Sylow-$2$ Subgroups of a simple group of order 168

Question

I'm stuck somewhere in the following claim, I would appreciate if you could help:

Claim: Let $G$ be a simple group of order $168(=2^3\cdot 3\cdot 7).$ Then all Sylow $2$-subgroups of $G$ are dihedral.

We can easily see that the number of Sylow-$7$ subgroups of $G$ must be $8,$ and each of them are cyclic. With a little effort we can also see that the number of Sylow-$3$ subgroups of such a group must be $28,$ and again each of them are cyclic. Also, counting the elements one can easily see that the number of Sylow $2$-subgroups of $G$ is $21.$ However, I could not see how to get the desired result.

Thanks in advance.

Answer 1

If you look at the number of Sylow $2$-subgroups of a simple group $G$ of order $168$, you know that it divides $21$. The group $G$ acts transitively on the set $\mathrm{Syl}_2(G)$ of Sylow $2$-subgroups by conjugation, so you get $|\mathrm{Syl}_2(G)|$ equals either $7$ or $21$, as $G$ simple cannot act transitively on a set of order $3$.

Now restrict this action to a Sylow $2$-subgroup $S$ of $G$. Its fixed points are the Sylow $2$-subgroups normalized by $S$. If $S$ normalizes $T$ then $S\cdot T$ is a $2$-subgroup containing $S$, implying $S=T$ as $S$ is a maximal $2$-subgroup of $G$. The other orbits of $S$ have order $2$, $4$ or $8$, but as neither $7$ nor $21$ equals $1\bmod 8$, there is an orbit of length $2$ or $4$. So there is an element $s\in S\setminus\{1\}$ and a Sylow $2$ subgroup $T\ne S$ such that $s$ normalizes $T$. As $\langle s, T\rangle$ is a $2$-subgroup and $T$ a maximal $2$-subgroup, we get $s\in S\cap T$.

We may assume that $s$ has order $2$ (otherwise take $s^2$), which implies that $s$ is central in both $S$ and $T$, UNLESS the Sylow $2$-subgroups are dihedral (which we now assume not to be). The centralizer $C:=C_G(s)$ of $s$ in $G$ contains both $S$ and $T$, so it has at least two Sylow $2$-subgroups. As $G$ is simple, the order of $C$ is $24$, and $C$ has three Sylow $2$-subgroups. The kernel $K$ of the transitive action of $C$ on the sets of its Sylow $2$-subgroups has order $4$, contains $s$, and $C/K = S_3$ the symmetric groups on $3$ elements. A Sylow $3$-subgroup $D$ of $C$ lies in the preimage of $A_3$ in $C$, centralizes $s$, and hence is normal in $C$.

$C$ is self-normalizing of index $7$, so look at the intersection of $C$ with one of its conjugates $C'\ne C$. If $C\cap C'$ contains an element of order $3$, hence $D$, then both $C$ and $C'$ normalize $D$ contradicting $G$ simple. If $C\cap C'$ contains an element of order $2$, than its centralizer contains all Sylow $2$-subgroups of $C$ and $C'$ (as they are assumed not to be dihedral), contradicting $G$ simple.

Hence $C$ intersects all its conjugates trivially, and the order of the union of all conjugates of $C\setminus\{1\}$ is $7\cdot 23 = 161$, showing that $G$ has a normal Sylow $7$-subgroup, contradicting the simplicity of $G$.

Answer 2

A $2$-Sylow of $G$ cannot be abelian by Burnside's transfer theorem, as noted by Derek Holt in the comments.

Say a $2$-Sylow of $G$ is $P = Q_8$.

The number of $7$-Sylows is $8$, which means that the normalizer of a $7$-Sylow has order $21$.

In particular, no element of order $2$ normalizes a $7$-Sylow, so $P$ acts regularly on the set of $7$-Sylows.

So $Z(P) = \langle z \rangle$ acts as $(12)(34)(56)(78)$ on the $7$-Sylows. The centralizer of $z$ in $S_8$ has order $2^7 \cdot 3$.

There is no element of order $3$ in $G$ that centralizes $z$, because otherwise a $3$-Sylow normalizer would be cyclic of order $6$, again a contradiction by Burnside's theorem.

Hence $C_G(z) = P$.

Any nontrivial subgroup of $P$ contains $z$, so if $P \cap P^x \neq 1$, then $z^x = z$ and thus $x \in C_G(z) = P$.

Therefore distinct conjugates of $P$ have trivial intersection. Counting elements gives you $21 \cdot (2^3 - 1) = 147$ elements of $2$-power order and $8 \cdot 6 = 48$ elements of order $7$, which is too many.

Answer 3

Here is another reasoning which uses the classification of groups of order $8$ and does not use Burnside's theorem.

1. We already know that in a simple group $G$ of order $168$ there are $8$ Sylow $7$-subgroups, $28$ Sylow $3$-subgroups, and $21$ Sylow $2$-subgroups. So there are exactly $48$ elements of order $7$, $56$ elements of order $3$, and no more than $64$ $2$-elements (i.e. elements whose order is a power of 2).

2. A simple group $G$ of order $168$ has no elements of order $6$. This has been discussed here and here.

3. Let a Sylow $2$-subgroup of $G$ be isomorphic to $Q_8$. If $P_1$ and $P_2$ are Sylow $2$-subgroups and $P_1\cap P_2\neq\{e\}$, then $P_1\cap P_2$ contains an element $a$ of order 2. Since $Q_8$ has a single element of order $2$ and it lies in its center, it follows that $P_1,P_2\subset C_G(a)$, that is, $|C_G(a)|>8$. Then $|C_G(a)|=56$ or $|C_G(a)|=24$. the first equality is impossible because $|G:C_G(a)|=3$. If $|C_G(a)|=24$, then $G$ contains an element of order 6, which contradicts point 2.

Hence for any two Sylow $2$-subgroups $P_1$ and $P_2$ we have $P_1\cap P_2=\{e\}$. But then the number of $2$-elements of $G$ must be $7\cdot21=147>64$. Contradiction.

4. It is proved in the same way that the Sylow $2$-subgroup cannot be abelian.

5. From the classification of groups of order $8$ it now follows that the Sylow $2$-subgroup of $G$ is dihedral.