How many isomorphisms between two abelian groups 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.
 A: 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.
A: Another way to think about the problem, since as you've stated the two groups are isomorphic, is to determine the size of the automorphism group of $C_2\times C_2.$  If we view $C_2\times C_2$ as a two dimensional vector space over the field of two elements, $\mathbb{F}_2$, then an automorphism is just an invertible $2\times 2$ matrix with entries in $\mathbb{F}_2$.  Hence $Aut(C_2\times C_2)\cong GL_2(\mathbb{F}_2)$.
To compute the order of $GL_2(\mathbb{F}_2)$ notice there are three possible choices for the first row ((1,0), (1,1), and (0,1)), and having chosen this row, we now have two choices for the second row (whichever two we didn't choose).  Thus the order of $GL_2(\mathbb{F}_2)$ is 6.
Following a similar procedure to what we did above, it can be shown that
$$
|GL_n{(\mathbb{F}_q)}|=(q^n-1)(q^n-q)(q^n-q^2)\ldots(q^n-q^{n-1})
$$
where $\mathbb{F}_q$ is the field of $q=p^m$ elements.  Our case follows from letting $n=q=2$.
A: Consider the isomorphisms going one way, all arrows to the same object. The comparison of those objects is an automorphism. The question is how you tell that those are two isomorphisms to the same object. So, the distinctness of isomorphisms is measured by needing a non-identity automorphism to convert the two. So, there can be no more isomorphisms than there are automorphisms of the target. However, the composition of isomorphisms is an isomorphism, so composing any isomorphism with an automorphism of the target will yield a new isomorphism to the target. Hence, the number of isomorphisms between two objects is the same as the number of automorphisms of either object.
