1
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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
4
$\begingroup$

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 $|H| = 8$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.