Sylow Theorems

Exceptional Embedding of S5 in S6

Bishal Deb

Jul 8, 2015 • 5 min read

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 $\{1\}$ .