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 !

  • 2
    $\begingroup$ 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 $A_n$? $\endgroup$ Jan 21, 2020 at 2:54
  • $\begingroup$ @ArturoMagidin Hmm. I guess I'm not sure. I'll have to think about that some more. Thanks for your comment. $\endgroup$ Jan 21, 2020 at 3:27
  • 3
    $\begingroup$ 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.... $\endgroup$ Jan 21, 2020 at 3:36

1 Answer 1


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 $H\le A_n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.