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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.

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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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  • $\begingroup$ $(12),(13),(23)$ can map only among themselves in $3!=6$ ways but, $(123)$ can map to either $(123)$ or $(132)$ in $2$ ways , hence there are $2.6=12$ ways? What could be wrong here? $\endgroup$ – MathMan Jun 5 '14 at 19:30
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    $\begingroup$ Ohh I see, what you mean. $(123) = (13)(12)$ So, $\Psi(123) = \Psi(13) \Psi(12)$ . Hence $\Psi(123)$ is fixed by $\Psi(13) \Psi(12)$ . $\endgroup$ – MathMan Jun 5 '14 at 19:36
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    $\begingroup$ Suppose (12) maps to (12) and (13) maps to (13). Let $\phi$ be the homomorphism. Then $\phi(132)=\phi((12)(13))=\phi(12) \phi(13)=(12)(13)=(132)$. Thus you cannot map (132) to (123). Hence you don't have a choice due to homomorphic property. $\endgroup$ – Anurag A Jun 5 '14 at 19:37

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