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Let $h:G \rightarrow G$ be a homomorphism, $G$ a group. Prove that $h^k(G)=\lbrace g \in G \mid ghg^{ -1}=h(g) \rbrace$
Let $G$ be a group. Let $h:G \rightarrow G$ be a homomorphism. Prove that $h^k(G)=\lbrace g \in G \mid ghg^{ -1}=h(g) \rbrace$.
I’ve managed to prove that the set above is closed under multiplication, and is a subgroup of $G$. But I’m not sure how to prove that $h^k(G)=h(h(…h(G))))$
A:
If $n$ is the order of $h$, $h^n(G)$ is the subgroup $\{g\in G:h^ng=h(g)\}$ and obviously $h^{nk}(G)\subseteq h^n(h^k(G))=h^n(\{g\in G: ghg^{ -1}=h(g)\})$.
If $g,h\in G$, then $g^{ -1}h=hg^{ -1}\in G$ and so $ghg^{ -1}=h(g)$ iff $g\in h^k(G)$ (since $g^{ -1}\in h^k(G)$). Therefore $G\subseteq h^k(G)$.
If $ghg^{ -1}=h(g)$ then $g^{ -1}h=h^{ -1}g=h$ and so $gh=hg$ and so $h$ is injective.
Let us give a very elementary proof of the result (now called the
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