Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 3 / 7D / Solution

Let $G$ act on its underlying set by left multiplication. The resulting homomorphismis injective because ${\lambda }_g(e)=g$. This is Cayley theorem. If $g\ne e$, then $gx=x$ would imply $g=e$ after right cancellation, so every nonidentity permutation in the image is fixed-point free.

If $|G|$ is even, pair every element with its inverse. Elements not equal to their inverses occur in pairs. Since the identity is self-inverse and the group has even size, there must be another self-inverse element $g\ne e$. It has order two.

A permutation is odd when its sign is $-1$. If a subgroup $H\le S_m$ contains an odd element, the restrictionis surjective. Its kernel is the set of even elements and has index two. Each coset has the same size, so precisely half of $H$ is odd.

Now let $n=4k+2$. An element of order two in the fixed-point-free regular representation is a product ofdisjoint transpositions, so it is odd. The image $H\cong G$ therefore has an index-two normal subgroup of even permutations. Its order is $2k+1>1$, so it is nontrivial and proper. Hence

Nonabelian simple groups of even order do exist; the smallest example is the alternating group $A_5$, of order sixty.

Solved by gpt-5.6-sol high.