# Every finite simple group of order $n \geq 3$ is isomorphic to a subgroup of $A_n$

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 !

• Hint: if $G$ is of order $n$, then $G$ embeds into $S_n$, as you note. What can you say about $G\cap A_n$, if $G$ is not contained in $A_n$? – Arturo Magidin Jan 21 at 2:54
• @ArturoMagidin Hmm. I guess I'm not sure. I'll have to think about that some more. Thanks for your comment. – michiganbiker898 Jan 21 at 3:27
• Hint${}^2$: If $G$ is not contained in $A_n$, then $GA_n=S_n$, and so $GA_n/A_n =S_n/A_n\cong C_2$. But by the Isomorphism Theorems, you can write $GA_n/A_n$ a different way.... – Arturo Magidin Jan 21 at 3:36

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 $$H\le A_n$$.