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I am finding the number of homomorphisms $f:S_3 \to S_3$ when $\ker(f)=(1)$.

By first isomorphism theorem, $$S_3/Ker(f)=S_3/(1) \simeq f(S_3)\subset S_3.$$i.e, $f:S_3 \to S_3$in this case. Now any element $x \in S_3$ goes to $f(x) \in f(S_3)=S_3$. Also $O(f(x)) |O(x)$.

The $identity$ permutation $(1)$ maps to $(1)$ as $1|1$;

order $2$ elements maps to order $2$ elements, i.,e., $\{(12),(13),(23) \} \to \{(12),(13),(23) \}$ in $3!=6$ ways.

But the problem occurs with the order $3$ elements, $(132)$ and $(123)$.

We can have $f((123))=(123)$ or $f((123))=(132)$ and $f((132))=(132)$ or $f((132))=(132)$.

So here total $2!$ ways.

Hence total $6 \times 2=12$ homomorphisms from $S_3 \to S_3$ if $\ker(f)=(1)$.

Is it true?

Edit: Why we cannot get $f((123))=(132)$ or $f((132))=(123)$ ?

J. W. Tanner
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MAS
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  • Do you seek the number of automorphisms of $S_3$? – lhf Apr 22 '20 at 17:37
  • Cf. this question; once you determine where the $2$-cycles are mapped, doesn't that determine where the $3$-cycles are mapped, since the $3$-cycles are products of $2$-cycles? – J. W. Tanner Apr 22 '20 at 17:37
  • Or this one. – Arnaud D. Apr 22 '20 at 17:38
  • @J.W.Tanner, Suppose $f(12)=(13)$ and $f(13)=(12)$,then $f(123)=f((13)(12))=f((13)(12))=(12)(13)=(132)$. Where is the error ? – MAS Apr 22 '20 at 17:47
  • @M.A.SARKAR: You said in OP we can have $f((123))=(123) $ or $(132)$, but I'm saying that once you have determined which one of the six ways $(12), (13)$, and $(23)$ are mapped by $f$, then there is only one possible result for $f((123))$, namely $f((13))f((12))$, which has already been determined – J. W. Tanner Apr 22 '20 at 17:53
  • @J.W.Tanner, there is subtle point which is making me confused. You said by homomorphism $f(123)$ is determined by $f(13)f(12)$. Now we have included the possibility $f(13)=(12)$ and $f(13)=(12)$ in the order $2$ case. So $f(123)=f(13)f(12)=(12)(13)=(132)$. I can not catch the error here . – MAS Apr 22 '20 at 17:58
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    I'm saying that if you have determined that $f((13))=(12)$ and $f((12))=(13)$, then you have determined $f((123))$ -- i.e., there is only one possibility for $f((123))$, not two – J. W. Tanner Apr 22 '20 at 18:00
  • @J.W.Tanner,that means if we have $f(132)=(123)$,then we can not have $f(132)=(132)$ and vice-versa – MAS Apr 22 '20 at 18:02
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    some of the homomorphisms will map $(132)$ to $(123)$ and some will map to $(132)$ to $(132)$, but, again, once you have determined where a homomorphism maps the $2$-cycles, you won't have a choice as to where it maps the $3$-cycles, if it's a bona fide homomorphism – J. W. Tanner Apr 22 '20 at 18:05
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    @J.W.Tanner, thanks for nice help.I got it finally – MAS Apr 22 '20 at 18:06

2 Answers2

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It seems that you are seeking the automorphisms of $S_3$

(homomorphisms from $S_3$ to itself with trivial kernel).

It is not true that there are $12$ of them; there are only $6$ of them.

That is because there are $6$ possible ways that the $2$-cycles could be mapped, as you pointed out,

and, once those are determined, there is only one possible way the $3$-cycles

(which are products of $2$-cycles) could be mapped.

J. W. Tanner
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By finiteness, such a map would be an automorphism. The set of automorphisms of $S_n,n\ne2,6$ is $S_n$. The natural map induces an isomorphism.