# Prove that the action of $H$ on $X$ has to be transitive and then explain why such a subgroup can't exist

Suppose $H \leq A_4$ is a subgroup containing $6$ elements. $H$ acts on $X=\{1,2,3,4\}$ by $\sigma \cdot i = \sigma (i)$.

First explain that an orbit with at least $3$ elements has to exist in order for $H$ to act on $X$. Then prove that the action of $H$ on $X$ has to be transitive. After this explain why such a group $H$ cannot exist.

So for the first part I just concluded that there has to be a $3$-cycle in $H$ since the only elements in $A_4$ are $3$-cycles and $2$-$2$-cycles. There are only three $2$-$2$-cycles in $A_4$ so there have to be at least three $3$-cycles in $H$. Let's call one of these $3$-cycles $g$, then we have $g \cdot a = b$, $g \cdot b = c$ and $g \cdot c = a$ for some $a,b,c \in X$ and $a\neq b \neq c$. Since $a,b,c$ are in the same orbit there has to be an orbit that contains at least $3$ elements.

The second part I struggled with. I thought maybe I can show that the $3$-cycles in $H$ have to contain all the integers. For example if $(123) \in H$ then there has to be another $3$-cycle containing the number $4$. But if that's even a correct thought how do I show from there that the action is transitive, i.e. that there is only one orbit containing all the elements of $X$? I could really use your help on this one. For the next part I just assumed that the action has to be transitive and so there is only one orbit containing all $4$ elements.

For the third part I used the following: the number of elements in the orbit of an element $a$ equals $|H : G_a|$ where $G_a$ is the stabilizer of $a$ in $H$. From the previous paragraph we know that there is only one orbit with $4$ elements. But $4$ doesn't divide $|H|$, so therefore such a subgroup cannot exist.

In short, these are my questions:

• Did I answer the first and third part correctly?

• How do I prove the second part?

First of all, $H$ must have the unit element, so you have at least two $3$-cycles. Without loss of generality, we may assume that $H$ contains $(123)$ (and $(132)$).
If you can use Cauchy's theorem, you know that there is an element of order $2$ in $H$, that is, a 2-2-cycle, which takes one of $1, 2, 3$ to $4$, so that $H$ is transitive.
If you cannot use Cauchy's theorem, then the alternative is that there is another $3$-cycle in $H$, which will then again take one of $1, 2, 3$ to $4$, so that $H$ is transitive.
But for $H$ to be transitive on a set with $4$ elements, as you correctly noted, it must have order divisible by $4$.