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We say that a subgroup $H\lhd G$ is normal iff it is closed under conjugation by $g \in G$, which implies that for a normal subgroup $gH = Hg$

After reading this definition I wondered, under what conditions (if any) does a subgroup $H\subset G$, $H\not\lhd G$ have the following property:

$\forall g\in G\exists a\in G (gHa^{-1} = H)$ or $...(gH = Ha)$

My intuition tells me that under no conditions does a subgroup $H\not\lhd G$ satisfy this condition, however, I am having a hard time proving my intuition, and thus have begun to doubt it. So I wonder if anyone can confirm my intuition and help me to get going on proving it, or can reject my intuition.

Thank you very much.

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up vote 5 down vote accepted

This condition implies that $H$ is normal. IOW, your intuition is correct. The underlying reason is that if two left (or right) cosets of the same subgroup intersect, then they necessarily are the same subset: L) If $x\in yH$, then $xH=yH$. R) If $x\in Hy$, then $Hx=Hy$.

Assume that $gH=Ha$. Observe that the element $g\in gH$, because $1\in H$. Therefore $g\in Ha$. By the part R) above we may conclude that $Hg=Ha$. In other words, if the condition $gH=Ha$ holds for some $a\in G$, then it will also hold for $a=g$.

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Ah, so basically if $gH = Ha$ then $g\in gH \implies g\in Ha \implies Ha = Hg \implies gH = Hg \implies gHg^{-1} = H \implies H\lhd G$. Thank you very much. –  Deven Ware Sep 10 '11 at 6:44
    
@Deven: Correct. –  Jyrki Lahtonen Sep 10 '11 at 6:46
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