Strange result of divisibility

Question

I have noticed experimentally that the following question has a positive answer.

Let $p>5$ and $H$ be a subgroup of $(\mathbb Z/p\mathbb Z) ^*$, with $a\in H$ and $a>2$.

Is it true that $$(a-1)\; | \; p\sum\limits_{h \in H} (-h/p \mod a) ?$$

Answer 1

I assume that in the calculations, you are identifying the elements of $\mathbb Z/p\mathbb Z$ with $\{0,\dots,p-1\}\subseteq\mathbb Z$, and likewise, that the $\bmod a$ operation takes values in $\{0,\dots,a-1\}$. Then the result follows from

Lemma: If $a,b>0$ are coprime and $-a<h<b$, then $$a(ha^{-1}\bmod b)-b((-hb^{-1})\bmod a)=h.$$

Proof: By negating $h$ and swapping the roles of $a$ and $b$ if necessary, we may assume $h\ge0$. Let $u=(ha^{-1}\bmod b)$. Then $au=h+bv$ for some $v$, and since $0\le au<ab$ and $0\le h<b$, we have $0\le v=\lfloor au/b\rfloor<a$. Also $bv\equiv-h\pmod a$, thus $v=((-hb^{-1})\bmod a)$. QED

Consequently, $$\begin{align*} p\sum_{h\in H}((-hp^{-1})\bmod a) &=a\sum_{h\in H}(ha^{-1}\bmod p)-\sum_{h\in H}h\\ &=(a-1)\sum_{h\in H}h, \end{align*}$$ as $\{ha^{-1}\bmod p:h\in H\}$ is just another enumeration of $H$.