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:   Regional Mathematical Olympiad – 2013

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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Bishal Deb

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

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