Why is the sum of all the elements in a subgroup generated by $10$ of $\Bbb Z_p$ is divisible by $p$ Consider the subgroup $\langle 10\rangle $ of  $\mathbb{Z}_p$. Let the order of the subgroup $\langle 10\rangle $ be $d$.
So $$\langle 10\rangle= \left\{10, 10^2, 10^3, \dots, 10^{d-1}, 10^{d} = 1\right \}.$$
Now is it true that sum of all the digits of $\langle 10\rangle $ is divisible by $p$? If so, then how is it true.
I have verified the same for $p=13$ where order of $10$ modulo $13$ is $6$. We have $$10+10^2+10^3+10^4+10^5+1 = 111111$$ and $$13 \times 8547 = 111111.$$
I am not able to see how this can be proved. Kindly help me out.
Thanks in advance!
Update:
Let $\langle 10\rangle $ be a subgroup of $\Bbb Z_p$ with ${\rm ord}(\langle 10\rangle) = n$ Thus if $x_1, x_2, x_3, \dots x_n$ are the elements of the subgroup $\langle 10\rangle $ then $x_{i}^n \equiv 1 \pmod{p}$ where $1 \le i \le n$. So, $x_i$'s are the $n^{th}$ roots of unity modulo $p$.
So, $x^n-1 = (x-x_1)(x-x_2)(x-x_3)\dots(x-x_{n-1})(x-x_n)$.
Now the coefficient of $x^{n-1}$ in $x^n-1$ is $0$. So $x_1+x_2+x_3+\dots+x_{n-1}+x_n \equiv 0 \pmod{p}$ This is where I am confused. How can I write $x_1+x_2+x_3+\dots+x_{n-1}+x_n \equiv 0 \pmod{p}$ ? For this to be true I must have $x^n-1 \equiv 0 \pmod{p}$. Right?
So, the sum of $x_i$'s which are the elements of $\langle 10\rangle $ is congruent to $0$ modulo $p$.
 A: Your difficulty is that the problem contains a red herring: This is true for any subgroup of $(\Bbb{Z}/(p))^*$ other than the identity subgroup $<1>$. The easiest way to see this is to remember that any subgroup is cyclic and thus is generated by a root of unity.  Thus, if $\xi_1,...,\xi_n$ ($n>1$) are the elements of the subgroup, we have $x^n-1=(x-\xi_1)\cdot\cdot\cdot (x-\xi_n)$.  (We know this because each of the $\xi_i$ satisfy $\xi_i^n=1$, and as there are $n$ of them, they are precisely the $n$ roots of $x^n-1$.)  But the coefficient of $x^{n-1}$ in this polynomial is both $0$ (since $n>1$) and also $-(\xi_1+\cdot\cdot\cdot +\xi_n)$.   Thus, the sum of the $\xi_i$ is $0$ in $(\Bbb{Z}/(p))$, i.e., it is divisible by $p$.
Note that this uses the well-known fact that $(\Bbb{Z}/(p))$ is a field.  Also, note that to solve your problem, you don't even need the fact that every multiplicative subgroup of a finite field is cyclic, since $<10>$ is obviously cyclic.
A: There is a symmetry you can exploit!  Let $S$ be your sum.
If you multiply it by 10, you will get the same sum modulo $p$; the only effect is to permute the order in which powers of 10 appear.
This means that $10 S \equiv S \pmod{p}$, which is $9 S \equiv 0 \pmod{p}$. If $p \neq 3$ this implies $S \equiv 0 \pmod{p}$ which is what we wanted to show.
If $p=3$ the claim is false: $10 \equiv 1 \pmod{3} $ so its order is 1 and 1 is not divisible by 3.
A: Hint:  It's well known that the $n $-th roots of unity add up to zero,  because  $1+\xi+\dots+\xi^{n-1}=\dfrac {\xi^n-1}{\xi-1}=0$.   But the elements of $\mathbb Z_p^*$ are $p-1$- th roots of unity (by Lagrange).
