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.

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