Problem 2B.4 in Finite Group Theory by Isaacs

Question

Suppose a finite group $G$ has more than one Sylow-2 subgroup and any two intersects trivially. Show $G$ contains exactly one conjugacy class of involutions.

Here are some of my thoughts: Suppose not, then there exist involutions $s,t$ such that they are not conjugate.

$s,t$ generates a dihedral group $D_{2n}$. $n$ is even since $s,t$ are not conjugate in $D_{2n}$.

If $n$ is not a power of 2, then $D_{2n}$ has more than one Sylow 2 subgroups and $D_{2n}$ are generated by them. But any two of the Sylow 2 subgroups of $D_{2n}$ has a nontrival intersection, and any two Sylow-2 subgroups of G intersects trivally, so all Sylow 2 subgroups of $D_{2n}$ must lie in the same Sylow-2 subgroup of $G$. So $D_{2n}$ lie in a Sylow-2 subgroup of $G$, this leads a contradiction.

But what if $n$ is a power of 2?

Answer 1

Thanks to a comment, I have solved it as follows.

My argument shows that any two non-conjugate elements generates a dihedral group whose order is a power of 2. Thus $s,t$ and $s,gtg^{—1}$ lies in a same Sylow-2 subgroup of G, for all $g\in G$. Thus it has only one Sylow-2 subgroup, contradiction.

Answer 2

Let's first consider the case where $G$ is dihedral with cyclic group $C$. Since $C\triangleleft G$, given $P,\,Q\in\text{Syl}_2(G)$ we have $C\cap P,\; C\cap Q\in\text{Syl}_2(C)$ (in any group the intersection of Sylow with normal is Sylow). But $C$ is abelian and therefore $C\cap P=C\cap Q$, which is trivial (if and) only if $|C|$ is odd.

In the general case take two different involutions $s$ and $t$. Let $P$ and $Q$ be the only Sylow $2$-subgroups of $G$ such that $s\in P$ and $t\in Q$. Since we can replace $s$ with $s^a$ for appropriate $a\in G$, we are allowed to assume that $P\ne Q$. Hence, $P\cap Q=\langle1\rangle$.

Consider the dihedral group $D=\langle s,t\rangle$ of order $2n$. The uniqueness of $P$ and $Q$ implies that $D\cap P$ and $D\cap Q$ are two different Sylow $2$-groups of $D$. Moreover, they have trivial intersection. As we saw above, this can only happen if $n$ is odd. Then, by Problem 2B.2, $s$ and $t$ are $D$- and hence $G$-conjugates.