# Let S be the subset of group G that contains identity element 1?

1. Sep 4, 2014

### Murtuza Tipu

1. The problem statement, all variables and given/known data

Let S be the subset of group G that contains identity element 1 such that left co sets aS with a in G, partition G .Probe that S is a subgroup of G.
2. Relevant equations

{hS : h belongs to G } is a partition of G.

3. The attempt at a solution
For h in S if I show that hS is S then that would imply that S is closed.
Now hS is a partition of S and contains h since 1 is in S and h is in S also.Hence h belongs to S intersection hS.
More over both these partition or two partition are disjoint or equal sets.
Hence h=hS which says that S is closed.

2. Sep 4, 2014

### Fredrik

Staff Emeritus
It was difficult to follow your argument, but I think you got it right. This is assuming that you meant to write S=hS where you wrote h=hS in the last line.

My version:
We will prove that S is closed under multiplication. Let $a,b\in S$ be arbitrary. Since $ab\in aS$, it will be sufficient to prove that $aS=S$. Since $e\in S$, we have $a=ae\in aS$. By assumption, we also have $a\in S=eS$. Since aS and eS are both cells of the partition, they are either equal or disjoint. So the fact that they both contain $a$ implies that $aS=eS=S$.

3. Sep 5, 2014

### Murtuza Tipu

The same I can show that it contains the inverses. WHy aren't they partitions of G?? As  ,h belongs to G , hS is a left coset.