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Let $G$ be the group of all $2 \times 2$ invertible matrices with elements in $\mathbb Z_5$. This group has order of $480=2^5\cdot 3 \cdot 5$.

I want to know if there is a quick and efficient strategy to find an example of each type of Sylow subgroup. For example, I know that there could be $1, 3, 5,$ or $15$ $2$-Sylow subgroups with $32$ elements.

emka
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1 Answers1

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You want matrices such that their order is $1\le2^n\le32$. So try finding "roots" of $I$:

$$\begin{align} \begin{pmatrix}a&b\\c&d\end{pmatrix}^2&\equiv I\pmod{5}\\ \begin{pmatrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{pmatrix}&\equiv I\pmod{5}\\ a^2+bc\equiv d^2+bc&\equiv 1\pmod 5\\ b(a+d)\equiv c(a+d)&\equiv 0\pmod 5 \end{align}$$

Since $a^2-d^2=(a+d)(a-d)\equiv0\pmod 5$, $5|(a+d)$ or $5|(a-d)$.

Case $1$: $5|(a+d)$

If $a\equiv d\equiv0\pmod 5$, then $bc\equiv 1$, so we have solutions $\begin{pmatrix}0&1\\1&0\end{pmatrix},\begin{pmatrix}0&2\\3&0\end{pmatrix},\begin{pmatrix}0&3\\2&0\end{pmatrix},\begin{pmatrix}0&4\\4&0\end{pmatrix}$

Otherwise, if $(a,d)=(1,4)\lor(4,1)\Longrightarrow bc\equiv0$ and if $(a,b)=(2,3)\lor(3,2)\Longrightarrow bc\equiv 2$. In the former case, we get solutions $(b,c)=(0,0),(0,1),(0,2),(0,3),(0,4),(1,0),(2,0),(3,0),(4,0)$ while the latter gets $(b,c)=(1,2),(2,1),(3,4),(4,3)$.

Case 2: $5|(a-d)$

We already covered $a\equiv d\equiv 0$, so we can move on

Since $a\equiv d\not\equiv0\pmod 5$, $a+d\equiv 2a\not\equiv 0$, so $b\equiv c\equiv 0\pmod 5$. We then have $a^2\equiv d^2\equiv 1$, so the only solutions in this case are $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $\begin{pmatrix}4&0\\0&4\end{pmatrix}$

So that's all $32$ roots of $I$, and these will generate more than one $2$-Sylow (you can check that the roots aren't closed). In general $n^{th}$ roots of $e$ for groups is a good start, but it depends on the group's structure.

Tim Ratigan
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  • Why are we taking the matrix to $\mod 5$ instead of $\mod 15$? Also, are we raising it to the second power because $32 \equiv 2 \mod 15$. – emka Dec 02 '13 at 22:40
  • mod 5 because you're talking about $\Bbb Z/5\Bbb Z$. Second power because we want elements with orders that are powers of $2$ to find elements of $2$-Sylows – Tim Ratigan Dec 02 '13 at 23:28
  • Why isn't it raised to $2^i$, $1 \leq i \leq 5$? – emka Dec 02 '13 at 23:53
  • It's the same thing – Tim Ratigan Dec 02 '13 at 23:54
  • Why are they all the same? I'm just curious because, $2^i\mod 5$ for $1 \leq i \leq 5$ is $2,4,3,2,2$. We found these roots, I'm not sure what to do with these. Are these the generators of a subgroup? Would I approach the 3-sylow and 5-sylow subgroups in a similar strategy. – emka Dec 03 '13 at 00:11
  • Inequalities don't make sense mod anything. When I gave that bound it was for the order, I wasn't concerning myself with modular arithmetic there. Exponents of two aren't elements of $GL_2(\Bbb Z/5\Bbb Z)$, which is the group you're talking about. Together, the roots will generate a subgroup, yes. Finding cube and fifth roots would be a lot more complicated, so I don't recommend it. – Tim Ratigan Dec 03 '13 at 08:29