The Problem:
Let $n$ be a positive integer, let $G$ be a finite group with an odd number of elements, and let $\phi: S_n \to G$ be a homomorphism. Prove that $\phi$ must be trivial, i.e., $\phi(\sigma) = e$ for all $\sigma \in S_n$.
My Approach:
I think I can go about it like this, but I keep getting stuck: We claim that, since $|G|$ is odd and $|S_n| = n!$ is even, it follows that $(|G|, |S_n|) = 1$; and so any homomorphism between them must be trivial (i.e., $|\phi(S_n)| = 1$). So, we'll prove this more general statement, which follows from the First Isomorphism Theorem (and Lagrange's Theorem). Indeed, we have that
$$ S_n/\ker\phi \cong \phi(S_n) \implies |S_n|/|\ker\phi| = |\phi(S_n)|. $$
Now, since $\phi(S_n) < G$, $|\phi(S_n)|$ must divide $|G|$; and so $|\phi(G)|$ cannot be even. But, since $|\phi(S_n)| = |S_n / \ker\phi|$, this means that $|S_n / \ker\phi|$ cannot be even. Note that, since $\ker \phi < S_n$, $|\ker\phi| = k$ must divide $|S_n| = n(n-1)\cdots2\cdot1$; and so $k \in \{n, n-1, …, 2, 1\}$. Moreover, since $|S_n / \ker\phi|$ must be odd, $k$ must be even… and here's where I'm stuck…
Best Answer
The order of the image of a transposition is $1$ or $2$, it cannot be $2$ since Lagrange implies $2$ divides the order of $G$ so the image of every transposition is trivial. Since $S_n$ is generated by transpositions, the morphism is trivial.