Elements of $\operatorname{SL}_2(\mathbb F_{p^n})$ of order $p^k$

Question

Let $p > 2$ be a prime number and $n\ge 1$ an integer, and consider the group $G = \operatorname{SL}_2(\mathbb F_{p^n})$ of order $p^n(p^{2n} - 1)$. Let us denote by $\operatorname{Inn}(G)$ (the "inner elements" of $G$; feel free to come up with a more creative name) the elements of $G$ whose order is some power of $p$. Do these elements constitute a subgroup?

Answer 1

As Quang Hong said right away, this does not hold. Making it concrete with the following example. The matrices $$ A=\left(\begin{array}{rr}1&1\\0&1\end{array}\right)\qquad\text{and}\qquad B=\left(\begin{array}{rr}1&0\\-1&1\end{array}\right) $$ both have order $p$. Their product $$ AB=\left(\begin{array}{rr}0&1\\-1&1\end{array}\right) $$ is of order six (unless $p=2$ in which case the order is three). I'm leaving the verification of that to you.

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The set of all the elements of order a power of $p$ is the union of $p$-Sylow subgroups. This is a subgroup if and only if there is exactly one $p$-Sylow subgroup $P$. But in that case $P$ would be a normal subgroup of $G=SL_2(\Bbb{F}_{p^n})$. But it is well known that when $p^n>3$ the only non-trivial normal subgroup of $G$ is the subgroup of scalar matrices $=Z(G)$.