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.)

  • $\begingroup$ 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

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 $u^{-1}$.]

  • $\begingroup$ 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 $e\in U$. $\endgroup$ Mar 15 '18 at 7:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.