# Finite group G=ST

1. Oct 28, 2008

### boombaby

1. The problem statement, all variables and given/known data
let G be a finite group, and let S and T be nonempty subsets.Prove either G=ST={st|s is in S, t is in T} or |G|>=|S|+|T|

2. Relevant equations

3. The attempt at a solution

So it is to prove G=ST, if |G|<|S|+|T|, which means also the intersection of S and T is nonempty (Note two smaller subsets can also have nonempty intersection, so |G|<|S|+|T| should have more properties than nonemptiness, but I fail to find one ). $$ST\subset G$$ is obvious. I want to prove the other direction by saying any g in G can be represented as g=st for some s,t, or $$s_{i}T$$ covers G.
This is what I 've done:
Let x be an element in both S and T. then x^2 is in ST. But x^2 can be outside S and T, So it is possible that $$x^{3}\notin ST$$, which seems to become useless...
Any hint would be appreciated....

Last edited: Oct 28, 2008
2. Oct 29, 2008

### Dick

This is really an annoying question. Did you get it? I didn't.

3. Oct 30, 2008

### boombaby

No. The only thing I proved is that it is true when G is abelian. But..when G is nonabelian, it is a headache.

4. Oct 30, 2008

### morphism

Let S={s_1,...,s_|S|} and T={t_1,...,t_|T|}. Suppose that there is some x in G but not in ST. Then x != s_i t_j. In particular, (s_i)^(-1) x is never in T. Use this to deduce that |S| <= |G|-|T|.

5. Oct 30, 2008

### boombaby

Ah, yea, it is true!
I only tried to make a 1-1 correspondence from those elements g in both S ant T to something outside S and T and got stuck since g^(-1)x can possibly drop in S again.
Thanks!

6. Oct 30, 2008

Nice!