**Problem**. Let be a subgroup of Show that if is even and then is not simple.

**Solution**. First notice that we are done if is abelian because then every subgroup of would be normal and since is even, has a subgroup of order two, which is clearly neither the trivial subgroup nor

So suppose, to the contrary, that is non-abelian and simple and consider the group homomorphism defined by for all Then since is a normal subgroup of and is not simple, either where is the identity matrix in or But we can’t have because then would be embedded into making abelian. Hence i.e. for all Now let be an element of order two. So which implies that is diagonalizable (because its minimal polynomial divides and so it splits into distinct linear factors). So

for some Since the set of eigenvalues of which is is a subset of Since we must have and so Since has order two, and hence Since is a central element of the subgroup is a normal subgroup of order two in and because So is not simple and that contradicts our assumption that is simple. So our assumption is wrong and is not simple indeed!