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I need to find the number of group homomorphisms from $S_4$ to $S_4$.

Let, $f$ be a group homomorphism. So, the $\ker(f)$ is a normal subgroup of $S_4$. The normal subgroups of $S_4$ are $\{(1)\}, V_4=\{(1), (12)(34), (14)(23), (13)(24)\}, A_4$ and $S_4$.

If $\ker(f)=\{(1)\}$ then we know that $\operatorname{Aut}(S_4)\cong S_4$. So, there are $24$ homomorphisms (isomorphisms).

If $\ker(f)=S_4$ there is only $1$ homomorphism.

If $\ker(f)=A_4$ then $|f(S_4)|=2$ by first isomorphism theorem. All elements outside $A_4$ must be sent to a single non-identity element of order $2$ because of transpositions that are present in $S_4$ which are of order $2$. So, $f(S_4\backslash A_4)$ has $6+3=9$ options corresponding to transpositions and product of two transpositions respectively. Hence, there are $9$ homomorphisms.

Now, adding all of these I get a total of $34$ homomorphisms. But, the answer to the problem given in the book is $34$. So, I am suspecting that there are no homomorphisms whose kernel is $V_4$ but I am not able to prove this.

So, can anyone please help me?

Thanks!

uuuuuuuuuu
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  • I'm not sure how you know that Aut$(S_4)=S_4$. This is true for most of the symmetric groups, but exceptionally Aut$(S_6)\neq S_6$ - just so you know. – Mark Bennet Sep 21 '19 at 21:12
  • @MarkBennet You can see that here - https://math.stackexchange.com/questions/880776/operatornameauts-4-is-isomorphic-to-s-4 – uuuuuuuuuu Sep 21 '19 at 21:14
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    I am counting 58 homomorphisms from $S_4$ to $S_4$, but many of them just have conjugate images. Up to conjugacy, there are 5 classes,with images respectively $\langle()\rangle$, $\langle(1,2)\rangle$, $\langle(1,2)(3,4)\rangle$, $S_3$, and $S_4$. – ahulpke Sep 21 '19 at 21:28
  • @ahulpke Do consider posting that as an answer. The OP is missing the homomorphisms with image $\cong S_3$. – Jyrki Lahtonen Sep 22 '19 at 13:52
  • Saikat, you are missing homomorphisms like this. I trust you can then inject the image of that homomorphism into $S_4$ :-) – Jyrki Lahtonen Sep 22 '19 at 13:54
  • @ahulpke I have just deleted an answer I wrote which was essentially the same so you can put yours as an answer. I think that he count of $34$ omits isomorphisms. It is possible that the wording of the original question from which OP was working excluded these - or perhaps it was just careless. – Mark Bennet Sep 22 '19 at 16:54
  • So, now it is established that there are $9$ homomorphisms corresponding to the case that the kernel is $A_4$. So, now I only need to prove that there are no homomorphisms for the case where kernel is $V_4$. Am I right? – uuuuuuuuuu Sep 23 '19 at 05:27
  • when kernel is K4 then I am getting 24 homomorphism – Raunit Singh Feb 05 '22 at 19:52
  • "Now, adding all of these I get a total of $34$ homomorphisms. But, the answer to the problem given in the book is $34$": why "but"? After all $34=34$ :) – citadel Jul 02 '23 at 08:07

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