_5\oplus H_7.$ Thus none of the $35$ elements in $H_5H_7$ is of order $17.$ Hence $G$ has at least $561+35=596$ elements!  Please tell me whether the proof is right?
122412
Suppose that $G$ is a group that has exactly one nontrivial proper subgroup. Then we have to show that $G$ is cyclic and order of $G$ is $p^2$ where $p$ is prime.   I tried as, if $a$ and $b$ two element of $G$ of order $p$ and $q$ then $\langle a \rangle$ and $\langle b \rangle$ are two proper subgroups – a contradiction. Thus every element of $G$ have order some power of $p$. After this I am stuck. Thanks for help.
36047
Given $G$, $H$, $G&#39;$, and $H&#39;$ are cyclic groups of orders $m$, $n$, $m&#39;$, and $n&#39;$ respectively.   If $G*H$ is isomorphic to $G&#39;* H&#39;$, I would like to show that either $m = m&#39;$ and $n = n&#39;$ or else $m = n&#39;$ and $n = m&#39;$ holds. Where * denotes the free product.  My approach:  $G*H$ has an element of order $n$, thus $G&#39; * H&#39;$ has one too.  But already the next step is not clear to me, should I show that there is an element of length $&gt; 1$ which has infinite order or what would be the right approach here?  Thank you.
17078
Question :   If $|G|=pq$ where $p$ and $q$ are primes that are not necessarily distinct.   Prove that the order of $Z (G) =1$ or $pq$.   Showing the order is $pq$ is trivial.   I unsure how to start with showing the order is $1$. Hints are appreciated.  Thanks in advance.
40122
Let $G=\langle a\rangle$ be a cyclic group of order $n$. Show that, for every divisor $d$ of $n$, there exists a subgroup of $G$ whose order is $d$.  This time I have no approach. I haven't found any relation between the divisors of $n$ and the order of the subgroups of $G$. How would approach this?
27711
Problem.  Let $G$ be a group of order $pq$ such that $p$ and $ q$ are prime integers.  I am to show that every proper subgroup of $G$ is cyclic.  My attempt.  What I know: Any element $a$ divides $pq$ and $a^{pq} = e$.  The order of subgroups $H$ divide $pq$ by Lagrange.  If I could show that $G$ is cyclic, then all subgroups must be cyclic.  If I can show that $G$ is a group of prime order, then I can show that it is cyclic.  I'm not sure what more I can do here...I've tried looking at Fermat's Little Theorem but I can't seem to properly understand it and how it could affect my problem..
115446
Find all groups of order $6$.   If there is element of order $6$ then group is cyclic. If not, I came to conclusion that there should be an element of group $G$ of order $3$ (otherwise the order of group is $2^{n},$ for some $n \in \mathbb{N} $). Let that element be $a$. Now, let $ b \in G \setminus\langle a\rangle$. It is easy to show that all elements $ e, a, a^{2}, b, ab, a^{2}b $ are different. Now I am stuck. I am sure that this isn't enough information about group $G$ but since I am a beginner in this scope I don't know what else should I write. Any hint helps!
52402
Let $G$ be a finite Group, let S=$\{s_1,s_2,s_3 ... |s_i\in G $ has odd order $\}$. Let $H=\langle S\rangle$ The question is:   Prove that $H$ is normal subgroup of $G$. prove that $G/H$ has order $2^k$ for $k \in \mathbb{N}$.   I tried look at $$gHg^{-1}$$ but can't understand why $gHg^{-1} \subseteq H$ probably because I cant understand what unique in H. Thanks much for help.
35738
Suppose that I have a group $G$ of order $p^{2}q$ for two distinct primes $p$ and $q$.  I need to first show that one of its Sylow subgroups is normal.  I start by letting $H$ be a Sylow $p$-subgroup and $K$ be a Sylow $q$-subgroup.  If $K$ is not normal, then letting $r$ denote the number of Sylow $q$-subgroups, I have that $r|p^{2}$ and $r\equiv 1$(mod $q$).  I also have by the second Sylow theorem that $r\neq 1$.  So $r = p$ or $p^{2}$.  If $r = p^{2}$, I can write out the Sylow $q$-subgroups, and count their combined elements to show that $H$ is uniquely determined and then again imply Sylow number $2$ to obtain that $H$ is normal.  But if $r = p$, then I cannot see how to proceed.  Can anyone give any advice?  Thank you.  ( I am happy to provide more detail for the $r = p^{2}$ case if it is desired or appropriate. )
121461
If $G = \begin{pmatrix} a &amp; b \\ 0 &amp; 1 \end{pmatrix} a,b\in (\mathbb{R}) : a \neq 0$)   and assume G is a group under matrix multpication  Assume that K = ($\begin{pmatrix} s &amp; 0 \\ 0 &amp; 1 \end{pmatrix}: s\in\mathbb{R}, s \neq 0$) is a subgroup of G. Prove that K is not a normal subgroup of G.  I have done some working out on this question which is the following:  $g^{-1}=\begin{pmatrix} a^{-1}&amp;-a^{-1}b\\0&amp;1\end{pmatrix}$   and so  $$gkg^{-1}=\begin{pmatrix} a&amp;b\\0&amp;1\end{pmatrix}\begin{pmatrix}sa^{-1}&amp;-a^{-1}bs\\0&amp;1\end{pmatrix}$$  then we get $$gkg^{-1}=\begin{pmatrix}s&amp;-bs+b\\0&amp;1\end{pmatrix}$$  What do I do from here, Why is $$gkg^{-1} \notin K$$ Can't I make b=0 and then it would be in K? or am I missing some key idea?
2919
How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?  Why isn't there other possibility?  Thanks :)