In $Z_p\times Z_p$, if the product of two group subsets of order $p$ and contain the identiy is the whole group, then one of them is a subgroup?

Question

This should be elementary, and I figured it should be true, but I can't find a good argument on it:

Let $$G=\mathbb Z_p\times \mathbb Z_p,$$ where $p$ is a prime number. Let $S,H$ be subsets of $G$ with order $$|S|=|H|=p,$$ and $$e\in H,S$$ i.e., both contains the identity element. Let us write the elements in $H$ as $$H=\{h_1,h_2,h_3,\ldots, h_p\},$$ now, if $$\{h_1S,h_2S,h_3S,\ldots, h_pS\},$$ is a partition of $G$, i.e., $$G=HS,$$ does it follow that either $H$ or $S$ must be a subgroup of $G$?

Answer 1

This is a question I asked in 2016, a few years ago I got to know the answer and it is not so elementary as it appears to be, so let me post it here and close this question.

The answer is Yes, and it follows from a classical result of Hajos and Redei, which says that if $G$ is a finite Abelian group and $G=A_1\times\ldots\times A_d$ where $A_1,\ldots, A_d$ are normalized (i.e., contains the identity) subset of $G$, and the product is direct (i.e., each $g\in G$ can be uniquely written as $g=a_1+\ldots+a_d$ with $a_i\in A_i$), then at least one of $A_1,\ldots, A_d$ is a subgroup.

Answer 2

For $p=3$ a counterexample (using additive notation) is

$S=\{(0,1),(0,2),(1,1)\}$, $H=\{(1,0),(0,2),(2,1)\}$.