# Determine the kernel of a homomorphism $θ: S_3 \rightarrow G$?

Let $G$ be a group and let $θ: S_3 \rightarrow G$ be a homomorphism such that $θ((1 2)) = e_G$.

I know that $θ((12)$ is in the kernel, as it gives me the identity. The same could be said about $θ(1)$ because $θ((1))θ((12)) = θ((1)(12)) = θ(12) = e_G$.

However there are supposedly more elements from $S_3$ that are in the kernel as well, and I'm not sure how to find them. I've tried using the homomorphism property that $θ(x)θ(y) = θ(xy)$, but that's only helped me find that $θ(1)$ is in the kernel.

• The kernel is a normal subgroup, so it's not only closed under multiplication, it's also closed under conjugation. What are the conjugates of $(12)$? – Qiaochu Yuan Oct 25 '17 at 5:13
• Do you mean the kernel is a normal subgroup of $S_3$? – Chris Oct 25 '17 at 5:44
• Yes. More generally, the kernel of any homomorphism $f : G_1 \to G_2$ is a normal subgroup of $G_1$ (and conversely every normal subgroup of $G_1$ is the kernel of some homomorphism to some other group, although you don't need to know this for this problem). – Qiaochu Yuan Oct 25 '17 at 5:48
• Ah ok. That's really useful to know and I think I'm able to solve the problem now. Thank you! – Chris Oct 25 '17 at 5:54

Hint: The Kernel is in fact a normal subgroup so try computing if $\langle(1,2)\rangle$ is normal and if not, what the smallest normal subgroup containing it has to be (which shouldn't be too hard since $S_3$ is a pretty small group.)