Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $a=g$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.