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Can somebody help me to find a $p$-Sylow-subgroup of GL$_2(\mathbb{Z}/p\mathbb{Z})$?

I actually dont even know how to start :/

Thank you!

TwoStones
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    Do you know what the size of such a subgroup must be? That'd be the first place I'd start. – Randall May 06 '19 at 14:52
  • Actually we have never dealt with $\mathbb{Z}/p\mathbb{Z}$, so I dont know of what order the GL$_2 (\mathbb{Z}/p\mathbb{Z})$ is. But in general the order of a $p$-Sylow-subgroup is $p^k$ where $k$ is maximal – TwoStones May 06 '19 at 14:56
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    $\DeclareMathOperator{\GL}{GL}$Once you know the size of the $p$-Sylow subgroup (as per the previous comment), you may want to note that if $a \in \GL_{2}(\mathbb{Z}/p\mathbb{Z})$ has order $p$, then $(a -1)^{p} = a^{p} - 1 = 0$, so that $a - 1$ is nilpotent, which may suggest something. – Andreas Caranti May 06 '19 at 14:57
  • $\DeclareMathOperator{\GL}{GL}$Following the comment of Randall, try and compute the order of $\GL_{2}(\mathbb{Z}/p\mathbb{Z})$, so that you will be able to see what your $k$ is. – Andreas Caranti May 06 '19 at 14:58
  • @TwoStones ok let me help. The group $\mathrm{GL}_2(\mathbb{Z}/p\mathbb{Z})$ has order $(p^2-1)(p^2-p)$. – Randall May 06 '19 at 14:59
  • @Randall Expanding then gives me the order $p^4-p^3-p^2+p$. Then the $k$ from my previous comment would be $1$, right? – TwoStones May 06 '19 at 15:02
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    It may be better to factor it as $p(p+1)(p-1)^2$. – Randall May 06 '19 at 15:03
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    If you want to know the order of your group, see https://math.stackexchange.com/questions/79047/on-the-order-of-elements-of-gl2-q – Julian Mejia May 06 '19 at 15:05
  • @Randall ok, but that would mean that my $p$-Sylow subgroup has $p^1 = p$ elements, right? – TwoStones May 06 '19 at 15:06
  • @JulianMejia Thank you! – TwoStones May 06 '19 at 15:06
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    @TwoStones yes, if you truly see why. Now, if you believe this, then the comment from Andreas should tell you how to finish. You are now in the advantageous position of being able to answer your own question, which, IMHO, is the best possible outcome on this site (or in mathematics, generally). – Randall May 06 '19 at 15:07
  • Yes! Let $A\in\text{GL}_2(\mathbb{Z}/p\mathbb{Z})$ with $A^p = E$. That means that $$ is a subgroup of order $p$ and therefore a $p$-Sylow-subgroup. – TwoStones May 06 '19 at 15:11
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    Now go find such an $A$.... – Randall May 06 '19 at 15:14
  • But I dont know how to do calculations in $\mathbb{Z}/p\mathbb{Z}$ – TwoStones May 06 '19 at 15:20
  • Sure you do. Add and multiply mod $p$, just like always. I have 100% faith you can do this. – Randall May 06 '19 at 15:25
  • What Andreas suggested was that if you want to find such an $A$, then $A=E+N$, $E$ being the identity and $N$ being nilpotent. Try to use the simplest nilpotent matrix you know. – Julian Mejia May 06 '19 at 15:25
  • Actually, at this point, since we only have $2 \times 2$ matrices, you can almost guess how to do it. – Randall May 06 '19 at 15:25
  • I would have tried $\begin{pmatrix} 1 & n \ 0 & 1 \end{pmatrix}$ because $\begin{pmatrix} 1 & n \ 0 & 1 \end{pmatrix}^m = \begin{pmatrix} 1 & mn \ 0 & 1 \end{pmatrix}$ – TwoStones May 06 '19 at 15:27
  • So what happens when $n=1$? – Randall May 06 '19 at 15:47
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    Ahhh, then it is $\begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}^p = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = E$ and we have found the $A$. – TwoStones May 06 '19 at 15:59
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    @TwoStones you should now write it all up here so that (a) you get good feedback on your complete solution and (b) get yourself some earned rep in the process. – Randall May 06 '19 at 16:40

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We have $|\text{GL}_2(\mathbb{Z}/p\mathbb{Z})|=p(p+1)(p-1)^2$.

Because the order of a $p$-Sylow-subgroup is the highest power of $p$ which divides the order of the group we get that every $p$-Sylow-subgroup has exactly $p$ elements.

Now let $A\in\text{GL}_2(\mathbb{Z}/p\mathbb{Z})$ with $A^p = E$, where $E$ denotes the identity matrix. From that we get that $|<A >|=p$.

So all thats left to do is to find a Matrix $A$ with $A^p =E$.

The Matrix $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ does the job, because $\begin{pmatrix}1&1\\0&1\end{pmatrix}^p=\begin{pmatrix}1&0\\0&1\end{pmatrix} =E$.

Therefore $<A>$ is a $p$-Sylow-subgroup of $\text{GL}_2(\mathbb{Z}/p\mathbb{Z})$.

TwoStones
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  • If I were grading this, I would want more explanation for your second paragraph. Why can't $p^2$ divide this? Also, if $A=E$ your third paragraph is wrong. – Randall May 06 '19 at 16:55