# Homomorphism from $C_4$ to $S_3$

Practicing for my upcoming exam and I have no idea how to do this:

Let $C_4$ be the cyclic group of order $4$ and $S_3$ the third symmetric group.

(a) Give an example of a homomorphism from $C_4$ to $S_3$ whose image contains more than just the identity element of $S_3$.

(b) Is it possible for such a homomorphism to be 1 to 1? Why or why not?

For (a), I know the subgroup of $Q = \{(), (1 2 3), (1 3 2)\}$ of $S_3$ is isomorphic to $C_3$ but I don't know how to make a homomorphism from $C_4$.

• Do you know of any subgroups of $S_3$ that are isomorphic to $C_2$? Does that help? – rogerl Dec 13 '16 at 16:42
• $C_4$ is cyclic, so has a generator. What can you say about the homomorphic image of that generator? – Eric Towers Dec 13 '16 at 16:43

(a) you can consider the subgroup $C_2 = (12)$ of $S_3$ and the projection $\pi$ of $C_4$ on $C_2$ given by $\pi([n]_4) = [n]_2$.
(b) In order to find a contradiction, suppose that there is an injection of $C_4$ in $S_3$. This means that there is a copy of the cyclic group $C_4$ in $S_3$. But you can check by hand that this is false because there is no-element of $S_3$ with order bigger or equal to $4$.
• OK I see! Can you explain what the projection $\pi$ does? Say $C_4 = <a>$. With $\psi : C_4 -> C_2$ would it be $\psi (0) = 0, \psi (a) = a, \psi (a^2) = 0, \psi (a^3) = a$ – Uq'''12wn1F12u2x3uW31H1JBk9m Dec 13 '16 at 17:39
• Yes. In other words, even powers of $a$ are sent to $0$ and odd powers to $1$. – Evian Dec 13 '16 at 18:43
• Where $C_2$ is identified with $\mathbb{Z}/2\mathbb{Z}$. – Evian Dec 13 '16 at 18:43