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?

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    $\begingroup$ Welcome to Stackexchange. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. $\endgroup$ – Shaun Mar 20 '19 at 13:27
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    $\begingroup$ Note that your arguments would work with any pair of non-conjugate involutions, and the conclusion so far is that they generate a subgroup of order $2^n$ for some $n$. So they are contained in the same Sylow $2$-subgroup. What does this tell you? $\endgroup$ – Tobias Kildetoft Mar 20 '19 at 14:24
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    $\begingroup$ I think the name of the author of the book in question is Isaacs not Issac, which is an acronym for International Symposium on Symbolic and Algebraic Computation. $\endgroup$ – Derek Holt Mar 20 '19 at 22:54

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.

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