# If $S$ and $T$ are nonempty subsets of a finite group $G$, then either $G=ST$ or $|G|\geq |S|+|T|$.

Exercise 2.29 from Rotman's book An Introduction to the Theory of Groups.

(H.B.Mann) Let $G$ be a finite group, and let $S$ and $T$ be (not necessarily distinct) nonempty subsets. Prove that either $G=ST$ or $|G|\geq |S|+|T|.$

I would appreciate any help in order to solve this question.

P.S. I've deleted the previuos post with the same question, because it was confuse. I am sorry for the inconvenience this might cause you.

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By deleting the previous post, you also deleted the comments, which should have helped you solve the problem! – Derek Holt Jan 29 '12 at 19:25

## 1 Answer

Here's a proof sketch:

Assume that $|S| + |T| > |G|$. We want to show that $G = ST$.

Fix $x \in G$, and focus the attention on the sets $S$ and $U = \{ x t^{-1} : t \in T \}$. First, these sets have cardinalities $|S|$ and $|U| = |T|$ respectively. At the same time, they are both subsets of $G$, whose size is strictly smaller than the sum of their cardinalities. This implies that $S$ and $U$ have a nonempty intersection. Can you take it from here?

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