Sum of all the elements of the coset of a subgroup A while ago I posted this question. I understood the case for the subgroup $\langle 10 \rangle$ in $\mathbb{Z}_p^*$. But, I think the same result holds for the cosets of the subgroup $\langle 10 \rangle$. So, my question is,

Why is the sum of all the elements of a coset of the subgroup $\langle 10 \rangle$ in $\mathbb{Z}_p^*$ is divisible by $p$ ?

 A: Let $a_1,\dots,a_k$ be the elements of a coset of the multiplicative group of $\mathbb Z_p$ of size other than $1$, assume the coset can be written as $a_1H$ where $H$ is a subgroup.
If we take all of the elements in the coset and multiply them by $h\in H$ with $h\neq 1$ we get the elements $a_1,\dots,a_k$ in a different order.
It follows $(a_1+ \dots + a_k) = h(a_1+ \dots + a_k)$, and so the sum is $0\bmod p$.
A: I like Onir's answer better since it proves the result directly, but since you already understoond the answer to your previous question I will show how this result follows easily from that one.
Let $h_1, h_2, \ldots, h_n$ be the elements of the subgroup $H = \langle 10 \rangle$, with $h_1 = 1$. Following Onir's notation the elements $a_1, \ldots, a_n$ of $H$ are all of the form $a_1 = a_1h_1, a_2 = a_1h_2, \ldots, a_n = a_1h_n$.
Now $a_1 + \ldots + a_n = a_1h_1 + \ldots + a_1h_n = a_1(h_1 + \ldots + h_n)$.
From your previous question we know that $(h_1 + \ldots + h_n)$ is divisible by $p$, so it follows that $a_1 + \ldots + a_n = a_1(h_1 + \ldots + h_n)$ is divisible by $p$ as well.
