Deeper question about Cayley's theorem Cayley's theorem states that there exists an injective homomorphism $\phi:G \rightarrow S_n$ for any finite group with $|G|=n$.
Let $\pi: \phi(G) \rightarrow \mathbb{Z}_2$ be the sign map, sending even permutations to $0$ and odd permutations to $1$. If I compose the functions to get $\pi \circ \phi$, then the kernel of the homomorphism is still a normal subgroup of $G$. So, $\ker(\pi \circ \phi) \trianglelefteq G.$
Does this normal subgroup have any significance in group theory? I'm not sure if there would be a connection to a normal subgroup of elements of even order (I can think of a counterexample to this) or if there's even any remote connection with elements of even order for this group.
I also have speculated this means that simple groups must be isomorphic to a generating set of only even permutations. The idea of the proof would be to assume that the kernel of the composition of functions is equal to the identity. Thus, the nonidentity elements of $G$ must map to odd permutations. However, multiplying two odd permutations gives an even permutation. So, there must be a nontrivial element mapping to an even permutation.
If there's anything I should consider or if there are any mistakes in my argument, I'd love to hear it.

Let $G$ be of order $n=2m$, $m$ odd, and $g\in G$ of order 2. then the left multiplication by $g$ is a product of $m$ transposition, so its signature is $(-1)^m=-1$.
In general if $g$ is of order $m$, and $k= {n\over m}$ the left multiplication is a product of $k$ cycles of length $m$, so its signature is $(-1)^k$ if $m$ is even, $1$ if $m$ is odd.
