Is the image of a cyclic group under a homomorphism $\phi : G → G'$ always
a normal subgroup of $G'$?
I believe the answer is yes as $G$ is abelian (since it is isomorphic to $Z$ or $Z_n$). So $\phi(ab) = \phi(a) \phi(b) = \phi(ba) = \phi(b)\phi(a)$. Thus the image is also abelian. However, I am not sure if this proves it is normal as $G'$ is not necessarily abelian. Still, we pick $j \in \phi(G)$ and $g \in G'$. We want to show that $g^{-1}jg \in \phi(G)$. I am stuck on this step as I feel like I am missing some key property to make it all work. Any help is appreciated, thank you
Best Answer
There is an injective homomorphism $Z_2\to S_3$ whose image is not a normal subgroup of $S_3$.
To be concrete, map the identity element of $Z_2$ to the identity permutation $(1)(2)(3)$ and map the nonidentity element of $Z_2$ to the permutation $(12)(3)$. The image is $\{(1)(2)(3),(12)(3)\}$ is not a normal subgroup of $S_3$.
All that you can say about the image of a cyclic group $G$ under a homomorphism $G\to G^\prime$ is that the image is cyclic subgroup of $G^\prime$. However, cyclic subgroups are not necessarily normal. The first counterexample is the one I gave above where $S_3$ has a non-normal cyclic subgroup of order $2$ (in fact, there are $3$ such subgroups).