How many group homomorphisms are there from $\Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z$ to $S_{3}$ There is a similar problem that $\Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z$ to $S_{4}$. But I still confuse,
This is what I tried:
Let $f$ be homomorphism from $\Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z \to S_{3}$
then $\Bbb Z_2\times \Bbb Z_2/\ker f \cong \text{Image}  f$
$|\text{Image}  f |=1,2,4$
If $|\text{Image}  f |=1$,
$f(a)=e$ ,for all  $a\in \Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z$, so that's 4 grooup homomorphisms, 
since there are 4 elements in $\Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z$ (Is that right?)
can anybody solve this problem more detail, I stuck on this problem for two hours
 A: Since $4$ does not divide $6$ there are no injections into $S_3$. Let $a,b$ be the generators of $\Bbb Z_2 \times \Bbb Z_2$. Then there are three possibilities for the cyclic kernel of order $2$: $\langle a \rangle, \langle b \rangle,$ and  $\langle ab \rangle$. There are three elements of order $2$ in $S_3$, combining these with the possible kernels (and mapping the other two elements to the same image) we obtain $9$ homomorphisms, together with the trivial homomorphism this gives us a total of  $10$ homomorphisms.
A: One way to count the group homomorphisms $f:\Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z \to S_3$ is to focus on the images of generators $a=(1,0)$ and $b=(0,1)$.
Since $a,b$ are both of order $2$, the images $f(a),f(b)$ must have orders that are divisors of $2$.  Therefore each image must be the identity in $S_3$ having order $1$, or (since $S_3$ is permutations of three items) a transposition (having order $2$).
Thus four choices in all for each generator image.
But the choices of $f(a),f(b)$, although they fully determine the map $f$, are constrained by the abelian property of $\Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z$ to satisfy $$f(a)f(b)=f(b)f(a)$$
One way to count these is to subtract from the $4\cdot 4 = 16$ those possibilities for which $f(a)$ and $f(b)$ would not commute.  There are six such non-commuting choices, picking one transposition for $f(a)$ and a different transposition for $f(b)$.
Thus we count $16-6=10$ possible homomorphisms, in agreement with the count given by Bogaerts Marc.
