Let $n>4$ be a natural number. Then the alternating group $A_n$ on $n$ elements is simple.

Proof

We go by induction on $n$. The base case of $n=5$ we treat separately.

For the inductive step: let $n\ge 6$, and suppose $H$ is a nontrivial normal subgroup of $A_n$. Ultimately we aim to show that $H$ contains the $3$-cycle $\left(123\right)$; since $H$ is a union of conjugacy classes, and since the $3$-cycles form a conjugacy class in $A_n$, this means every $3$-cycle is in $H$ and hence $H=A_n$.

Lemma: $H$ contains a member of $A_{n-1}$

To start, we will show that at least $H$ contains something from $A_{n-1}$ (which we view as all the elements of $A_n$ which don't change the letter $n$). ^[1] (This approach has to be useful for an induction proof, because we need some way of introducing the simplicity of $A_{n-1}$.) That is, we will show that there is some $\sigma \in H$ with $\sigma \ne e$, such that $\sigma \left(n\right)=n$.

Let $\sigma \in H$, where $\sigma$ is not the identity. $\sigma$ certainly sends $n$ somewhere; say $\sigma \left(n\right)=i$, where $i\ne n$ (since if it is, we're done immediately).

Then if we can find some ${\sigma }^{\prime }\in H$, not equal to $\sigma$, such that ${\sigma }^{\prime }\left(n\right)=i$, we are done: ${\sigma }^{-1}{\sigma }^{\prime }\left(n\right)=n$.

$\sigma$ must move something other than $i$ (that is, there must be $j\ne i$ such that $\sigma \left(j\right)\ne j$), because if it did not, it would be the transposition $\left(ni\right)$, which is not in $A_n$ because it is an odd number of transpositions. Hence $\sigma \left(j\right)\ne j$ and $j\ne i$; also $j\ne n$ because if it were,

Now, since $n\ge 6$, we can pick $x,y$ distinct from $n,i,j,\sigma \left(j\right)$. Then set ${\sigma }^{\prime }=\left(jxy\right)\sigma (jxy{)}^{-1}$, which must lie in $H$ because $H$ is closed under conjugation.

Then ${\sigma }^{\prime }\left(n\right)=i$; and ${\sigma }^{\prime }\ne \sigma$, because ${\sigma }^{\prime }\left(j\right)=\sigma \left(y\right)$ which is not equal to $\sigma \left(j\right)$ (since $y\ne j$). Hence ${\sigma }^{\prime }$ and $\sigma$ have different effects on $j$ so they are not equal.

Lemma: $H$ contains all of $A_{n-1}$

Now that we have shown $H$ contains some member of $A_{n-1}$. But $H\cap A_{n-1}$ is normal in $A_n$, because $H$ is normal in $A_n$ and it is _intersect _subgroup _is _normal the intersection of a subgroup with a normal subgroup. Therefore by the inductive hypothesis, $H\cap A_{n-1}$ is either the trivial group or is $A_{n-1}$ itself.

But $H\cap A_{n-1}$ is certainly not trivial, because our previous lemma gave us a non-identity element in it; so $H$ must actually contain $A_{n-1}$.

Conclusion

Finally, $H$ contains $A_{n-1}$ so it contains $\left(123\right)$ in particular; so we are done by the initial discussion.

Behaviour for $n\le 4$

• $A_1$ is the trivial group so is vacuously not simple.
• $A_2$ is also the trivial group.
• $A_3$ is isomorphic to the cyclic group $C_3$ on three generators, so it is simple: it has no nontrivial proper subgroups, let alone normal ones.
• $A_4$ has the following normal subgroup (the Klein four-group): $\{e,(12\left)\right(34),(13\left)\right(24),(14\left)\right(23\left)\right\}$. Therefore $A_4$ is not simple.

1. ^{^︎}

   Recall that $A_n$ acts naturally on the set of "letters" $1,2,\dots ,n$ by permutation.