Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 3 / 1D (Groups)

The second component must have order $4$ and the first is arbitrary, giving $2\cdot 2=4$ elements.

Exactly the six $4$-cycles have order $4$: $(4-1)!=6$.

The cycle types are $(4)(1)(1)$ and $(4)(2)$. Each contributes $\left(\genfrac{}{}{0ex}{}{6}{4}\right)(4-1)!=90$, for a total of $180$.

Only type $(4)(2)$ is even, so the answer is $90$.