1. There is a non-trivial group homomorphism from $S_3$ to $\mathbb Z/3\mathbb Z.$

My Attempt:

True: Choose $a,b\in S_3$ such that $|a|=3,|b|=2.$

Then $S_3=\{1,a,a^2,b,ba,ba^2\}.$

Define $f:S_3\to\mathbb Z/3\mathbb Z:b^ia^j\mapsto j+3\mathbb Z$

Then $f$ is a nontrivial homomorphism.

Is the attempt correct?

  • $\begingroup$ You must verify that the map you defined is indeed a homomorphism. In order to do so, you must first compute the multiplication table of $S_3$, and then calculate $f(xy)$ and $f(x)+f(y)$ for all $x,y\in S_3$ to see if they match. You will see that $f$ fails. The reason it fails, is that any transposition must be sent to $0$, yet the transpositions generated $S_3$. $\endgroup$ – Olivier Bégassat Sep 16 '13 at 6:50
  • 4
    $\begingroup$ It seems, you are just guessing. Hint: consider the kernel of any such homomorphism. $\endgroup$ – Dune Sep 16 '13 at 6:50


  1. The image of a homomorphism is a subgroup of co-domain. Does $\mathbb{Z}_3$ has any subgroups except $\{\bar{0}\}$ and itself?

  2. Since $|S_3|=3!=6$ and $|\mathbb{Z}_3|=3$ then any function $\varphi: S_3 \to \mathbb{Z}_3$ will be not one-to-one. So, the kernel must be non-trivial.

  3. If $\varphi: S_3 \to \mathbb{Z}_3$ is a homomorphism that doesn't send everything to $\bar{0} \in \mathbb{Z}_3$ then it must be surjective (According to what I said in 1). Now, what happens if you use the first isomorphism theorem? Note that $\operatorname{ker{\varphi}}$ is a normal subgroup of $S_3$ but $S_3$ has no normal subgroups of order 2.

I have actually given you more information that you need, but to sum it up, there are no non-trivial homomorphisms from $S_3$ to $\mathbb{Z}_3$. In a fancy way, they write this as $\operatorname{Hom}(S_3,\mathbb{Z}_3)=\{e\}$ where $e: S_3 \to \mathbb{Z}_3$ is defined as $e(\sigma)=\bar{0}$ for any $\sigma \in S_3$.


Hint: Observe that $(ba)(b)=a^2$. Apply your $f$ to both sides. 

Hint for the original problem: The kernel of a homomorphism must be a normal subgroup.


Reference: p. 1 https://www.math.okstate.edu/~mantini/4613info/ps04.pdf

Proof: We must use the result of exercise 4.3.19 in this problem. Note that in $\mathrm{Z}_{3}=\{0,1,2\}$, the orders of the elements are $|0|=1$, $o(1)=o(2)=3$.\ In $S_{3}= \{(1),$\ (12)\ ,\ (13)\ ,\ (23)\ $,\ (123), (132)\}$,the orders of the transpositions are each 2.

Transpositions in $S_3$.
For any transposition $\tau$, $o(f(\tau))$ must divide $o(\tau)=2$, and this says that $o(f(\tau))=1$, since 3 does not divide 2. Therefore $f(\tau)=0$, the identity, for any transposition $\tau$.

3-cycles in $S_3$.
Note that $(123)=(13)(12)$. This says that $f(123)=f((13)(12))= f(13)+f(12)=0+0=0$.
Similarly, $f(132)=0$, so $f(g)=0$ for any $g\in S_{3}$, if$f$ : $S_{3}\rightarrow \mathrm{Z}_{3}$ is a homomorphism.

On the whole, there is only the trivial homomorphism from $S_3$ to $\mathbb{Z_3}$: $f(g) = 0$.♥

  • 2
    $\begingroup$ Why are there backslashes everywhere? $\endgroup$ – Potato Feb 15 '14 at 18:16

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