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:

  • 1
    $\begingroup$ I vaguely recall having done this once. Looks like $7$-Sylows might be a good starting point. There are eight of them (as you pointed out). These should probably be in bijective correspondence with the 8 lines the group $PSL_2(7)$ is acting on. $\endgroup$ – Jyrki Lahtonen Mar 7 '13 at 13:04
  • 3
    $\begingroup$ And yes, I recommend doing it for 60 first. Much easier IIRC :-) $\endgroup$ – Jyrki Lahtonen Mar 7 '13 at 13:05
  • 1
    $\begingroup$ A hint for 60: If $G$ has a subgroup $H$ of index $<5$, then the permutation action of $G$ on $G/H$ gives a non-trivial homomorphism from $G$ to $S_{[G:H]}$, which is a contradiction. That allows you to exclude some possibilities, and then you will find a subgroup of index exactly five... $\endgroup$ – Jyrki Lahtonen Mar 7 '13 at 13:11
  • 1
    $\begingroup$ Similarly you can exclude the possibilities $n_3=4$, $n_2=3$ in the 168 case. The group conjugates the Sylows transitively, so if $n_3=4$ we have a non-trivial homomorphism from $G$ to $S_4$, which is a no-no. $\endgroup$ – Jyrki Lahtonen Mar 7 '13 at 13:16
  • 3
    $\begingroup$ If you want to see this problem worked out, check Dummit and Foote's Abstract Algebra, 3rd Ed. pages 207-212. $\endgroup$ – Josh B. Mar 7 '13 at 14:20

To get you started some rather basic (or even trivial, depends) observations, putting $\,n_p=$ number of Sylow $p-$subgroups , $\,G:=\,$ simple group of order $\,168\,$:

$$168=2^3\cdot 3\cdot7\Longrightarrow n_7=8\;,\;n_3=7,28$$

It can't be $\,n_3=4\,$ as then $\,G\,$ has a subgroup of index 4, which is impossible since this would mean $\,G\,$ is isomorphic with a subgroup of $\,S_4\,$ . For the same reason, it must be $\,n_2=7,\,21\,$...

Perhaps reading here you'll have have the whole view. Don't worry about the number of pages as the first ones are basic results listed.

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