Is it necessary to determine a definite homomorphism here?

Let $S_{3}$ be the symmetric group of degree $3$ and let $\Bbb Z_{15}$ be the cyclic group of order $15.$ Given that $\theta: S_3 \to\Bbb Z_{15}$ is a group homomorphism, what can you say about $\ker\theta$ and ${\rm im}\theta= \theta(S_{3})$? Clearly justify your answer.

What I've done so far:

The image of $S_{3}$ under $\theta$ is a subgroup of $\Bbb Z_{15}$, by Lagrange theorem, the order of this subgroup must be $1,3,5$, or $15.$ It can’t be $15$ since otherwise the mapping is not well-defined.

For the $\ker\theta$, possible order is $1, 2, 3, 6.$ It cannot be $1$ cause then the mapping is injective and $\Bbb Z_{15}$ has no subgroup of order $6$.

Also $|S_{3}|/|\ker \Phi| = | \Phi (S_{3}) |$ tells us that possible sizes for the image are $1$ and $3$ implying the possible values for $|\ker \Phi|$ are $6$ and $2$.

Is it possible to narrow down two remaining candidates for a homomorphism between the two groups?

Solution 1:

From what you have already, possible sizes of images of these homomorphisms are $1,3$ (this can be proved easier because by Lagrange these sizes must divide both $6$ and $15$). There is of course a homomorphism with image of size $1$: the one that maps everything to $0$. If there was a homomorphism with image of size $3$, then its kernel $H$ would have order $2$. But $S_3$ does not have normal subgroups of order $2$, so this option is not possible.