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Here I use the term coset more loosely to mean the sets obtained by composition with any subset (not necessarily a subgroup).

Let $G$ be a group and $U$ a subset. If $U$ is a normal subgroup then its cosets satisfy $gUhU$=$ghU$. I call this the "natural" composition of cosets (which is well defined for a normal subgroup).

Is the converse statement also true? I.e. if the cosets of a subset, $U$, satisfy $gUhU$=$ghU$ is $U$ necessarily a normal subgroup?

I've already attempted to solve this. It seems that one need only show $uU=U$. After which it is easy to show $U$ partitions $G$ and must therefore be a subgroup. It then follows by the "natural" composition condition that $U$ is normal.

Thanks in advanced for any help.

Addendum:

Here is the full proof:

For infinite groups, this is in general false. Take a submonoid which is not a subgroup of a group (e.g. $(\mathbb{N}_0,+)$ in $(\mathbb{Z},+)$) and verify that it satisfies the natural composition.

For finite groups, this is true. Take a subset, $U$, of a finite group, $G$, which satisfies the natural composition. Since $gUhU$=$ghU$, setting $g=h=e$, one gets $UU=U$ and with this $uU\subseteq{}U$, $u\in{}U$.

Left composition (i.e. $\phi_g:h\mapsto{gh}$) is an isomorphism and, in particular, bijective. This implies $U$ partitions $G$, since for $gU\cap{hU}\neq{\emptyset}$ there exists $u,u'\in{U}$ such that $gu=hu'$ and as such $gU=g(uU)=(gu)U=(hu')U=hU$.

Finally, since $uU=U$, it must be that $e\in{U}$ and, consequently, $U$ contains inverses. Therefore $U$ is a subgroup. To show U is normal, notice that since $e\in{U}$, then $gUg^{-1}\subseteq{gUg^{-1}U}=U$.

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This is for finite $G$ and non-empty $U$; $e$ denotes the identity of $G$.

(a) $eU=U$.

(b) With $g=h=e$ we have $UU=U$.

(c) Let $u\in U$. Then $uU=U$. [$uv=uw$ implies $v=w$ and so use $uU\subseteq U$ and $|uU|= |U|$.]

(d) For some $f\in U$ we have $uf=u$.

(e) $f=e$. [$uf=u=ue$, pre-multiply by $u^{-1}$.]

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  • $\begingroup$ Sorry. I looked back at my work. I only need uU=U as you show. Thanks. Any ideas as to how the infinite case may be approached? $\endgroup$ – MrHolmes Mar 14 '18 at 18:45
  • $\begingroup$ @verret has put paid to that; although his c'example has the identity in $U$ which was what I was aiming for as per the original question. I still can't get a c'ex to $e\in U$. $\endgroup$ – ancientmathematician Mar 15 '18 at 7:45
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It does not hold in general. For example, if $G$ is an abelian group, then the defining equation becomes $UU=U$. Just take $U$ to be a sub-monoid that is not a subgroup. (This cannot happen in a finite group.)

(For example, take $G$ to be the integers with addition, and $U$ the non-negative integers.)

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  • $\begingroup$ Do you mean the non-negative integers? Since U+U=/=U (1 is not the sum of two positive integers). $\endgroup$ – MrHolmes Mar 15 '18 at 4:30
  • $\begingroup$ Right, I'll edit that. $\endgroup$ – verret Mar 15 '18 at 6:33
  • $\begingroup$ I wish I could choose both your answers since together they make up what for me would be the "complete" solution. $\endgroup$ – MrHolmes Mar 15 '18 at 13:58

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