## 08 Jul Exceptional Embedding of S5 in S6

We first note that there are several subgroups of \$S_n\$ isomorphic to \$S_{n-1}\$. Infact there is one for each \$1leq ileq n\$
\$H_i={sigmain 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_5rightarrow 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 \$ngeq 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.

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