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I'd like to find all of the homomorphisms $ \varphi: S_3 \to \mathbb{Z}_{4} $.

What I've tried so far:

I tried to do $ \varphi(Id) = \bar{0} = \bar{4} $ (as somebody used here). But then I realized that it was because the identity element coincide with the generator of domain group $ \mathbb{Z}_{15} $.

And, as far as I know, $ S_3 $ doesn't have a generator.

I really have no idea of how to do it.

** Corrected: $ S_3 $ has two generators: $ d_1 $ and $ d_2 $. Even though, I can't use the same trick that in the other post.

$$ S_3 = \left \{ Id=\begin{pmatrix}1&2&3\\1&2&3\end{pmatrix}, d_1=\begin{pmatrix}1&2&3\\3&1&2\end{pmatrix}, d_2=\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}, t_1=\begin{pmatrix}1&2&3\\1&3&2\end{pmatrix}, t_2=\begin{pmatrix}1&2&3\\2&1&3\end{pmatrix}, t_3=\begin{pmatrix}1&2&3\\3&2&1\end{pmatrix} \right \} $$

Community
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francolino
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3 Answers3

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The group $S_3$ has only $\{\mathit{Id}\}$, $A_3$ (the alternating subgroup) and $S_3$ as normal subgroups. If $\ker\varphi=S_3$, then we have the trivial homomorphism. We can rule out $\ker\varphi=\{\mathit{Id}\}$, because $S_3$ is not abelian.

Let's assume $\ker\varphi=A_3$. Then $\varphi$ induces an injective homomorphism $\hat\varphi\colon S_3/A_3\to\mathbb{Z}_4$ and there's only one of them, because $\mathbb{Z}_4$ has just one (cyclic) subgroup of order two, namely $\{\bar{0},\bar{2}\}$.

Finish up proving that indeed there is only one homomorphism $\varphi$ such that $\ker\varphi=A_3$.

egreg
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Hint: Every 3-cycle must map to some $a\in\mathbb Z_4$ with the property $a+a+a=0$, and there is only one possibility for that.

Since the product of two different transpositions is a 3-cycle, this leaves very few possibilities for what the image of each of the three transpositions can be.

hmakholm left over Monica
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Suppose you already know that:

$S_3 \cong \langle a,b: a^3 = b^2 = e;ba = a^2b\rangle$.

Then any homomorphism $\phi:S_3 \to \Bbb Z_4$ is completely determined by $\phi(a)$ and $\phi(b)$.

Now the order of $\phi(a)$ divides the order of $a$, which is $3$ (a prime). So $\phi(a)$ must have order $1$ or $3$. However, $\Bbb Z_4$ has no elements of order $3$ ($3$ does not divide $4$), so we know $\phi(a) = \overline{0}$.

Thus $\phi$ is completely determined by $\phi(b)$, which must have order $1$, or order $2$ (again, $2$ is a prime number, which is rather convenient for us).

If $\phi(b)$ has order $1$, $\phi$ must send everything to $\overline{0}$. This is the trivial (zero) homomorphism.

On the other hand, if $\phi(b)$ has order $2$, it must be that $\phi(b) = \overline{2}$, as this is the only element of $\Bbb Z_4$ of order 2. This allows us to write the sole non-trivial homomorphism explicitly as:

$\phi(e) = \overline{0}\\ \phi(a) = \overline{0}\\ \phi(a^2) = \overline{0}\\ \phi(b) = \overline{2}\\ \phi(ab) = \overline{2}\\ \phi(a^2b) = \overline{2}$

(If you have trouble following this, substitute $(1\ 2\ 3)$ for $a$, and $(1\ 2)$ for $b$).

David Wheeler
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  • I don't know the notation that you use: $ (1\ 2\ 3) $ and $ (1\ 2) $. Which I use for is: $ Id=\begin{pmatrix}1&2&3\1&2&3\end{pmatrix} $. Mind to clarify this?

    I've tried to translate what you wrote to my "language". Then: the only possible non-trivial homomorphism is the one determinated by: $ \varphi(d_i) = \bar{0}, \forall i \quad \varphi(t_i) = \bar{2}, \forall i \quad \varphi(Id) = \bar{0} $. I updated the post according to the notation I used here. Is this what you said?

     – francolino Jul 03 '15 at 02:34
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    $(1\ 2\ 3)$ is shorthand ("cycle notation") for what you would write in matrix form as $\begin{pmatrix}1&2&3\2&3&1\end{pmatrix}$, and likewise $(1\ 2)$ means$\begin{pmatrix}1&2&3\2&1&3\end{pmatrix}$. One reads $(1\ 2\ 3)$ as: "1 goes to 2, 2 goes to 3, 3 goes to 1". – David Wheeler Jul 03 '15 at 03:53