Let G be a finite group, and let S and T be (not necessarily distinct) nonempty subsets. prove that either $G=ST$ or |$G|\geq|S|+|T|$
That's my thougt, I am thinking suppose $G$ does not equal to $ST$, then there exist a $h$ in $G$, but $h$ $\notin ST$. Thus, I suppose T={$x_1,x_2......x_N$}and $S=\{{t_1,t_2,.......,t_m,x_{k+1},......,x_{N}}\}$, where $x_k+1,...x_N$ are the common elements between two sets S and T. So, I want to construct two totally different sets A and B so that $|A|+ |B| > |G|$, firstlt,if $1$ is not in $S$ and $T$, then suppose T=A, and then suppoese B={$h,t_1,...,t_m, hx_{k+1}^{-1},...,hx_N^{-1}$}. I think if we can prove B$\neq$A(orT) and B$\neq$S, then I can know the question will be proved since $|A|+|B|>|G|$, but I only can prove B$\neq$S, I don't know how to prove B$\neq$T. Maybe my thoughts are wrong, can someone tell me how to solve this question? I think my methods are not correct, I hope someone can help me solve this question.
it is enough to show that when $|S|+|T|>|G|$ then $TS=G$.
Define $S^{-1}=\{s^{-1}| s\in S\}$ and let $g\in G$. Notice that $|gS^{-1}|=|S^{-1}|$ thus $gS^{-1}$ and $T$ must intersect with each other.
Thus, $gs^{-1}=t\implies g=ts \implies G=TS$ .