Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 3 / 5D (Groups)

For $\sigma \in S_n$, let $P_{\sigma }$ be its permutation matrix and define the sign of a permutation bySince $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$,so this is a homomorphism. A transposition interchanges two columns of the identity matrix and therefore has determinant $-1$. If $\sigma$ is expressed as a product of $r$ transpositions, then $\mathrm{sgn}(\sigma )=(-1{)}^r$; because the determinant depends only on $\sigma$, the parity of $r$ is independent of the chosen expression.

The conjugacy classes in $A_5$ consist of the identity, the $15$ double transpositions, the $20$ three-cycles, and two classes of $12$ five-cycles. Thus their sizes areA normal subgroup is a union of conjugacy classes containing the identity, and its order must divide $60$ by Lagrange's theorem. Checking sums of a subset of $15,20,12,12$ shows that no numberdivides $60$. Hence the only possible normal-subgroup orders are $1$ and $60$. This proves the Simplicity of the alternating group A5.

If a homomorphism $A_5\to \{\pm 1\}$ were surjective, its kernel would be a normal subgroup of index two and hence order $30$. This contradicts the simplicity of $A_5$. Therefore no such surjective homomorphism exists.

Let $\varphi :A_5\times A_5\to \{\pm 1\}$ be a homomorphism. Its restrictions to $A_5\times \{1\}$ and $\{1\}\times A_5$ cannot be nontrivial, because any nontrivial subgroup of $\{\pm 1\}$ is the whole group and part (iii) rules out such a surjection. Both restrictions are therefore trivial. Since$\varphi$ itself is trivial and in particular is not surjective.