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Let $G$ be a group and $H$ a subgroup of $G$.

I have proven in an exercise that $gH = Hg$ for all $g \in G$ implies $(g_1H)(g_2H) = (g_1g_2)H$, $g_1, g_2 \in G$ where $(g_1H)(g_2H) = (xy | x\in g_1H, y\in g_2H)$.

Can one prove $(g_1H)(g_2H) = (g_1g_2)H$, $g_1, g_2 \in G$ if and only if $gH = Hg$ for all $g \in G$ ?

I have been sitting trying to come up with a counter example. Can anyone help ? Thanks in advance.

/Nicolas

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Yes, it is. Put $g_2=g_1^{-1}$, then $g_1Hg_1^{-1}H=H$, hence $g_1Hg_1^{-1}\subseteq H$.

Addendum: More details:

1) Putting $g_1=g_2=1$ we get $H$ is a subgroup (so the condition "$H$ is a subgroup" is unnecessary).

2) Put $g_2=g_1^{−1}$, then $g_1Hg_1^{−1}H=H$. Hence $g_1Hg_1^{−1}\subseteq H$, i.e. $g_1H\subseteq Hg_1$. From here Hg_1^{−1}\subseteq g_1^{−1}H. This is true for all $g_1$, so $xH\subseteq Hx$ and $Hx\subseteq xH$ for all $x$.

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  • $\begingroup$ Sorry, are you proving $(g_1H)(g_2H) = (g_1g_2)H$, $g_1, g_2 \in G$ implies $gH = Hg$ for all $g \in G$ ? - or is this some counter example ? $\endgroup$ – Shuzheng Oct 11 '13 at 20:48
  • $\begingroup$ It follows from $gHg^{-1}\subseteq H$ that $gH\subseteq Hg$ and $Hg^{-1}\subseteq g^{-1}H$ for ALL $g\in G$. So $xH=Hx$ for all $x\in G$. $\endgroup$ – Boris Novikov Oct 11 '13 at 20:56
  • $\begingroup$ But as this proof shows, $gHg^{-1} \leq G$ is just a subgroup in $G$ not neccesarily a subset of $H$ for arbitary $g \in G$ ? proofwiki.org/wiki/Conjugate_of_Subgroup_is_Subgroup $\endgroup$ – Shuzheng Oct 12 '13 at 9:11
  • $\begingroup$ But your condition implies that $H$ is a subgroup: take $g_1=g_2=1$. $\endgroup$ – Boris Novikov Oct 12 '13 at 9:35
  • $\begingroup$ Sorry, my mistake. I've edited. What i'm interested in is: $(g_1H)(g_2H)=(g_1g_2H)$ implies $\forall g \in G (gH = Hg)$, then I have proven both ways in the bicondition. Also, thanks for your help so far Boris. $\endgroup$ – Shuzheng Oct 14 '13 at 11:04

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