Exceptional Embedding of S5 in S6


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In this article we shall show a peculiar property of S_5.

We first note that there are several subgroups of S_n isomorphic to S_{n-1}. In fact there is one for each 1\leq i\leq n
H_i=\{ \sigma \in S_n |\sigma (i)    = i \}

Let X=\{1,2,3,4,5,6\}.

All of the above subgroups have two orbits in X. But for S_6 we show that there is a seventh embedding of S_5 which acts transitively on X.

We now show that there is a transitive action of S_5 on X.

We look at the 5-Sylow subgroups of S_5. They will be of order 5 each as S_5 is of order 120 and 5 is the highest power of 5 dividing it. Hence, the only possibility for the 5-Sylow subgroups is to contain 5 cycles. Now, the number of r cycles of S_n is \binom{n}{r}(r-1)!. Hence there are 24 5-cycles. As each 5-Sylow subgroup consists of 4 cycles and as any two 5-Sylow subgroups has only identity in their intersection so number of 5-Sylow subgroups is 24/4=6. Let the subgroups be P_1, \ldots,P_6. The number of subgroups is compatible with the result of Sylow’s theorem which says that the number of 5-Sylow subgroups should be of the form 5k+1 and should divide 120.

Let Syl_5(S_5):=\{P_1,\ldots, P_6\}. We define an action of S_5 on this set by conjugation. By Sylow’s theorem this is an action. Infact it is a transitive action. As Syl_5(S_5) and X are bijective, we have established an action of S_5 on X.

Let \phi : S_5\rightarrow Perm(X) be the action homomorphism. If we show that \phi is injective then we see that there is an embedding of S_5 in S_6 which acts transitively on X.

Now we show that \phi is injective.

Showing \phi is injective is equivalent to showing that Ker(\phi) is \{1\}. Now Ker(\phi) is a normal subgroup of S_5. Hence Ker(\phi) is either \{1\}, A_5 or S_5. Here we use the fact that A_n is simple for all n\geq 5. We know that S_5/Stab(P_1) is isomorphic to orb(P_1). As Ker(\phi)\subset Stab(P_1) so if Ker(\phi) is either A_5 or S_5 then |orb(P_1)|\leq 2 but |orb(P_1)|=6 as the action is transitive. Hence Ker(phi) is \{1\}.

It can be further showed that only n for which S_n has a subgroup isomorphic to S_{n-1} which acts transitively on \{1,\ldots,n\} is 6.

The above article is based on a lecture given at Chennai Mathematical Institute by Prof P. Vanchinathan.

I am a masters student at Chennai Mathematical Institute. I am also an editor of the English section.

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