About alternating group $A_4$ This is a simple exercise telling that $A_4$ cannot have a subgroup of order $6$. Here in my way:
Obviously, for any group $G$ and a subgroup $H$ of it with index $2$; we have $∀$$ g\in G$ ,$g^2\in H$. I suppose that $A_4$ has such this subgroup, named $H$, of order 6. Then for any $\sigma\in A_4$; $\sigma^2\in H$. I think maybe the contradiction happens when we enumerate all $\sigma^2$. May I ask if there is another approach for this problem? Thanks.
 A: Suppose $A_4$ has a subgroup of order 6. Then that subgroup which we call $H$ must be a normal subgroup because the index of $H$ in $A_4$ is 2. Now we know that for any group $G$, if  a subgroup say $K$ of $G$ is normal then it must be a union of conjugacy classes. So in our case, $H$ must be a union of conjugacy classes of $A_4$. What are the conjugacy classes in $A_4$?
You already know that the identity element is always in the conjugacy class of itself. Alternatively you could use the fact that the only groups up to isomorphism of order 6 are the cyclic group of order 6 or $S_3$. 
Suppose we now view  $A_4$ as a subgroup of $S_4$. If it has a subgroup order $6$ sitting inside of it, it cannot be the cyclic group of order 6 because $A_4$ has no element of order $6$. It is also impossible for $S_3$ to be contained inside of $A_4$ because $S_3$ has odd cycles. It follows that $A_4$ has no subgroup of order 6.
A: Your approach and many more can be found in this article, where $11$ different proofs are given.

Michael Brennan. Des Machale. Variations on a Theme: $A_4$ Definitely Has no Subgroup of Order Six!,  Mathematics Magazine, Vol. $73$, No. $1$ (2000) JSTOR

A: Suppose there exists a subgroup $H$ of order $6$, so $[A_4:H]=2$. Now there are $8$ $3$-cycles in $A_4$, so there exists a $3$-cycle $g\notin H$. Then consider the cosets $H, gH, g^2H$. So two must coincide. Since $H\neq gH$, $g^2H$ must equal one of the others, but either case implies $g\in H$, a contradiction.
A: Well, I am wondering if I could use the characters... If so, then here as follows:
As an exercise, show that, if G quotient its center is abelian, and if its commutator subgroup is of prime order p, then every non-linear irreducible character must be of degree n such that n²=|G:Z(G)|.
However, as H is of index 2, it is normal and its quotient group is abelian, that is, the commutator subgroup G' of G is either of prime order or equal to H. If the former case, then there is a non-linear character whose degree² is = 2, not possible. Hence G'=H. But then G/G' is cyclic, so G is abelian, which is not. Therefore, no such H can exist. Maybe there is some gap, as this argument shows that no group of order 12 can have subgroups of order=6...
Inform me if the gap is detected, thanks. 
