Show that $A_{4}$ has no subgroup of order $6$ I have a few questions on the following proof:
We claim there are no subgroups of order 6. Any such subgroup must contain some elements of order $3$ (since there are only $4$ other elements)$^{*1}$, and these come in inverse pairs. So any subgroup $H$ of order $6$ must contain the identity$^{*2}$, either $2$ or $4$ elements of order $3$, and either $3$ or $1$ elements of order $2$.$^{*3}$ After renumbering$^{*4}$, we may therefore assume that $(1 2 3) \in H$ and $(1 2)(3 4) \in H$. But then H must contain $$(1 2 3)(1 2)(3 4) = (1 3 4);$$ together with its inverse $(1 4 3)$, and must also contain $$(1 2)(3 4)(1 2 3) = (2 4 3);$$ together with its inverse $(2 3 4)$. This gives at least $3$ pairs of elements of order $2$, together with the identity, so that $H$ contains at least $7$ elements. This is impossible, so $A_{4}$ has no subgroup of order $6$.
$^{*1}$ Why must any such subgroup contain some elements of order 3?
$^{*2}$ I'm assuming the order doesn't effect it any way?
$^{*3}$ Where are these numbers from (I understand they must add up to the order - 6)?
$^{*4}$ What does it mean by renumbering?
 A: $1$. Suppose there were a subgroup of order $6$, say $H$. $H$ clearly cannot be generated by a single element. So there are no elements of order $6$. Since the elements in $H$ cannot all have order $1$ and it still must have $6$ elements, the only possibility is that the elements of $H$ have order $2$ or order $3$. Since $A_4$ has only $3$ elements of order $2$, clearly $H$ must contain at least one element of the form $(abc)$ .
$2$. Not quite sure what you are asking but if I understand your question correctly, no, the order does not matter.
$3$. Yes, it must total to $6$ elements. In any case, it must include the identity. That leaves $5$ elements. Then you have two cases depending on which type of element you assume $H$ has. The two cases are $2$ elements of order $3$ and $3$ elements of order $2$ or $4$ elements of order $3$ and $1$ element of order $2$.
$4.$ What it means is there is no way of knowing that $H$ contained $(123)$, maybe it was $(234)$. But just call what you originally called $2$ instead $1$, what you called $3$ instead $2$ and what you called $3$ instead $4$ (ie renumbering), and $H$ contains $(123)$ instead. So what you are doing is saying it doesn't matter what element in particular $H$ contains, i.e. $(123)$ or $(234)$ or $(142)$ etc, just so long as it contains one that 'looks like' $(abc)$. They just pick $(123)$ and run through the argument because if it were any other $3$-cycle, you could just renumber everything and it works just the same.
EDIT: There is a nice simple proof here (bottom of page $4$).
