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I know that there exist no nontrivial homomorphism from $S_3$ into $Z_5$ as they are groups of co-prime order. I am not looking for an explanation of this but for an explanation concerning the obvious misunderstanding I have between group actions and homomorphisms. As to me $S_3$ acting on $Z_5$ does not seem trivial.

This is how I have been thinking of group actions.

$\begin{array}{ l | c |c|c|c|r } &0 & 1 & 2 &3 &4\\id&1&1&2&3&4\\(12)&0&2&1&3&4\\(13)&0&3&2&1&4\\(23)&0&1&3&2&4\\(123)&0&2&3&1&4\\(213)&0&3&1&2&4 \\ \end{array}$

The left hand column represent an element in $S_3$ and the upper row are element of $Z_5$. How does this now relate to homomorphism? And why is the homomorphism trivial?

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3 Answers 3

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You have defined a successful group action on the set $\{0,1,2,3,4\}$, but this does not give a homomorphism to $\mathbb Z_5$. To do this, you would need to assign an element of $\mathbb Z_5$ to each element of $S_3$. So for example if your homomorphism is called $\phi\colon \mathbb S_3\to \mathbb Z_5$, then $\phi((12))$ would have to be an element of $\mathbb Z_5$. Say $\phi((12))=k$. But then $0=\phi(id)=\phi((12)^2)=2k$. But the only element of $\mathbb Z_5$ satisfying $2k=0$ is $0$. Hence $\phi((12))=0$. You can argue similarly for all the other elements of $S_3$.

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Thank you. This answer really helped clear up my misunderstandings! –  fafddf Nov 10 '13 at 2:03
    
I think this misses the point since acting on a group is different than giving a homomorphism into a group. –  James Cameron Nov 10 '13 at 2:12
    
@JamesCameron: actually I think this illustrates that point. –  Grumpy Parsnip Nov 10 '13 at 2:24

I think the difference is that you are thinking of group actions of $G$ on $S$ as mapping elements of $G$ to elements of $S$, thus inducing a homomorphism of $G$ into $S$. What is actually happening is that elements of $G$ are being mapped to permutations of S. The following should help understand this difference.

You can think of finite group actions as homomorphisms from the acting group to the symmetric group on the set being acted on. To be explicit, let $G$ act on $S$, where $G$ is finite with order $|G|=n$, and $S$ is finite with cardinality $m$. Then $\operatorname{Sym}(S)$ is the group formed by (arbitrarily) numbering the elements of the finite set $S$, and then associating the permutations in $S_m$ with the corresponding numbers. A group action is a homomorphism $\phi:G\rightarrow \operatorname{Sym}(S)$. (Note that this action need not be injective, i.e. some elements of $G$ may not do anything to $S$ at all. If the action is injective, we call it faithful.)

Now, when you're talking about groups acting on other groups, you add in additional information - generally, you want the action $\phi$ to be compatible with the group operation on $S$. In other words, you don't want the permutations of $S$ induced by the action of $G$ to be inconsistent with how $S$ works as a group. This means that defining an action requires you instead to make a homomorphism from $G$ to $\operatorname{Aut}(S)$, the subgroup of $\operatorname{Sym}(S)$ of permutations such that the group structure of $S$ remains unchanged. (This, by the way, is the beginning of making a semidirect product.) Things get more complicated in this situation. In your example, $\operatorname{Aut}(\mathbb{Z}_5)\cong \mathbb{Z}_4$. Can you find a homomorphism $\phi:S_3\rightarrow \operatorname{Aut}(\mathbb{Z}_5)$?

Finally, note that you can still have $S_3$ act nontrivially on $\mathbb{Z}_5$ as a set, simply by permuting its elements, which is exactly what you've done in the question. It just doesn't give rise to any group homomorphism $S_3\rightarrow \mathbb{Z}_5$.

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If you want to define an action of a group $G$ on another group $H$, you need a homomorphism from $G$ to the automorphism group of $H$. The automorphism group of the cyclic group of order 5 is the cyclic group of order 4, so you can define a nontrivial action since there are maps from $S_3$ to $\mathbb{Z}/4$.

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