Exceptional Embedding of S5 in S6

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 $S_{n-1}$. 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 ${1}$.

READ:   জহান্স কেপলাৰ

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.

Print Friendly, PDF & Email
The following two tabs change content below.

Bishal Deb

Bishal Deb is an undergraduate student at Chennai Mathematical Institute. He is an editor of Gonit Sora.

Latest posts by Bishal Deb (see all)

No Comments

Post A Comment