How would I find the number of distinct homomorphisms and isomorphisms mapping the Klein four group to the Klein four group? How would I find the number of distinct homomorphisms and isomorphisms from the Klein four group to the Klein four group?
Thank you 
 A: Here’s an approach that perhaps is more advanced, maybe even too advanced.
Your group is a two-dimensional vector space $V$ over the field $k=\Bbb F_2$ with two elements. Every homomorphism $V\to V$ is automatically a $k$-linear map, so to count these, we need only count the $2\times2$ matrices over $K$, so sixteen in number.
For automorphisms, we need to count the nonsingular matrices, or what is the same thing, the number of distinct ordered $k$-bases of $V$. To get a basis of any two-dimensional vector space, you choose first a nonzero vector (three choices, in our case), and then a vector not in the space spanned by your previous choice. This spanned space has cardinality $2$ in our case, so there are only two possible choices for the second vector. Thus six different bases in all, and six automorphisms of $V$.
A: Note that $K_4=\mathbb{Z}_2\times \mathbb{Z}_2$  (thinking of $\mathbb{Z}_2$ as $\{0,1\}$ with addition) is generated by the two elements $(0,1)$ and $(1,0)$. Now let $f: K_4\to K_4$ be a homomorphism. If I determine what $f(0,1)$ and $f(1,0)$ is the map $f$ is fully known. Now $f(0,1)$ has four choices and so does $f(1,0)$. So there are $16$ homomorphisms.
For isomorphisms you need $f$ to be injective. So $f(1,0), f(0,1)\neq (0,0)$ and also $(0,0)\neq f(1,1)=f(1,0)+f(0,1)$. Let $f(1,0)=(a,b)$ and $f(0,1)=(c,d)$. Then the last condition means either $a+c\neq 0$ or $b+d\neq 0$. The possibilities for both being non-zero (with $(a,b)\neq (0,0)$ and $(c,d)\neq (0,0)$) are


*

*$a=1$, $b=0$, $\Longrightarrow c=0$, $d=1$. (This gives the identity map)

*$a=0$, $b=1$, $\Longrightarrow c=1$, $d=0$. (This gives the swap map)


The possibilities for $a+c=0$ but $b+d\neq 0$ are


*

*$a=1$, $b=0$, $\Longrightarrow c=1$, $d=1$.

*$a=0$, $b=1$, $\Longrightarrow c=0$, $d=0$ (Impossible).

*$a=1$, $b=1$, $\Longrightarrow c=1$, $d=0$.


Similarly for $a+c\neq 0$ but $b+d= 0$


*

*$a=1$, $b=0$, $\Longrightarrow c=0$, $d=0$ (Impossible).

*$a=0$, $b=1$, $\Longrightarrow c=1$, $d=1$ 

*$a=1$, $b=1$, $\Longrightarrow c=0$, $d=1$.


So we have found a total of 6 candidates for isomorphisms. It is easy to check that all indeed give an isomorphism (need to check surjectivity).
A: The lazy man's proof:
$V \cong \Bbb Z_2 \times \Bbb Z_2$ is generated by (any) two (non-identity) elements, call them $a$ and $b$.
We know that an automorphism has to send generators to generators. This gives a possibility of $6 = 3!$ such maps (since these are all the bijective maps on the set $\{a,b,ab\}$).
Thus we can regard $\text{Aut}(V)$ as (that is, isomorphic to) a subgroup of $S_3$ by considering how it permutes the (possible) generators.
We of course have the identity automorphsim (that's one), and the automorphism (verify it is one!) that swaps $a$ and $b$ (that's two). Note that this automorphism has order $2$ (it's its own inverse).
We also have the automorphism (again, verify this is a bona-fide automorphism) that sends:
$a \mapsto ab\\
b \mapsto a.$
Call this automorphism $\gamma$. We see that $\gamma^2 = \gamma \circ \gamma$ sends:
$a \mapsto ab \mapsto \gamma(ab) = \gamma(a)\gamma(b) = (ab)a = a(ba) = a(ab) = a^2b = b$,
$b \mapsto a \mapsto \gamma(a) = ab$.
Finally, $\gamma^3$ sends:
$a \mapsto ab \mapsto b \mapsto a\\
b \mapsto a \mapsto ab \mapsto b.$
We conclude $\gamma$ has order $3$, and thus $\text{Aut}(V)$ is (isomorphic to) a subgroup of $S_3$ with elements of order $2$ and $3$, that is: $6\mid|\text{Aut}(V)|$.
Hence $S_3 \cong \text{Aut}(V)$.
For general homomorphisms, we have a bit more freedom, we can send a generator (or more) to the identity. We've already found $6$, so what else is left?
Well, there's the trivial homomorphism which sends $a \mapsto e$ and $b \mapsto e$. We've already counted those homomorphsims that send both of $a,b$ to distinct generators, so there are possibilities left which are homomorphisms that send a single generator to $e$. Clearly there are three possibilities for the choice $b \mapsto e$:
$a \mapsto a\\
a \mapsto b\\
a \mapsto ab.$
Similarly, there are three more where $a \mapsto e$. So far, we're up to $13$ homomorphisms.
Have we missed any? Yes, precisely $3$: these are the homomorphisms where neither of $a$ nor $b$ maps to the identity, but both map to a common element of order $2$, and we have just three of those to choose from:
$a \mapsto a\\
b \mapsto a$
$a \mapsto b\\
b \mapsto b$
$a \mapsto ab\\
b \mapsto ab.$
This gives a total of $16$, as a previous answer indicates.
