TITLE: How many homomorphisms $\Psi : S_3 \rightarrow S_3$ exist?
QUESTION [9 upvotes]: How many homomorphisms $\Psi : S_3 \rightarrow S_3$  exist?
Attempt: I found $16$ homomorphisms in total.
$S_3 ={(1).  (12),(13),(23),   (123),(132)}$
There are three normal subgroups in $S_3 = \{(1),~ S_3 ~, ~A_3 = \langle (123)\rangle\}$
Let $\Psi : S_3 \rightarrow S_3$ be a homomorphism. then $Ker~ \Psi$ can be any of the normal subgroups in $S_3$. When :
(a) $Ker ~\Psi= S_3 \rightarrow$ then this is the trivial homomorphism with all mappings converging to identity. There is one homomorphism in this case
(b) $Ker ~\Psi= (1) \rightarrow \Psi(x) = y \implies$ then $O(y)$ must divide $O(x)$. Hence, $(123)$ can get mapped only to $(132)$ or $(123)$.
$(12), (13) ,(23)$ can be mapped among themselves in $3! = 6$ ways.
Hence $6 .2 =12$ homomorphisms are possiblie in this case
(c) $Ker ~\Psi=(\langle 123 \rangle) \rightarrow$  now $(\langle 123 \rangle) = \{ (1), (123),(132) \}$ . All elements in this set get mapped to $(1)$ in just $1$ way. We have to think of how the other elements get mapped 
in this case, the image of $\Psi$ is a subgroup of order $2$ since $S_3/ Ker~\Psi \approx \Psi(S_3) \implies |\Psi (S_3)| = 2$
$ \implies \Psi (S_3) = \{1, (12)\}$ or $\{1,(13)\}$ or $\{1,(23)\}$
There are $3$ possible homomorphisms in this case
Hence total number of homomorphisms possible = $16$. 
Is my attempt correct?
Thank you for your help.

REPLY [5 votes]: Your analysis is ALMOST correct. Except that in part (b) it should only give you $6$ homomorphisms. For example, when you know where ( 1 2) maps to and where (1 3) maps to then you no longer have the luxury to choose where (1 2 3) maps to.
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