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 $n \geq 6$, 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 $n$). %%note:Recall that $A_n$ acts naturally on the set of "letters" $\{1,2,\dots, n \}$ by permutation.%% (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 $\sigma \not = e$, such that $\sigma(n) = n$.

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

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

$\sigma$ must move something other than $i$ (that is, there must be $j \not = i$ such that $\sigma(j) \not = j$), 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 $\sigma(j) \not = j$ and $j \not = i$; also $j \not = n$ because if it were, [todo: this]

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

Then $\sigma'(n) = i$; and $\sigma' \not = \sigma$, because $\sigma'(j) = \sigma(y)$ which is not equal to $\sigma(j)$ (since $y \not = j$). Hence $\sigma'$ 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 $(123)$ 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 Klein four-group]): $\{ e, (12)(34), (13)(24), (14)(23) \}$. Therefore $A_4$ is not simple.