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.