Let $G$ be a finite group, $S\in\text{Syl}_{2}(G)$, and $T=\{g_{1},\ldots,g_{n}\}$ be a transversal of $S$ in $G$. (This can be a left or right transversal, it shouldn't matter.) Thus, $G=Sg_1 \cup\ldots\cup Sg_{n}$, a disjoint union. I want to demonstrate that all the involutions in $G$ are contained in the set $S^{g_1}\cup\ldots\cup S^{g_n}$. That is, given any involution $x\in G$, there exists $g_i \in T$ for which $x\in S^{g_i}$. I'm fairly sure this is true, but I can't quite get there.
It is true that if $x$ is an involution in $G$ then $x$ belongs to some Sylow-2 subgroup of $G$, say, $S'$. By Sylow's Theorems, $S'\sim S$. Say $S'=S^h$ for $h\in G$. Then $x\in S^h$ but I want to show that this $h$ can be selected from the transversal, $T$.
We have $h = sg_i$ for some $g_i \in T$ and $s \in S$, and then $S^h = S^{g_i}$.