## Image of a subgroup under conjugation

Posted: May 16, 2010 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , ,

Problem. By giving an example, show that there exist a group $G, \ g \in G$ and $H < G$ such that $gHg^{-1} \subsetneq H.$

Solution. Let $G=\text{GL}(2,\mathbb{Q})$ and put $g=\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix} \in G.$ Now we can give a sequence of subgroups of $G$ having the required property: for a given $k \in \mathbb{N}$ we choose $H_k=\left \{ \begin{pmatrix} 1 & kn \\ 0 & 1 \end{pmatrix} : \ n \in \mathbb{Z} \right \}.$ It is easy to see that each $H_k$ is a subgroup of $G$ and $gH_kg^{-1} = \left \{ \begin{pmatrix} 1 & 2kn \\ 0 & 1 \end{pmatrix} : \ n \in \mathbb{Z} \right \} \subsetneq H_k.$