I'm trying to follow a sketch proof about the abstract characterization of $S_5$, by Walter Feit.

Suppose $G$ is a finite group with exactly two conjugacy classes of involutions, with $u_1$ and $u_2$ being representatives. Suppose $C_1=C(u_1)\simeq \langle u_1\rangle\times S_3$ and $C_2=C(u_2)$ be a dihedral group of order $8$. The eventual result is that $G\simeq S_5$. Also, $C(u)$ denotes the centralizer of $u$ in $G$.

I don't understand the observation some involution is in the center of a Sylow subgroup, and that $C_2$ is a Sylow $2$-subgroup. I do know that $C_2$ is contained in a Sylow $2$-subgroup at least, from the Sylow theorems, but without knowing the actual order of $G$, I don't see why it necessarily a Sylow subgroup itself.

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    $\begingroup$ A p-group has a non-trivial centre. The centre of a Sylow 2-subgroup is therefore non-trivial, hence has even order, and therefore contains an element of order 2. $\endgroup$ – Mark Bennet Jun 5 '12 at 9:25

Let $H$ be a Sylow 2-subgroup of $G$. Since $H$ is a $p$-group, $Z(H)$ is a non-trivial $p$-group, hence contains an involution, say $u$.

Since $u$ is central in $H$, we see that $H \leq C(u)$. Since we have just two conjugacy classes of involutions in $G$, we have either $u$ is conjugate to $u_1$, or $u_2$.

Suppose $u$ is conjugate to $u_1$, so $u_1 = gug^{-1}$ for some $g \in G$. If $x \in G$ commutes with $u$, then $gxg^{-1}$ commutes with $u_1$. This shows that $C(u_1) = gC(u)g^{-1}$, in particular, these groups have the same order.

Since $G$ contains a subgroup of order $8 = 2^3$, $H$ contains a subgroup of order $8$, whence $C(u)$ contains a subgroup of order $8$. But if $u$ is conjugate to $u_1$, then $\langle u_1 \rangle \times S_3$ contains a subgroup of order $8$, but $8$ does not divide $12$.

Therefore, $u$ must be conjugate to $u_2$, in which case we have that $H$ is completely contained in a subgroup of order $8$, thus $|H| = 8$.

  • $\begingroup$ Many thanks David! $\endgroup$ – Adelaide Dokras Jun 5 '12 at 20:37

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