= Solution
Let $G$ act by left multiplication on the coset set $G/H$. Composing the resulting homomorphism
$$
G\longrightarrow S_{[G:H]}
$$
with the sign homomorphism gives a homomorphism $G\to\{\pm1\}$. If it were nontrivial, its kernel would have index two, contrary to the hypothesis. Thus every element of $G$, and in particular $x$, induces an even permutation.
Since $P$ is a Sylow $2$-subgroup, $[G:P]$ is odd. Therefore
$$
[G:H]=[G:P][P:H]=2[G:P]\equiv2\pmod4.
$$
An involution acts as disjoint transpositions and fixed points. If $x$ fixed no coset, it would be a product of $[G:H]/2$ transpositions, an odd number, contradicting evenness. Hence
$$
\boxed{x\text{ fixes at least one coset of }H}.
$$
This is the <even-involution coset fixed-point lemma>.
Back to article page
