Let 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 $n \geq 6$ , and suppose $H$ is a nontrivial normal subgroup of $A_{n}$ . Ultimately we aim to show that $H$ contains the $3$ -cycle $(123)$ ; 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 $σ \in H$ with $σ \neq e$ , such that $σ (n) = n$ .

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

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

$σ$ must move something other than $i$ (that is, there must be $j \neq i$ such that $σ (j) \neq j$ ), because if it did not, it would be the transposition $(n i)$ , which is not in $A_{n}$ because it is an odd number of transpositions. Hence $σ (j) \neq j$ and $j \neq i$ ; also $j \neq n$ because if it were,

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

Then $σ^{'} (n) = i$ ; and $σ^{'} \neq σ$ , because $σ^{'} (j) = σ (y)$ which is not equal to $σ (j)$ (since $y \neq j$ ). Hence $σ^{'}$ and $σ$ 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 $(123)$ in particular; so we are done by the initial discussion.

Behaviour for $n \leq 4$

$A_{1}$ 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): ${e, (12) (34), (13) (24), (14) (23)}$ . Therefore $A_{4}$ is not simple.

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

The alternating groups on more than four letters are simple