## The Outer Automorphism

We construct an outer automorphism of .

First we prove a crucial lemma: the only proper normal subgroups of is . Since we have that is normal in . Suppose that is normal in . Then is normal in . Since is simple, either or . If then . But the largest divisor of is sixty and by Lagrange’s theorem, . Thus , a contradiction. Thus . Then . It follows that and is a transposition. But all the the transpositions of are conjugate and thus is not normal. It follows that the only normal subgroup of is . This claim holds, *mutatis mutandis*, for all .

Consider the set of 5-Sylow subgroups of . Let denote the number of subgroups of . Then . We have that since the only proper normal subgroup of is . We have that by testing divisibility. Thus . We write the Sylow 5-subgroups of :

We then let act on by conjugation. This defines a homomorphism . Consider , it must be a normal subgroup of . Thus is either a non-proper subgroup of or it is equal to . We consider . Calculation shows that since . Thus and thus . It follows that . Let . We claim that: is acts transitively and contains no transpositions. Since all the Sylow subgroups are conjugate, it follows that is transitive. Since five divides we have that it divides , it must contain a five cycle, without loss of generality we may take the 5-cycle to be . Suppose that contains a transposition . Then by transitivity there is an such that and we have that for . But then generates , which is strictly larger that . Thus contains no transposition.

We then let by left multiplication. Note that . This induces a homomorphism . We have that . But is a normal subgroup of and thus . By considering magnitudes it follows that . Thus is an automorphism. We claim that is outer. If were inner then it would preserve cycle structure and would be a transposition, but then would fix at least one point. Suppose . Then . It follows that , contradicting the fact that contains no transpositions. Thus is a non-outer automorphism of .

Alex Fsaid, on 2010/05/09 at 22:32In this connection I can hardly not mention Rick’s Tricky Six puzzle.