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 \(n\)). noteRecall 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, 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): \(\{ e, (12)(34), (13)(24), (14)(23) \}\). Therefore \(A_4\) is not simple.