Finite group and its proper subsets

Question

Let G be a finite group and S, T are two non empty subsets of G.
If G#(not equal ) ST . Show that |G|>=|S|+|T| . I try it

But not succeed

Answer 1

If $G \ne ST$ there is a $g \in G; g \not \in ST$.

So for every $s\in S$, then $s^{-1}g\not \in T$. (Other wise $s*(s^{-1} g) = g\not \in ST$ but $s \in S$ so $s^{-1} g\not \in T$

These $s^{-1}g$ are distinct (i.e. if $s,t \in S; s \ne t$ then $s^{-1}g \ne t^{-1}g$).

So if $T^c = \{h \in G| h \not \in T\}$ then for every $s \in S$, $s^{-1}g \in T^c$ so $|S| \le |T^c|$.

So $|G| = |T| + |T^c| \ge |T| + |S|$.