Can a Finite Group Be Expressed as the Product of Two Subsets?

[boombaby]

Homework Statement

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|

Homework Equations

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...

 

[Dick]

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

 

[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.

 

[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|.

 

[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!

 

[Dick]

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|.

Nice!