Questions about $\mathrm{SL}_2(\mathbb{F}_7)$

Question

Let $G=\mathrm{SL}_2(\mathbb{F}_7)$, which has order $336=2^4\cdot 3\cdot 7$. And I may assume that $G$ is generated by the two matrices $$\begin{pmatrix}1&1\\0&1\end{pmatrix}, \begin{pmatrix}1&0\\1&1\end{pmatrix}.$$

Suppose $N\leq G$ is a normal subgroup not contained in $\{\pm I\}$.

I am trying to show the following assertions:

1. $|N|$ is divisible by 7 $\implies$ all Sylow 7-groups of $G$ lie in $N$ and thus $N=G$
2. $|N|$ is divisible by 3 $\implies$ $N$ has an element of order $7$
3. $|N|$ is not a power of $2$

My thoughts:

1. Let $n_7$ be the number of Sylow 7-groups in $G$. By the Sylow theorem, we know $n_7\equiv 1 \mod 7$ and also that $n_7$ divides $2^4\cdot 3=48$. Therefore $n_7=1$, i.e. there is exactly one Sylow 7-group $S_7$. In particular it is necessarily normal. But now I don't know how to use the fact that $7$ divides $|N|$ to show $S_7\subset N$.
2. Here I tried using the generators that we're given to construct something explicit. But again I wasn't sure how to use that $3$ divides $|N|$.
3. I didn't see immediately how to do this so I postponed it.

Answer 1

This solution is long and somewhat computational, but hopefully some of the ideas can be made more elegant by others:

1. Check here that $$ H := \left\{\begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix} : x\in \mathbb{F}_p\right\} $$ is a $7$-sylow subgroup. Also, if $$ A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \text{ and } B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix} $$ then $BAB^{-1} \notin H$ so $H$ is not normal. Thus, the number of $7$-Sylow subgroups $n_7$ must be $>1$. However, $n_7 \equiv 1\pmod{7}$ and $n_7 \mid 48$, so $n_7 = 8$ must hold.

Now if $N$ is a normal subgroup, as mentioned in the comment, if $7\mid |N|$, then $N$ contains one and hence all the $7$-sylow subgroups. In particular, $N$ must contain both the generators mentioned in the question, and so $N = G$.

2. Consider $$ C = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix} $$ and the subgroup $K = \langle C\rangle$ that it generates. Note that $K$ is a 3-Sylow subgroup. Clearly the diagonal matrices normalize $K$. Also, one can check that the matrix $B$ mentioned above normalizes this subgroup, so $B \in N_G(K)$. Since $B^2 = -I$, we have $o(B) = 4$, and so $4\mid |N_G(K)|$. Since $3 = |K| \mid |N_G(K)|$, we have $12\mid |N_G(K)|$.

Now conjugating $C$ with a general matrix $$ D = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$ with $ad-bc = 1$, we obtain $$ E = \begin{pmatrix} 2(1-bc) & 2ab \\ -2cd & 2(ad+1) \end{pmatrix} $$ Taking cases, $E = C$ or $E = C^2$, we can conclude that there are no matrices other than these that normalize $K$, and so $|N_G(K)| = 12$. Hence, $$ n_3 = [G:N_G(K)] = 336/12 = 28 $$ Now suppose $3\mid |N|$, then as before, $N$ must contain all the 3-Sylow subgroups. Since any two of these subgroups must intersect trivially, it follows that $|N| \geq (2\times 28) + 1 = 57$. So $|N| \in \{84, 112, 168, 336\}$. All these numbers are divisible by 7, and we are done.

3. Now if $N$ is a $2$-Subgroup with $N \neq \{\pm I\}$, then $|N| \{4,8,16\}$, and so $|G/N| \in \{84, 42, 21\}$ respectively. In each case, one can check (simply by the sylow theorems) that the number 7-Sylow subgroups in $G/N$ must be one. Thus, the 7-Sylow subgroup of $G/N$ is normal. This subgroup lifts (by the correspondence theorem) to a non-trivial normal subgroup $L$ of $G$ with $7\mid |L|$. This is impossible by (1), and this completes the proof.

To answer the questions in the comments:

(a) Any two 7-Sylow subgroups have order 7, so their intersection must have order 1 or 7.

(b) The same holds for 3-Sylow subgroups, of which there are 28. Thus, there are $28\times 2$ elements of order 3 in the group. The identity is in $N$ as well, giving at least 57 elements.

(c) This is just the Sylow (or Cauchy theorems) applied to $N$

(d) The correspondence theorem is quite a strong theorem: If $f: G\to H$ is an epimorphism, then it not only gives a one-to-one correspondence between subgroups of $H$ and subgroups of $G$ that contain $\ker(f)$, it even controls the orders. If $L < G$ such that $\ker(f) \subset L$, then $L/\ker(f) \cong f(L)$ so $|L| = |\ker(f)||f(L)|$. In particular, if $7\mid |f(L)|$, then $7\mid |L|$.