Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose G is a group and for every $H_1,H_2\le G$, $H_1H_2=H_2H_1$. Is $G$ abelian?

share|improve this question

1 Answer 1

up vote 12 down vote accepted

Not necessarily.

Let $H_1$ and $H_2$ be subgroups of $G$. Then $H_1H_2 = H_2H_1$ if and only if $H_1H_2$ is a subgroup.

Thus if for example every subgroup of $G$ is normal, then $H_1H_2$ is always a subgroup. The quaternion group of order $8$ is a nonabelian group where every subgroup is normal.

share|improve this answer
    
Oh my, yes! Can you clarify the blunder i commited in my answer! Thank you. :) –  Swapnil Tripathi Jul 13 at 13:19
3  
@SwapnilTri: Your mistake was thinking that $HK = KH$ implies that $hk = kh$ for all $h \in H$ and $k \in K$. This is not true. What $HK = KH$ implies is that for all $h \in H$ and $k \in K$, we have $hk = k'h'$ for some $k' \in K$, $h' \in H$ (not necessarily $h = h'$, $k = k'$) –  Mikko Korhonen Jul 13 at 13:23

Your Answer

 
discard

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.