# 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.