Understanding homomorphism and non-homomorphism from $A_4$ to $S_6$

Question

Could anyone help me how to find an example of a function $f: A_4\to S_6$ that is not a homomorphism and one $f$ that is a homomorphism? And is $A_4\times S_6$ isomorphic to $S_6\times A_4$? Thanks so much for your help.

Answer 1

Most functions $f: A_4 \to S_6 $ will not be homomorphisms. It's quite a restrictive property. Recall that for $f$ to be a homomorphism we require $$ f(\sigma \tau) = f(\sigma) f(\tau) \text{ for all } \sigma, \tau \in A_4.$$ You can show that homomorphisms must map the identity to the identity. So simply define any $f : A_4 \to S_6$ which doesn't map the identity to the identity, and your function will fail to be a homomorphism.

If you now want a homomorphism $f : A_4 \to S_6$, there's a really simple way to construct a homomorphism between any two groups. Let $G$ and $H$ be groups with $e_H$ the identity in $H$. Then the map $f : G \to H$ given by $f(g) = e_H$ for all $g \in G$ is always a homomorphism.

For something more interesting in this particular case, it's helpful to note that there is a copy of $S_4$ sitting inside $S_6$. For any $\sigma \in S_4$, we can define a $\tau_{\sigma} \in S_6$ simply by acting with $\sigma$ on $\{1,2,3,4\}$ and leaving $\{5, 6\}$ alone, i.e. $$\tau_{\sigma}(i) = \begin{cases} \sigma(i) & \text{if } i \in \{1,2,3,4\}, \\ i & \text{if } i \in \{5,6\}. \end{cases} $$ Since $\sigma(i) \in \{1,2,3,4\}$ for any $i \in \{1,2,3,4\}$, we then have $$\tau_{\sigma_1 \sigma_2} = \tau_{\sigma_1} \tau_{\sigma_2} \text{ for any } \sigma_1, \sigma_2 \in S_4,$$ which demonstrates that the map $f : S_4 \to S_6$ given by $f(\sigma) = \tau_{\sigma}$ is a homomorphism. This map is injective, since $$\tau_{\sigma}(i) = i \text{ for all } i \in \{1,2,3,4,5,6\} \iff \sigma(i) = i \text{ for all } i \in \{1,2,3,4\}.$$ So this $f$ is in fact an isomorphism onto its image, which is what is meant by "there is a copy of $S_4$ sitting inside $S_6$". You can think of this $f$ as an inclusion map, even though strictly speaking $S_4$ is not a subset of $S_6$. Now, $A_4$ is a subgroup of $S_4$, so $f$ restricts to a homomorphism $A_4 \to S_6$, which is still injective and so we also get a copy of $A_4$ inside $S_6$.

Finally, $A_4 \times S_6$ is isomorphic to $S_6 \times A_4$. Indeed, $G \times H$ is isomorphic to $H \times G$ for any groups $G, H$ via the map $(g, h) \mapsto (h, g)$. This is a good example of the idea that $$\text{isomorphism} = \text{relabelling things}.$$