Conjugate subgroup strictly contained in the initial subgroup?

Let G be a group, HG a subgroup and aG an element of the group. Is it possible that aHa1H, but aHa1H?

If H has finite index or finite order, this is not possible.

Answer

Consider the group of matrices G={[xy01]:xQ×,yQ}=AGL(1,Q) and its subgroup H={[1y01]:yZ}Z and of course the single element a=[2001] A direct calculation gives aHa1={[12y01]:yZ}<H is a proper subgroup of H.

Similar issues showed up in this question.

Attribution
Source : Link , Question Author : Sasha , Answer Author : principal-ideal-domain

Leave a Comment