Homomorphism from Q8 to V

Question

Question: find the number of group homomorphism from Quaternion group $Q_8$ to Klein four group $V$?

My attempt:

When $kerf = \{e\}$ then there is no homomorphism.

When $kerf = <-1>$ where <-1> denotes the subgroup generated by -1 in $Q8$, In this case we get, $\frac{Q8}{<-1>} ≈ V$ hence in this case there will be $6$ homomorphism.

and so on by considering different kernels, finally we get,

Total number of homomorphism from $Q8$ to $V$ is = $25$ homomorphism.

I didn't have solution of this problem. I don't know my answer is correct or not, please verify it & provide the answer. Thank you.

Answer 1

There remain just 3 other subgroups which are order of 4 and because $|G:H|=8:4=2$, they are normal and $G/H=\mathbb Z/2$

So for each of them there is 3 cases and totally 9 cases,and overall $9+6=15$ homomorphisms

So for example, if $kerf=\{1,i,-1,-i\}$ ,

then cosets are $K,Kj\quad where \quad(Kj)^2=1$

Hence there is 3 cases depending on where $Kj$ goes

EDİT: I've forgotten the case $H=Q_8 $

So, total number is $16$