Parker Glynn-Adey

The Outer Automorphism

Posted in Math by pgadey on 2010/05/09

We construct an outer automorphism of S_6.

First we prove a crucial lemma: the only proper normal subgroups of S_5 is A_5. Since [A_5  : S_5 ] = 2 we have that A_5 is normal in S_5. Suppose that H is normal in S_5. Then A_5 \cap H is normal in S_5. Since A_5 is simple, either A_5 \cap H = \{e\} or =A_5. If A_5 \cap H = A_5 then 60 \leq |H|. But the largest divisor of |S_5| is sixty and by Lagrange’s theorem, |H| = 60. Thus H = A_5, a contradiction. Thus H \cap A_5 = \{e\}. Then |HA_5| = |H||A_5| / |H \cap A_5| = |H|60 \leq 120. It follows that |H| = 2 and H is a transposition. But all the the transpositions of S_5 are conjugate and thus H is not normal. It follows that the only normal subgroup of S_5 is A_5. This claim holds, mutatis mutandis, for all 5 \leq n.

Consider the set of 5-Sylow subgroups of S_5. Let n_5 denote the number of 5 subgroups of S_5. Then n_5 = 1, 6, 11, 16, 21. We have that n_5 \neq 1 since the only proper normal subgroup of S_5 is A_5. We have that n_5 \neq 11, 16, 21 by testing divisibility. Thus n_5 = 6. We write the Sylow 5-subgroups of S_5:
\begin{array}{ccccccc}& A & B & C & D & E & F\\Syl_5(S_5) =  & \left\{ \langle (12345) \rangle, \right.  &  \langle (12435) \rangle, &  \langle (12543) \rangle,  &  \langle (12354) \rangle,  &  \langle (12534) \rangle, \ & \left. \langle (12453) \rangle  \right\}\end{array}

We then let S_5 act on Sly_5(S_5) by conjugation. This defines a homomorphism \phi : S_5 \rightarrow S_6. Consider \ker \phi, it must be a normal subgroup of S_5. Thus \ker \phi is either a non-proper subgroup of S_5 or it is equal to A_5. We consider \phi((123)). Calculation shows that (123)(12345)(123) = (23145) \in B since (23145)^3 = (12435). Thus \phi((123))[A] = B and thus \ker \phi \neq A_5, S_5. It follows that \ker \phi = \{e\}. Let R = im\ \phi \leq S_6. We claim that: R is acts transitively and contains no transpositions. Since all the Sylow subgroups are conjugate, it follows that R is transitive. Since five divides |S_5| we have that it divides |R|, it must contain a five cycle, without loss of generality we may take the 5-cycle to be (1\ 2\ 3\ 4\ 5). Suppose that R contains a transposition (i\ j). Then by transitivity there is an r \in R such that r[j] = 6 and we have that r(i\ j)r^{-1} = (r(i)\ 6) for r(i) \neq 6. But then \{(1\ 2\ 3\ 4\ 5), (r(i)\ 6)\} generates S_6, which is strictly larger that R. Thus R contains no transposition.

We then let S_6 : R by left multiplication. Note that |S_6 / R| = |S_6| / |S_5| = 6. This induces a homomorphism \Phi : S_6 \rightarrow S_6. We have that \ker \Phi \leq R. But \ker \Phi is a normal subgroup of S_6 and thus \ker \Phi = \{e\}, A_6, S_6. By considering magnitudes it follows that \ker \Phi = \{e\}. Thus \Phi is an automorphism. We claim that \Phi is outer. If \Phi were inner then it would preserve cycle structure and \Phi((1\ 2)) would be a transposition, but then \Phi((1\ 2)) would fix at least one point. Suppose sR = (1\ 2)sR. Then R = s^{-1}(1\ 2)sR. It follows that s^{-1}(1\ 2)s \in R, contradicting the fact that R contains no transpositions. Thus \Phi is a non-outer automorphism of S_5.

One Response

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  1. Alex F said, on 2010/05/09 at 22:32

    In this connection I can hardly not mention Rick’s Tricky Six puzzle.

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