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Find the number of group homomorphisms between $C_6$ and $S_3$.

For all the group theory buffs this is probably a piece of cake, but how does one generally go about a question like this. Is there a general way to figure this out, or do you need to make use of case specific counting arguments, or group specific characteristics? Any tips would be very helpful!

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    Homomorphic images of a cyclic group is cyclic (why?). What are the possible homomorphic images of $C_6$? (You have four). Which of these images correspond to subgroups of $S_3$? This is the same as "does there exist an element of order $x$", as $C_x\leq G$ if and only if $G$ contains an element of order $x$. Then, how many such elements are there? For example, $S_3$ has one subgroup of order $3$, but this subgroup has two generators. This corresponds to two homomorphic images.2012-06-26

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Suppose $G$ is any finite cyclic group--generated by $g$, say--and $H$ is any other group. If $\varphi:G\to H$ is a homomorphism, then the order of $\varphi(g)$ must divide the order of $g$.

Now, given some appropriate $h\in H$ (in the finite $|G|$ case, such that the order of $h$ divides the order of $g$), we can define a function $\varphi_h:G\to H$ by $\varphi_h(g^k)=h^k$. This will be a homomorphism.

In fact, all such homomorphisms $\varphi:G\to H$ must have this form, for one can readily show by induction that $\varphi(g^k)=\varphi(g)^k$ for all integers $k$. Thus, the situation is completely described.

Note that in the particular case you've described, all $h\in H$ are appropriate, so fixing $g$ as one of the two possible generators of $C_6$, we may describe $6$ homomorphisms $C_6\to S_3$ as described above.

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Any such homomorphism is completely determined by what it does to a generator of $C_6$. Can you take it from there?

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    It's clear that $C_6$ is generated by a single element of order 6, say $x \in C_6$. From how I understand it then if $\phi$ is our homomorphism, then $\phi(x^6) = \phi(x)^6$, so it must be sent to an element of order $6$. But $S_3$ doesn't have any elements of order $6$. Anything I can do with this?2012-06-26
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    @BallzofFury: That argument would seem to conclude that there can be no homomorphisms at all, but there is always at least one homomorphism, namely the one that sends everything to the identity in $S_3$. Can you use that example to figure out where your argument goes wrong?2012-06-26
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    Other than that I also have that $C_6$ has 1/1/2/2 elements of order 1/2/3/6 respectively. $S_3$ then has 1/3/2 elements of order 1/2/3. This must uniquely determine the number of homomorphisms, or is more information needed? @Henning, yes we at least have the trivial homomorphism, but if I'm not mistaken the 2 elements of order 6 can't be "used" by the homomorphism. This leaves us with only a few remaining options for the remaining 3 elements (since identities must be matched).2012-06-26
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    @BallzofFury: A homomorphism always "uses" all elements of its domain, so I'm not sure what you mean here. Further hint: you argued that because $\phi(x^6)=\phi(x)^6=e$, $\phi(x)$ must have order $6$. How does that work out when $\phi$ is the trivial homomorphism?2012-06-26
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    I think I've got it, clearly the identity must be sent to the identity. Also our 2 order 3 elements in $C_6$ must go to the 2 order 3 elements in $S_3$, as it's image has to be order 3, and order 2 doesn't suffice. Therefore we are left with 1 order 2 and 2 order 6 elements which may be freely sent to the remaining 3 order 2 elements in $S_3$. This leaves $3! = 6$ possible homomorphisms. This along the right lines?2012-06-26
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    @BallzofFury: Now you're overcomplicating it. Go back to Gerry's answer: The only thing that matters is where the generator is sent, which determines everything. Your first guess was that the image of the generator must have order 6 -- that is wrong, because the image of the generator under the trivial homomorphism has order 1, not order 6. So which _correct_ thing can you say about the possible images of a generator?2012-06-26
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    You're right, I was counting the isomorphisms. The image of the generator must be a divisor of 6, so 1,2,3 or 6. So any element of $S_3$ is eligible as an image?2012-06-26
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    @BallzofFury: Any element of $S_3$ whose order is a divisor of $6$ (which, as it happens, is all of them); yes.2012-06-26