Structure of simple group of order 168 just via Sylow theory (fix for Math Doctor Bob)

There are a lot of questions on here about the simple group of order 168, e.g., this, this, this, this, this, this, and this. There is also plenty of info on the internet about this group, e.g., this. I am about to ask something very specific that I think is different from these, but please let me know if I've missed something.

A student of mine recently showed me this pair of videos (part 1, part 2) by Robert Donley (aka Math Doctor Bob), which attempts to derive the number and structure of the Sylow subgroups of the simple group of 168, along with its class equation, using only Sylow theory, the classification of groups of order 8, and elementary counting arguments.

I buy the arguments in the first video, but the second video seems to me to contain a significant gap in reasoning. My question is about how to patch the reasoning in the video using only the types of tools used in the video. (The closest thing I could find was these notes, which use the same types of tools and obtain the same results, but what I want is specifically to rescue Donley's argument itself.) Here are the details:

Up til about 2m,15s into the second video, the argument has established that:

• There are 8 Sylow 7-subgroups, for a total of 48 elements of order 7.
• There are 28 Sylow 3-subgroups, for a total of 56 elements of order 3, and they are all conjugate.
• There are no elements of orders 6 or 14.

At this point, Donley turns attention to the Sylow 2-subgroups, arguing first that they are nonabelian, and then using this to conclude that there are 21 of them and they are self-normalizing, and proceeding from there to use counting arguments to deduce that they are isomorphic to $D_4$. His argument that they are nonabelian appears to me to have a big gap. Donley says, consider one Sylow 2-subgroup; call it $H_8$. Look at its order 2 elements. If $H_8=C_2^3$, there are 7 of them; let a Sylow 3-subgroup act by conjugation (this claim is what I have a problem with); since $3\nmid 7$, there is a singleton orbit, so an order 2 element is centralized by an order 3 element, and there is an order 6 element, contradiction. If $H_8=C_4\times C_2$, he lets a Sylow 7-subgroup act and concludes in the same way (since $7\nmid 3$) that there is a singleton orbit and thus an element of order 14. If $H_8=C_8$, he lets either one act because there is only one element of order 2.

My objection is that you can't let a Sylow 3-subgroup act on the order 2 elements in a specific Sylow 2-subgroup unless you have already established, or are at least explicitly provisionally assuming, that the Sylow 2 normalizer contains a Sylow 3. (And similarly for a Sylow 7 in place of a Sylow 3.) In fact, Donley concludes shortly thereafter that the Sylow 2's are self-normalizing, by reasoning that the nonabelian groups of order 8 each have a center of order 2, thus a factor of 3 or 7 dividing the order of the Sylow 2 normalizer would imply an order 2 element centralized by an order 3 or 7 element, and thus an element of order 6 or 14, so the Sylow 2 normalizer's order must only be divisible by 2. So he wants to conclude this is actually false; it certainly shouldn't be assumed implicitly.

What I would like to ask your help with is reorganizing just this specific part of the argument, beginning from bulleted information above, to obtain the conclusion that the Sylow 2 is nonabelian. I see how to do it using more powerful tools (specifically, Burnside's transfer theorem), but I would like to see how to do it using only Sylow theory, the classification of groups of order 8, and counting. If it requires to access information about the structure of $S_4$, that's fine too. Really I'm thinking in terms of what my student knows.

To begin with, I think it's better tactically to consider the number of Sylow 2's before asking about their structure, since this controls whether they are normalized by Sylow 3's etc. Therefore, I think the argument should begin:

The number of Sylow 2's is, by Sylow theory, either 1, 3, 7 or 21. It can't be 1 because the group is simple, and similarly 3 can be ruled out because this would imply a nontrivial (and therefore injective, by simplicity) homomorphism to $S_3$, which is impossible because $168>6$. So there are 7 or 21 Sylow 2's.

Suppose there are 7. Then the Sylow 2 normalizer is order 24, and it contains a Sylow 3, which therefore acts on the Sylow 2. Now Donley's exact reasoning can be used to rule out the cases that $H_8$ is isomorphic to $C_2^3$, $C_8$, $D_4$, or $Q_8$, and a minor adjustment can be used to rule out $C_4\times C_2$. Specifically, $C_2^3$ has 7 elements of order 2, so (since $3\nmid 7$) the action on these has a fixed point, and Donley's exact reasoning then yields an element of order 6, a contradiction. Meanwhile, all four of $C_4\times C_2$, $C_8$, $D_4$, and $Q_8$ have a characteristic subgroup of order 2, so this too is a fixed point, leading to the same contradiction. (For $C_4\times C_2$, it's generated by the unique order 2 element that is a square. For the other three, it was identified by Donley, see above: the unique subgroup of order 2 in $C_8$, and the centers of $D_4$ and $Q_8$.) These contradictions now rule out the possibility that there are 7 Sylow 2's; there must be 21, and they must be self-normalizing.

The real question I have is:

From here, how do we conclude that the Sylow 2's are not abelian?

Donley's line of argumentation is a nonstarter because we know that the Sylow 3's and Sylow 7's do not act on a Sylow 2.

The high-tech answer is the Burnside transfer theorem. If $H_8$ is abelian and self-normalizing, then certainly it is central in its normalizer, and the Burnside transfer theorem then gives us a normal 2-complement, which is impossible since the group is simple.

But how would you do it using only Sylow theory, the classification of groups of order 8, counting, and, if you need it, the structure of $S_4$? (And the bulleted facts above?)

Solution 1:

A common technique is to look at intersections of Sylow subgroups. If you allow that in your bag of tools, then you can do the following.

1. Assume that all the $21$ Sylow $2$-subgroups are abelian.
2. Those $21$ groups of order $8$ can only occupy the available space of $168-48-56=64$ elements outside the union of Sylow $3/7$-subgroups (save for $1_G$), so we can locate at least two, say $P_1$ and $P_2$, that intersect non-trivially.
3. As a $2$-group the intersection $P_1\cap P_2$ must contain an element $x$ of order two. Let's fix one.
4. Let's look at the centralizer $H=C_G(x)\le G$. The group $H$ has at least two Sylow $2$-subgroups, namely $P_1$ and $P_2$ (we assumed they are both abelian, so they are both contained in $H$!). Therefore $|H|>8$.
5. Therefore $H$ has an element $y$ of order $3$ or of order $7$.
6. As $y\in C_G(x)$, the order of $xy$ is either $6$ or $14$ contradicting the result of an earlier step.