For groups $G,\bar{G},H$; $H$ abelian; and $\phi:G\to\bar{G}$, a homomorphism, if $H≤G$ with $\ker(\phi)⊆H$, then $∀g∈G,∀h∈H,ghg^{-1}∈H$.

Question

I have the following question:

Let $G ,\bar{G}$ and $H$ be groups, suppose $H$ is abelian. Suppose that $\phi: G \rightarrow \bar{G}$ is a homomorphism. If $H \leq G $ such that $\ker(\phi) \subseteq H$, then prove $\forall g \in G, \forall h \in H, ghg^{-1}\in H$.

My attempt at the above question is that the statement is false. The counter example I have is:
Take $G = \bar{G} = S_3, H = \{e, (1~2)\}$. So $H$ is abelian. $\phi: S_3 \rightarrow S_3 $ where $\phi(\alpha) = \alpha, \alpha \in S_3$. Then $\ker(\phi) = \{e\} \subseteq H \leq G$. But $(1~2~3)(1~2)(1~2~3)^{-1} = (2~3) \not\in H$.

Is my example correct? Or did I do something wrong? Because I think the statement is suppose to be true.

Answer 1

If so, then any abelian subgroup would be normal. That would be a surprise. And indeed counterexamples abound. For instance, any symmetric group ($n\ge3$), has a non-normal subgroup which is abelian.

(Note: as @lulu remarked, the homomorphism barely comes into play, since you can always take an injective one, including, say, the identity)