I'm trying to work out how many isomorphisms there are from $V_4$ (Klein-4) to $C_2 \times C_2$ (2nd cyclic group)
Firstly, I've already shown they are indeed isomorphic and clearly there are $4!$ bijections from $V_4$ to $C_2 \times C_2$, however only $4!/4 = 6$ of these map the identity to the identity, so there are at most $6$ isomorphisms from $V_4$ to $C_2 \times C_2$.
Now, I'm slightly confused at this point, mostly because both are abelian. The map I used to show they are isomorphic I 'constucted' by comparing the Cayley tables of both, and showing they are 'in essence' exactly the same, and I mapped element to element in the order they appeared on the Cayley table. My confusion comes in because I'm not sure of the consequences of mapping the non-identity elements differently, so ultimately don't really know whether there are $6$ isomorphisms or just $1$. Having said that, I did map the non-identity elements differently and saw that it's still an isomorphism, that is $\phi (g_1 g_2) = \phi (g_1) \phi(g_2)$ for $g_1, g_2 \in V_4$. So I'm thinking that there are $6$ isomorphisms, but because I'm doubtful I thought I'd post it for some insight.
Thanks a lot.
Best Answer
We need precisely two non-identity elements in $C_2\times C_2$ to generate the group. Fixing two non-identity generators for $V_4$, we need only figure out how many ways we can map those two to two distinct non-identity elements of $C_2$ (since a homomorphism is determined by what is done to generators, and we'll need to map the fixed generators of $V_4$ to two distinct non-identity elements of $C_2\times C_2$ to get an isomorphism).
Indeed, the answer is $6$. There are $3$ options for where we send the first generator, and regardless of which of those $3$ we choose, there $2$ remaining options for where we send the second. Thus, there are $3\cdot 2=6$ ways to make such a map.
See also this related answer.