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).