ALG2_Q2

Prove or disprove: Let $G=A_4$. There exist proper subgroups $H$ and $K$ such that $A_4\cong H\times K$.

WLOG, the orders of $H$ and $G$ must be $2$ and $6$ or $4$ and $3$. Since $A_4$ does not have a subgroup of order $6$, it must be the latter case. Say $|H|=4$ and $|G|=3$. $A_4$ has exactly one subgroup of order $4$, which is isomorphic to $V_4$, which is abelian. Also, all groups of order 3 are cyclic, and hence also abelian. It follows that $H\times G$ is also abelian. But, $A_4$ is not abelian.

An alternate argument is as follows: $H$ is forced to be $\{e,(01)(23),(12)(13),(03)(12)\}$, and $G$ can be one of four cyclic order 3 subgroups of $A_4$. WLOG, say $G=\{e,(132),(123)\}$. Note that $((01)(23),(132))\in H\times G$ is of order $6$. Since $A_4$ does not have an element of order 6, $H\times G$ cannot be isomorphic to $A_4$.

A flawed argument: If $H$ and $G$ are subgroups of $A_4$, then for the map $(h,g)\mapsto hg$ to be an isomorphism it is necessary that $H$ and $G$ are normal subgroups of $A_4$. Since there are no normal subgroups of $A_4$ of order 3, there must not exist $H$ and $G$ such that $H\times G\cong A_4$! --- While it is true that there cannot exist maps of the form $(h,g)\mapsto hg$, there might exist other maps!