Let $f: S_{3} \to \mathbb Z_{6}$ be a group homomorphism. Then the number of elements in $f(S_{3})$ is...? Let $f: S_{3} \to \mathbb Z_{6}$ be a group homomorphism. Then the number of elements in $f(S_{3})$ is
$a$.  1
$b$.  1 or 2
$c$.  1 or 2 or 6
$d$.  1 or 2 or 3.
Since $G/k(f)  ≈ f(G)$
So, possible $f(G)$ is isomorphism to $S_{3}$ or $Z_{2}$ or ${e}$. So possible orders are $6,2$ and $1$. So, d is incorrect. Now, how to choose correct option?
 A: Hints: 


*

*$S_3$ is non-abelian and $\Bbb Z_6$ is abelian. Hence  $|f(S_3)|\neq
   6$.

*$\text{ker}(f)$ is a normal subgroup and the only proper non-trivial normal subgroup of $S_3$ is of order $3$.
A: If you know what Normal Subgroups are (I suppose since you are talking about quotients), this is a possible solution to your problem.
First of all, I make the following observations:


*

*The kernel of a homomorphism is a normal subgroup of the "domain" group.

*$\operatorname{S}_3$ has only three normal subgroups. In fact, they are $\{ \{1\}, \operatorname{A}_3, \operatorname{S}_3 \}$.

*As Thomas Shelby noted in the comments, the fact that $\operatorname{S}_3$ is not abelian gives the non existence of an isomorphism between these two groups.
Starting with the first and second observation, we know that if a homomorphism between these two groups exists, then the kernel must be one of the normal subgroups listed above.
The first isomorphism theorem will be useful, as you observed. The fact that $\operatorname{S}_3/\operatorname{Ker}(f)$ is isomorphic with $f(\operatorname{S}_3)$, gives $$\frac{|\operatorname{S}_3|}{|\operatorname{Ker}(f)|}=|f(\operatorname{S}_3)|.$$
Then, in a preliminary solution, $|f(\operatorname{S}_3)|$ can have a value of $6$, $2$, or $1$.
Now, $|f(\operatorname{S}_3)|=6$ is impossible since this means that the homomorphism is injective, but an injection between two sets with the same number of elementes is a bijection, and this will give us an isomorphism.
Sure, $|f(\operatorname{S}_3)|=1$ is possible, since this sais that $f$ send every element of $\operatorname{S}_3$ to the identity element of $\mathbb{Z}_6$, and this is a possible homomorphism.
To see that $|f(\operatorname{S}_3)|$ could take the value $2$ is a bit more complex. I will give you the definition of one homomorphism that verifies this, but I wont give you the details (prove it is, in fact, an homomorphism, that has kernel $\operatorname{A}_3$, etc..).


*

*Define $f : \operatorname{S}_3 \rightarrow \mathbb{Z}_6$ as $f(\sigma) := 0$ if $\sigma$ is even, and $f(\sigma) = 3$ if $\sigma$ is odd. About the definition of this homomorphism you must read Symmetric group - Transpositions, and what is a really useful observation, the comment and the answer made by mathcounterexamples.net about how to define this homomorphism when the image is a subset of the additive group $\mathbb{Z}_6$.


So the correct anwer to your qestion is b) 1 or 2.
A: As mentioned by Thomas Shelby, you can’t have $|f(S_3)| =6$ as $\mathbb Z_6$ is commutative while $S_3$ isn’t.
$\ker f$ is a normal subgroup of $S_3$. It can’t be the trivial subgroup $\{Id\}$ as this will force $|f(S_3)| =6$. The kernel can be $S_3$ itself and in that case $|f(S_3)| =1$ and $f(x) =1$ for all $x \in S_3$.
Now if $f$ is the signature, $\ker f $ is the subgroup generated by the $3$-cycle $(1 \ 2 \ 3)$ and $|f(S_3)| =2$. More precisely, $f(S_3)=\{0,3\} \simeq \mathbb Z_2$.
Finally the proper answer is b.
