Let $S$ be a subset of a group $G$ that contains the identity element $1$ and such that the left cosets $aS$ with $a$ in $G$, partition $G$.Prove that $S$ a is a subgroup of $G$.

My try:

For $h$ in $S$, If I show that $hS=S$, then that would imply that $S$ is closed.

Now $hS$ is a partiton of $S$ and contains $h$ since $1$ is in $S$. Also $h$ is in $S$. Hence $h \in S\cap hS$. Moreover both of these are partitions and two partitions are either disjoint or equal. Hence $S=hS$ which says that $S$ is closed.

Does this seem alright??


  • $\begingroup$ $hS$ is a translate of $S$, not a partition of $S$. One doesn't call $S$ and $hS$ partitions, one calls them cells or parts. Anyway, you've shown that $S$ is closed under multiplication and contains the identity - what about closure under inverses? $\endgroup$ – whacka Aug 22 '14 at 2:12
  • $\begingroup$ @whacka The same I can show that it contains the inverses. WHy aren't they partitions of $G$?? As $h \in G$ , $hS$ is a left coset. $\endgroup$ – tattwamasi amrutam Aug 22 '14 at 2:19
  • $\begingroup$ A partition of $G$ is a collection of disjoint subsets of $G$. Each of the subsets is not a partition. $\endgroup$ – Quang Hoang Aug 22 '14 at 2:21
  • $\begingroup$ @Quang... The question says they partition $G$. Else I cann't argue that either they are disjoint or equal $\endgroup$ – tattwamasi amrutam Aug 22 '14 at 2:24
  • 1
    $\begingroup$ The problem says $\{hS:h\in G\}$ is a partition of $G$. We don't call a particular subset $hS$ a partition. $\endgroup$ – Quang Hoang Aug 22 '14 at 2:31

Let $x,y\in S$, and consider the sets $x^{-1}S$ and $y^{-1}S$, now remark that $1\in x^{-1}S\cap y^{-1}S$ as $\{aS\}_{a\in G}$ is a partition of $G$, then $x^{-1}S=y^{-1}S$, since $x^{-1}\in x^{-1}S$ we get $x^{-1}\in y^{-1}S$, there exists $s\in S$ such that $x^{-1}=y^{-1}s$, so $yx^{-1}=s\in S$. It follow that $S$ is a subgroup of $G$.

  • $\begingroup$ what else do u think I did?? $\endgroup$ – tattwamasi amrutam Aug 22 '14 at 2:40
  • $\begingroup$ @tattwamasiamrutam You actually didn't show the same thing. Hamou shows that $S$ satisfies the subgroup criterion. All your argument does is establish that $S$ is closed. When $G$ is infinite, this does not guarantee that $S$ is a subgroup. For instance, take $G= \mathbb Z$ and $S = \mathbb N$. Though I will grant that in the comments you say you also show that $S$ is closed under inversion. $\endgroup$ – Shawn O'Hare Aug 22 '14 at 3:22
  • $\begingroup$ @ShawnO'Hare.. I meant by the same arguement, I can show that the inverses belong to the set. $\endgroup$ – tattwamasi amrutam Aug 22 '14 at 3:45
  • $\begingroup$ @Hammou Do you use that to conclue that $x^{-1} \in S$ ? $\endgroup$ – Carpediem Sep 1 '14 at 16:23
  • $\begingroup$ @Hamou Your last formula should be $yx^{-1}=s\in S$ (which doesn't ruin the argument). $\endgroup$ – egreg Sep 1 '14 at 16:33

It was difficult to follow your argument, but I think you got it right. My try: We will prove that S is closed under multiplication. Let a,b∈S be arbitrary. Since ab∈aS, it will be sufficient to prove that aS=S. Since e∈S, we have a=ae∈aS. By assumption, we also have a∈S=eS. Since aS and eS are both cells of the partition, they are either equal or disjoint. So the fact that they both contain aimplies that aS=eS=S.


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