# The alternating groups on more than four letters are simple

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.

Parents:

• Alternating group

The alternating group is the only normal subgroup of the symmetric group (on five or more generators).

• A question about the requisites for this page: should the alternating group on five elements is simple be a requisite? It’s necessary for the base case of the induction, but one can probably understand the proof without it, simply referring to it as a known fact.