$A_4$ has no subgroup of order $6$?

Can a kind algebraist offer an improvement to this sketch of a proof?

Show that $A_4$ has no subgroup of order 6.

Note, $|A_4|= 4!/2 =12$.
Suppose $A_4>H, |H|=6$.
Then $|A_4/H| = [A_4:H]=2$.
So $H \vartriangleleft A_4$ so consider the homomorphism
$\pi : A_4 \rightarrow A_4/H$
let $x \in A_4$ with $|x|=3$ (i.e. in a 3-cycle)
then 3 divides $|\pi(x)|$
so as $|A_4/H|=2$ we have $|\pi(x)|$ divides 2
so $\pi(x) = e_H$ so $x \in H$
so $H$ contains all 3-cycles
but $A_4$ has $8$ $3$-cycles
$8>6$, $A_4$ has no subgroup of order 6.

Solution 1:

Consider the group $A_4/H$. Let $x$ be a $3$-cycle, not in $H$, and consider the cosets $H$, $xH$, and $x^2H$ in $A_4/H$. Since this is a group of order $2$, two of the cosets must be equal. But $H$ and $xH$ are distinct, so $x^2H$ must be equal to one of them.

If $H=x^2H$, then $x^2=x^{-1}\in H$, so $x\in H$, contradiction. If $xH=x^2H$, then $x\in H$, same problem. So $H$ doesn't exist.

Solution 2:

By looking at the possible cycle types, we see that $A_4$ consists of the identity element (order $1$), $3$ double transpositions (order $2$) and $8$ $3$-cycles (order $3$).

Assume that $A_4$ has a subgroup $H$ of order $6$. Since $A_4$ does not contain elements of order $6$, $H$ cannot be cyclic. Therefore $H \cong S_3$, implying that $H$ contains $3$ elements of order $2$. So $H$ contains the identity element and the $3$ double transpositions. Since those $4$ elements form a subgroup of $A_4$, $H$ contains a subgroup of order $4$. Contradiction.

Solution 3:

I know this post is old, but there's another elegant way to prove this - a subgroup of order 6 has index 2. We prove the following statement: Any subgroup of index 2 of a finite group must contain all elements of odd order.

Let $G$ be finite and $H\subseteq G$ a subgroup of index 2. Any subgroup of index 2 is normal, so $G/H$ is a group and we write $\bar g:=gH$. Let $g$ be an element of odd order. Now we have $$g^{\operatorname{ord}g}=e\ \Rightarrow\ \bar g ^{\operatorname{ord}g}=\bar e\ \Rightarrow\ \operatorname{ord}\bar g\mid\operatorname{ord}g.$$ On the other hand, by Lagrange's theorem, we know that $\operatorname{ord}G/H =2$ so $$\bar g^2=\bar e\ \Rightarrow\ \operatorname{ord}\bar g\mid2.$$ Since $ \operatorname{ord}g $ is odd, it follows that $$\operatorname{ord}\bar g=1\ \Rightarrow\ \bar g=\bar e=e_{G/H}=H\ \Rightarrow\ g\in H.$$

Now since $A_4$ contains 9 elements of odd order, a subgroup of index 2 would, by the above statement have at least 9 elements, but by Lagrange's theorem has exactly 6 elements, which is a contradiction.