How would I prove that there is at most one simple group of order 168?

I've already seen that $GL_3(2)$ and $PSL_2(7)$ are simple groups of order 168, and I have seen direct proofs that they are equal.

Now I'd like to show there is only one simple group of that order.

What should I think about to start on doing that? How would you actually prove it?

The only thing I know that gives structural results about groups given only its order is Sylow's theorem. So I apply that. Since $168 = 2^3 \cdot 3 \cdot 7$ I know from Sylow's theorem there would be:

- $n_2 | 3 \cdot 7$ and $n_2 \equiv 1 \pmod 2$ so there could be $1,3,7$ or $21$ Sylow 2-subgroups.
- $n_3 | 2^3 \cdot 7$ and $n_3 \equiv 1 \pmod 3$ so there could be $1, 4, 7$ or $28$ Sylow 3-subgroups.
- $n_7 | 2^3 \cdot 3$ and $n_7 \equiv 1 \pmod 7$ so there are $1$ or $2^3 = 8$ of these.

I proved a lemma that says if a Sylow subgroup is normal iff it's unique, that proves there cannot be only $1$ 2-Sylow subgroup.

*edit*: So I think I got another useful lemma from the comments and here:

**Lemma** Let $G$ be a simple group of order $N$ with $p|N$, then either $|G| \le n_p!$.

*proof*: Let $G$ act by conjugation on its $n_p$ $p$-Sylow subgroups, because they are all conjugate this gives a surjective map $G \to S_{n_p}$, if $n_p$ is not 1 and $|G| > |S_{n_p}|$ the kernel of this map would be a normal subgroup $1 \not = N \unlhd G$.

this leaves us with the possibilities:

- $7$ or $21$ 2-Sylow subgroups.
- $7$ or $28$ 3-Sylow subgroups
- exactly $2^8$ $7$-Sylow-subgroups.

Relevant questions:

- Prove there is no element of order 6 in a simple group of order 168 - This question starts off with more knowledge about the Sylow subgroups than I am able to deduce.
- Sylow 7-subgroups in a group of order 168 - Shows that 7-Sylow subgroups are normal or have maximal normalizers (I don't understand this)
- Why $PSL_3(\mathbb F_2)\cong PSL_2(\mathbb F_7)$? - Direct proofs of the special isomorphism.