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How do I find the number of group homomorphisms from the symmetric group $S_3$ to $\mathbb{Z}/6\mathbb{Z}$?

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What have you tried? And if this is homework (which it really looks like), please add that tag. –  Tobias Kildetoft Dec 18 '12 at 17:36
    
By the way, "the options I am given..." isn't really a part of the question from the mathematical point of view, so it probably shouldn't be posted here. If I think I can prove that the answer is 7, I will answer that way regardless of any instructions that may have been given to you by a third party for whatever reason :) –  Trevor Wilson Dec 18 '12 at 22:21
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marked as duplicate by robjohn May 11 '13 at 9:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3 Answers

Hint: The group $\mathbb{Z}/6\mathbb{Z}$ is Abelian, and $S_3$ is non-Abelian. What does this tell us about the kernel of any homomorphism from $S_3$ to $\mathbb{Z}/6\mathbb{Z}$? Can we guarantee that certain elements inside $S_3$ must lie in the kernel? How many should there be? Now use this to count the total number of homomorphisms.

Added: $S_3$ consists of three elements of order $2$, two elements of order $3$, and the identity. The elements of order $2$ and $3$ do not commute. What possible places in $\mathbb{Z}/6\mathbb{Z}$ could I send an element of order $2$? What about the two elements of order $3$?

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Can you plz explain how to find out the kernel of any homomorphism from S3 to Z/6Z? –  Alka Goyal Dec 18 '12 at 18:01
    
@AlkaGoyal I added some more information –  Eric Naslund Dec 18 '12 at 18:05
    
It is still confusing for me,, can u plz further elaborate it? –  Alka Goyal Dec 18 '12 at 18:15
    
@AlkaGoyal The kernel of any homomorphism is defined as the set of elements which it maps to the identity. As for how the added hint helps: if, e.g., $s \in S_3$ is a permutation which swaps two of the three elements, then no homomorphism can map $s$ to 2. Why? Because $s^2$ is the identity in $S_3$, and $2^2$ is not the identity in Z/6Z. –  Lopsy Dec 18 '12 at 18:24
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In case you do not know, kernel of a homomorphism must be a normal subgroup of the inverse image. Simply $S_6$ has 3 normal subgroups which are {$e$}, $A_3$, and $S_3$.

Let $\phi : S_3 \rightarrow Z_6 $

Then possible kernels are {$e$}, $A_3$, and $S_3$.

Firstly, try {$e$}. By First Isomorphism Theorem; $S_3/{e}$ which is $S_3$ itself, $S_3$$\simeq \phi(S_3)$. The order of $S_3$ is 6 and observe that $Z_6$ has the same order. Thus, it yields $\phi(S_3)=Z_6$. However, $S_3$ is not abelian although $Z_6$ is. It is contradiction. Therefore, ker$\phi$ cannot be {$e$}.

Secondly, let's check for $S_3$, then $S_3/S_3$ is identity, so you can map every element of $S_3$ to the identity of $Z_6$. $\phi(s)=0$ , $\forall s \in S_3$ and $0$ is the identity of $Z_6$.

The last option is Ker$\phi$ = $A_3$, therefore the order of the factor group $S_3/A_3$ is $2$. First isomorphism theorem gives us: $S_3/A_3 \simeq \phi(S_3)$, then $\phi(S_3)$ is {$3,0$}.

$\phi(s)= 0$ if $s \in A_3$

Otherwise, $\phi(s)= 3$

As a conclusion, the answer is $2$.

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bt in solution set answer is 1, thats why i am confused. –  Alka Goyal Dec 19 '12 at 7:03
    
can it be possible that the trivial one should not be taken into account, otherwise I dont see any explanation, I'm quite what I've explained. –  Ada Dec 21 '12 at 13:55
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Let $\phi: G\rightarrow H$ be a surjective homomorphism. Suppose that $H$ is abelian. Can you prove that the commutator subgroup of $G$ must be contained in $\text{ker}\phi$? (Recall: the commutator subgroup of $G$, denoted $G'$, is the subgroup generated by $\{g^{-1}h^{-1}gh : g,h \in G \}$.)

By the first isomorphism theorem, $G/\text{ker}\phi\cong H$. Write $\text{ker}\phi$ as $K$. Since $H$ is abelian, so is $G/K$, and thus $(gK)(hK)=(hK)g(K) \Leftrightarrow (gh)K=(hg)K \Leftrightarrow (g^{-1}h^{-1}gh)K= K$ we have $g^{-1}h^{-1}gh\in K$. Since this is true for all $g,h\in G$, we have that $G'\leqslant K$. (Of course, we can simply do these steps backwards to show that the converse is true, so in fact $G'\leqslant K$ if and only if $G/K$ is abelian.)

By a simple computation, $S_3'$ is the subgroup of rotations of order $3$. I claim that $\phi:S_3\rightarrow \mathbb{Z}_6$ cannot be surjective. Why?

Since $[S_3:S_3']=2$ and $S_3'\leqslant \text{ker}\phi$, $S_3/\text{ker}\phi$ has order at most $2$.

With this information, you should be able to deduce which subgroups of $\mathbb{Z}_6$ can be an image of a homomorphism from $S_3$. From this you can easily count the number of homomorphisms.

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