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How many Homomorphisms are there from S3 to the Klein 4 Group?

I am only thinking of constructing a group table. Can someone explain the problem, this is NOT a homework problem, but for additional practice for final.

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up vote 1 down vote accepted

One of the fundamental isomorphism theorems states that, if $\phi: A \rightarrow B$ is a homomorphism, then $A/\ker \phi \cong \operatorname{im}\phi$. This means that the homomorphisms from $S_3$ to $Z_2\times Z_2$ must correspond to isomorphisms $S_3/N \cong M\subset Z_2\times Z_2$. Any two isomorphic groups must have the same order, so we must have $6/|N| = |M|$, and by Lagrange's theorem $|N| | 6$ while $|M| | 4$. It should be clear that the only time this holds is when $|N| = 3$ and $|M| = 2$. There is one subgroup of $S_3$ with order $3$, the alternating group $A_3$, and three subgroups $0\times Z_2, Z_2\times 0,\{(0,0),(1,1)\}\subset Z_2\times Z_2$ of order $2$. These correspond to the two isomorphisms from $S_3$ to $Z_2\times Z_2$.

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$\mathbb{Z}_2 \times \mathbb{Z}_2$ has three subgroups of order 2. – Chris Eagle May 1 '11 at 8:15
@Chris: Thanks. – Alex Becker May 1 '11 at 8:20
you've changed the two subgroups to three, but there's still a "two" in your last sentence. Also, you write "isomorphisms" in that last sentence, but of course you mean "homomorphisms". Finally, you've missed a fourth homomorphism. – Gerry Myerson May 1 '11 at 12:52

First figure out where the elements of order $3$ have to go. Then pick a place for some element of order $2$ to go, and see what this implies for where everything else goes.

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The six elements of $S_3$ are: 1, (12), (13), (23), (123), and (132). If $\phi :S_3\rightarrow K$ is the homomorphism, then we must have that $\mathrm{ord}\left[ \phi (x)\right] \mid \mathrm{ord}[x]$. As the order of every nonidentiy element in $K$ is $2$, it follows from this that $\phi (x)=1$ for $x=(123),(132)$ (because these elements have odd order).

If we write $(123)=(13)(12)$, we see that $\phi \left( (13)\right) =\phi \left( (12)\right) ^{-1}$, and hence, WLOG, $\phi \left( (13)\right) =\phi \left( (12)\right) =a$, for some $a\in K$ (as each element in $K$ is its own inverse). Similarly, because $(132)=(13)(23)$, we see that $\phi \left( (13)\right) =a=\phi \left( (23)\right)$. As there were four choices of $a$, there are four such homomorphisms.

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