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I denote by $\leq$ subgroup. If $H\leq G$ and $g \in G$, how does it follow that if $gHg^{-1} \leq H$ then $H \leq g^{-1}Hg$?

I look first at an element $gh_ig^{-1}\in gHg^{-1}$, $h_i \in H$. We have that $gh_ig^{-1} = h_j$, for a unique $h_j \in H$ ($h_j$ is unique because $ghg^{-1} = gh'g^{-1}$ implies that $h=h'$), which gives $h_i = g^{-1}h_jg$. But then we get an equality, $H = g^{-1}Hg$, since every element $h_i \in H$ is also in $g^{-1}Hg$, and $|H| = |g^{-1}Hg|$.

What am I doing wrong?

jock.marshall
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    Your argument is OK if $H$ is finite, but it does not work if $H$ is infinite, and there are examples in which it is not true. See https://math.stackexchange.com/questions/107862/ – Derek Holt Jan 26 '18 at 20:37
  • I meant to say that your proof that $gHg^{-1} \le H \Rightarrow H \le g^{-1}Hg$ is fine, but this does not imply that $H = g^{-1}Hg$ when $H$ is infinite. – Derek Holt Jan 26 '18 at 21:02

1 Answers1

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Notice that $H = g^{-1}Hg$ is just a special case of $H \leq g^{-1}Hg$.

Nonetheless, you could argue as follows:

$$gHg^{-1} \leq H \implies gH \leq Hg \implies H \leq g^{-1}Hg$$

RGS
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  • I do not see why the implications follow directly. Could you clarify that? – jock.marshall Jan 26 '18 at 20:40
  • @jock.marshall I am just doing what you did, but instead of doing it element by element, I am multiplying the hole set at once $ gHg^{-1} \leq H \implies gHg^{-1}g \leq Hg \iff gH \leq Hg $ – RGS Jan 27 '18 at 07:07