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So, I know that every abelian (commutative) group $G$ is such that, for any subgroup $H$, the left cosets of $H$ in $G$ are the right cosets. I guess this is true even if $G$ is not abelian, but $H$ is (but I'm not sure). Is this enough to characterise the subgroup as abelian, or are there examples of non-abelian subgroups with this property?

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up vote 5 down vote accepted

For a subgroup of a group, the condition of being abelian is neither necessary nor sufficient for being normal:

If $G$ is non-abelian, then $H=G$ is a normal but non-abelian subgroup. As another example, the alternating group is a normal but non-abelian subgroup of the symmetric group.

For a counterexample in the converse direction, consider the free group with two generators. The subgroup generated by one of the generators is abelian (isomorphic to $\mathbb Z$) but not normal. (This means that your guess above is wrong.)

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I would have given a smaller example of "abnormal" abelian subgoups, like the order 2 subgroups of S_3. – Pierre-Yves Gaillard Aug 29 '10 at 17:52

The left cosets of $H$ are the same as the right cosets of $H$ if and only if $H$ is a normal subgroup of $G$. There are non-abelian groups all subgroups of which are normal, for example the quaternion group of order $8$.

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They are called "Hamiltonian groups": – Pierre-Yves Gaillard Aug 29 '10 at 15:52
@Robin, the "if" direction is obvious, but (at least to me) the "only if" is not. Could you elaborate? – Qiaochu Yuan Aug 29 '10 at 16:20
Qiaochu Yuan: $gH=Hg\Leftrightarrow gHg^{-1}=H$ – Pierre-Yves Gaillard Aug 29 '10 at 16:39
@Pierre-Yves Gaillard: yes, that shows the "if" direction. My interpretation of the left cosets being "equal" to the right cosets is that they determine the same partition of G, and I do not see how this argument addresses the "only if" direction. – Qiaochu Yuan Aug 29 '10 at 16:41
Consider the partition given by the gH. Then gH is THE member of the partition containing g. (Similarly for the other partition.) – Pierre-Yves Gaillard Aug 29 '10 at 16:57

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