Let $G$ be a finite simple group of order $n$. Prove that if $n \geq 3$, then $G$ is isomorphic to to a subgroup of $A_n$, the alternating subgroup of the symmetric group $S_n$.
My idea here was to use the First Isomorphism Theorem. If we could construct a homomorphism $\phi: G \longrightarrow H$, where $H$ is a subgroup of $A_n$ ($n \geq 3$), such that $\phi$ is surjective (so that $im(\phi)$ is exactly $A_n$) and $\ker(\phi) \neq G$ (so that, since $\ker(\phi)$ is normal in $G$ and $G$ is simple, $\ker(\phi)$ would have to be trivial), it would follow that $G/\ker(\phi) = G \cong im(\phi) = H$.
How can one construct such a homomorphism ? Can we use the fact that every finite group is isomorphic to a subgroup of $S_n$, and go from there ?
Thanks !
Best Answer
We have an injective homomorphism $f:G\to S_n$. Let $H$ be the image of $f$. By the first isomorphism theorem we have that
$$G\cong G/\ker(f)\cong\text{Im}(f)=H.$$
Let $g:S_n\to\mathbb{Z}_2$ be the sign homomorphism. $g$ sends $\sigma\mapsto0$ if $\sigma$ is an even permutation and $\sigma\mapsto1$ if $\sigma$ is an odd permutation. Let $K=\ker(g\circ f)$.
We have that $G$ is simple, and that $G\cong H\le S_n$. We want to show that $H\le A_n$. So let's suppose that $H\nleq A_n$.
Since $H\nleq A_n$, there is a $y\in H$ with $y\notin A_n$. Since $y\in H$, $y=f(x)$ for some $x\in G$.
Note that $x\notin K=\ker(g\circ f)$, since $g(f(x))=g(y)$ and $g(y)=1$ since $y\notin A_n$.
Since $G$ is simple, and $K$ is normal in $G$, and $K\ne G$, we must have that $K$ is trivial. Also, note that $g\circ f$ is surjective since $g\circ f$ sends $1_G\mapsto0$ and $x\mapsto1$.
Hence, by the first isomorphism theorem we have that
$$G\cong G/K=G/\ker(g\circ f)\cong\text{Im}(g\circ f)=\mathbb{Z}_2.$$
But the fact that $G\cong\mathbb{Z}_2$ implies that $n=|G|=2$. So if $n\ge3$, then there is no $y\in H$ with $y\notin A_n$. So we must have $H\le A_n$.