Proof that $A_4$ is the unique group of order $12$ with no subgroup of order six

Question

Is there a simple proof that $A_4$ is the only group of order $12$ containing a subgroup of order six? (i.e. if $G$ is a group of order $12$ not having a subgroup of order six, then $G \cong A_4$?)

Answer 1

Claim 1: $A_4$ has no subgroup of order $6$.

Assume $H$ be a subgroup of order $6$ then $H$ is normal in $A_4$. Let $K$ be subgroup of $H$ eith order $3$ then $K$ must be uniqe subgroup of $H$ with order $3$ so $K$ is normal in $A_4$ contradiction.

Claim 2: $A_4$ is only subgroup of $S_4$ with order $12$.

Assume $H$ be another subgroup of order $12$ then we must have $HA_4=S_4$ which means that $$\dfrac{|H||A_4|}{|H\cap A_4|}=24 \implies |H \cap A_4|=6$$

which is impossible by claim $1$.

Claim 3: Let $H$ be any subgroup of $12$ with has no subgroup of $6$. By calim $1$, we know that it can not have normal subgroup $K$ of order $3$. Let $H$ act on left coset of $K$ by left multiplication then $ker=core(K)=e$ so we have injection from $H$ to $S_4$. By claim $2$ we are done.