Surjective group homomorphism $f:A_4\times \mathbb{Z}_2 \to S_3$

Question

Is there a surjective group homomorphism $f:A_4\times \mathbb{Z}_2 \to S_3$?

I need to find one if the answer is "yes" or explain why the answer is "no".

With $\mathbb{Z}_n$ I mean the quotient group $\mathbb{Z} / n\mathbb{Z}$.

Answer 1

The answer is no.

Suppose you have such a morphism $f$, and look at the restriction $g = f_{|A_4 \times 0}$. It's a morphism from (a group isomorphic to) $A_4$ to $S_3$.

We know explicitly the normal subgroups of $A_4$: besides the trivial ones, there is only one of them: the Klein subgroup $V_4 = \{ \mathrm{id}, (12)(34), (13)(24), (14)(23)\} \lhd A_4$. So we have three cases:

• $\ker g = \{\mathrm{id}\}$. That would force $g$ to be one-to-one, which is absurd, because $|A_4| = 12 > 6 = |S_3|$.

• $\ker g = A_4$. Then, the kernel of $f$ would contain $A_4 \times \{0\}$. The factorisation theorem then implies the existence of a morphism $\bar f : \mathbb Z_2 \to S_3$ such that $f = \bar f \circ \pi$, where $\pi$ is the canonical projection $A_4 \times \mathbb Z_2 \to \mathbb Z_2$. In this case, $\bar f$ cannot be surjective (because the source group is too small), and the same goes for $f$.

• $\ker g = V_4 \times \{0\}$, which implies $V_4 \times \{0\} \subset \ker f$. Through the same reasoning as in the second case, there is a morphism $\bar f : A_4/V_4 \times \mathbb Z_2 \to S_3$ such that $f = \bar f \circ \varpi$, where $\varpi$ is the canonical projection $A_4 \times \mathbb Z_2 \to A_4/V_4 \times \mathbb Z_2$. If $f$ was to be surjective, $\bar f$ would also have to be surjective. Now, the quotient $A_4/V_4$ is a cyclic group of order $3$, so $\bar f : A_4/V_4 \times \mathbb Z_2 \to S_3$, if surjective, would have to be an isomorphism. But $S_3$ is not isomorphic to the product of a cyclic group of order three and a cyclic group of order 2 (for example, it's not Abelian), so this case is also excluded.