Let $H\leq G$. If $g\in G$, show that $gHg^{-1}=\left \{ g^{-1}hg:h\in H,g\in G \right \}$ is a subgroup of $G$.

Question

Let $H$ be a subgroup of $G$. If $g\in G$, show that $gHg^{-1}=\left \{ g^{-1}hg:h\in H,g\in G \right \}$ is a subgroup of $G$.

Proof: Since $e,e^{-1}\in G$, then $e^{-1}he\in gHg^{-1}$, and so $gHg^{-1}\neq \varnothing$.

Let $g^{-1}h_{1}g,\;\; g^{-1}h_{2}g\in gHg^{-1}$. Since $H$ is a subgroup, then $h_{1}h_{2}^{-1},\;\; h_{2}^{-1}\in H$. Then $g^{-1}h_{1}g\cdot (g^{-1}h_{2}g)^{-1}=g^{-1}h_{1}g\cdot g^{-1}h_{2}^{-1}g=g^{-1}gh_{1}h_{2}^{-1}g^{-1}g=eh_{1}h_{2}^{-1}e=h_{1}h_{2}^{-1}\in gHg^{-1}$. Then $gHg^{-1} \leq G$. $\square$

Answer 1

The equality $g^{-1}h_{1}g\cdot g^{-1}h_{2}^{-1}g=g^{-1}gh_{1}h_{2}^{-1}g^{-1}g$ is wrong for non-abelian groups in general.