Showing a semidirect product group is isomorphic to A4 Let $V$ be the Klein four-group and $f : V \rightarrow V$ map the identity to itself, (12)(34) to (14)(23), (13)(24) to (12)(34), and (14)(23) to (13)(24).
Let $C$ be a cyclic group of order 3 generated by $c$. Let $\phi : C \rightarrow Aut(V)$ be a group homomorphism such that $\phi_c=f$. Show that $V \rtimes_\phi C $ is isomorphic to $A_4$.
I've shown that $f$ is a group homomorphism, $f \circ f \circ f$ is the identity mapping, and $f \circ f$ is a similar mapping to $f$, mapping the identity itself but mapping any other element to whatever $f$ didn't map to (but not itself). Since $\phi_c=f$ and $\phi_1=id =f \circ f \circ f$, and since $\psi$ is a group homomorphism, then I figured $\phi_{c^2}=f \circ f$ in order to maintain that homomorphism property. The order of $V$ is 4 and the order of $C$ is 3, and the order of $A_4$ is 12, so it makes sense that at least the orders between the twice groups are the same.
I was going to simply draw a group table for $A_4$ and $V \rtimes_\phi C$ and then define a bijective mapping between each element based on the tables, but for 144 combinations this is tedious although it definitely would work. In any case, I'm almost certain it's not what I'm supposed to learn from this exercise. Is there a simpler method or an obvious group isomorphism between the groups that's I'm not seeing?
 A: I would approach this as follows.


*

*Observe that $f$ has the same effect on $V$ as does conjugation by $(243)$ (check this, it's late here and I may have made a mistake).

*So you should be able to construct an isomorphism from $V\rtimes_\phi C_3$ to $A_4$ that is the identity on $V$ and sends a generator of $C_3$ to $(243)$.


Another way of looking at it could be to construct $A_4$ as the (internal) semidirect product of $V$ and $\langle(243)\rangle$. Then you could use any other 3-cycle in place of $(243)$, but you need to be prepared to replace it with its inverse, if the conjugation 3-cycle runs in the opposite direction.

Adding a bit extra. Recall that the group operation in a semi-direct product $H\rtimes_\phi K$ is that of $H$ and $K$ on the individual subgroups together with the rule that conjugation of $H$ by an element $k\in K$ gives exactly the automorphism $\phi(k)$ of $H$. This implies the following. To get a homomorphism $f:H\rtimes_\phi K\to G$ from $H\rtimes_\phi K$ to a third group $G$ you need to:


*

*Specify a homomorphism $i:H\to G$,

*specify a homomorphism $j:K\to G$, and

*verify that for each $k\in K$ and $h\in H$ you have
$$j(k)i(h)j(k)^{-1}=i(\phi(k)(h)).$$

*Then
$$f(h,k)=i(h)j(k)$$
is automatically a homomorphism.


In the present case (always the case when you have an internal semi-direct product) the homomorphism $i$ is meant to be the inclusion mapping $V\to A_4$ and the homomorphism $j:C_3\to A_4$ was described by telling you where the generator of $C_3$ goes.
