Transitive group actions and transitive subgroups

abstract-algebragroup-actionsgroup-theorysylow-theorysymmetric-groups

I am given the following problem:

Show that $S_6$ has a subgroup $H$, which is isomorphic to $S_5$ and acts transitively on $\{1,2,3,4,5,6\}$.

Before this I proved that $S_5$ has six $5$-sylow groups. Working on this question I soon got pretty lost, so I looked up the solution. Trying to understand it, I came up with three questions.

Here is the part of the solution I struggle with:

$S_5$ acts transitively on the six $5$-Sylow groups. This gives a homomorphism $\sigma: S_5 \to S_6$. Therefore $H:= \sigma(S_5)$ is transitive. Since $H \cong S_5/K$, where $K$ denotes the kernel of $\sigma$, is transitive of order $6$ we have $|S_5/K| \geq 6$. (The solution than continues to show that $K$ is trivial.)

I am not familiar with the concept of transitive groups. I looked it up and found the following definition of transitive groups: "Let $M$ be a set and $G:=S_{|M|}$ the permutation group of $M$. A subgroup $H$ of $G$ is called transitive if for each $x,y \in M$ there is a $\sigma \in H$ such that $\sigma(x) = y$." So far, so good.

Regarding the solution above, I struggle to understand three points:

  1. Why is $H:= \sigma(S_5)$ transitive?
  2. Why is $H$ transitive of order $6$?
  3. Why does this mean that $|S_5/K| \color{red}{\geq} 6$?

Best Answer

  1. $H = \sigma(S_5)$ is a transitive subgroup of $S_6$ precisely because the action of $S_5$ on the $5$-sylow subgroups is transitive.

To say more, we know that $S_5$ acts on a set of size $6$, namely the $5$-sylow subgroups, by conjugation. Moreover, since we know any two $5$-sylow subgroups are conjugate, this action carries any subgroup to any other subgroup (that is, the action is transitive). But what is an action on a set of size $6$? It's exactly a homomorphism $\sigma : S_5 \to S_6$. If we look at $H$, the image of $\sigma$, it encodes the same action (modulo any two elements that act the same way). Now since $S_5$ was acting transitively on those $6$ elements, we see that $H$ acts transitively too, so it's a transitive subgroup of $S_6$.

  1. I'm not sure what "$H$ is transitive of order $6$" even means here... I imagine it means it's acting transitively on a set of size $6$, because

  2. If $H = S_5 / K$ is acting transitively on a set of size $6$, then it has to have at least $6$ elements! Indeed, we must have $6 \ \big | \ |H|$, basically because of the orbit stabilizer theorem (since the action has one orbit).

For more information about all this, you might be interested in Keith Conrad's notes on transitive group actions, which you can find here.

I hope this helps ^_^

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