# The alternating groups on more than four letters are simple

https://arbital.com/p/alternating_group_is_simple

by Patrick Stevens Jun 17 2016 updated Jun 17 2016

The alternating groups are the most accessible examples of simple groups, and this fact also tells us that the symmetric groups are "complicated" in some sense.

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.