# Subset is normal subgroup iff its cosets inherit natural composition

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.

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 $$gUg^{-1}\subseteq{gUg^{-1}U}=U$$.

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.)

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

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 $u^{-1}$.]

• 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? Mar 14 '18 at 18:45
• @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 $e\in U$. Mar 15 '18 at 7:45