How many number of group homomorphisms are there from $S_3$ to $A_3$ ??

In case of $Z_m$ to $Z_n$, I can map 1 to an element whose order divides order of both $Z_m$ and $Z_n$ (m and n).

What method should be applied to non abelian groups??

$S_3$ is of order 6 , $A_3$ is of order 3 Identity should be mapped to identity. And how can we conclude about mapping of other elements in $S_3$ to $A_3$..


Note that if $\phi : G \to H$ is a homomorphism, then $|\phi (a)|$ divides $|a|$ for all $a \in G$.

Now, let $\phi : S_3 \to A_3$ be a homomorphism.

What does this imply on the order of the image of $2$-cycles under $\phi$? (What are the possible orders in $A_3$?)

Then, recall that the $2$-cycles generate $S_3$, and conclude that the only homomorphism $S_3 \to A_3$ is trivial.

  • $\begingroup$ Possible orders in A3 are 1 and 3 .. As 3 doesn't divide 2 , All 2 cycles should be mapped to identity and therefore, 3 cycle elements should also be mapped to identity to preserve homomorphism $\endgroup$ – Bharadwaj Rcr Feb 6 '18 at 5:20
  • $\begingroup$ @BharadwajRcr exactly. $\endgroup$ – idok Feb 6 '18 at 5:51

$S_3$ is generated by ??

It does not help to consider generating set as full $S_3$. What is the smallest generating set can you think of for $S_3$?

Any morphism is determined by generators of its domain. Once you know generators you will know possible images in $A_3$ depending on order.

  • $\begingroup$ Elements (123) and (132) generates A3 Elements (12) ,(13), (23) generates respective subgroups.. $\endgroup$ – Bharadwaj Rcr Feb 6 '18 at 5:03
  • $\begingroup$ It was not clear what respective groups you are talking about. I was asking for generators of $S_3$ and not of $A_3$.. Suppose you know $a$ is a generator of $S_3$ and is of order $3$, any homomorphism $\phi:S_3\rightarrow A_3$ should be such that $\phi(a)^3=1$. So, you have to check what are elements of $A_3$ that satisfy the property that its cube is identity. Does this help? $\endgroup$ – user312648 Feb 6 '18 at 5:07
  • $\begingroup$ I meant subgroups generated by (12),(13) and (23) . That is {(12),e} , {(13),e}{(23),e} S3 doesn't have any generators, otherwise it would become cyclic $\endgroup$ – Bharadwaj Rcr Feb 6 '18 at 5:12
  • $\begingroup$ I think you misunderstood the word generators.... I did not mean single generator. If a group has a generating set with single element, you call it cyclic group. I can write down the definition but I want you to read some book and see what does it mean to say generators of a group. you can google what is it mean by generating set of a group. $\endgroup$ – user312648 Feb 6 '18 at 5:18

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