Can a non-trivial subgroup have negative elements

Question

Lets say we have a cyclic group $\Bbb Z_{13}$ and we have found the following subgroup $\langle 3\rangle$.

Now we calculate the following:

\begin{align}3^1 \pmod{13} &= 3,\\ 3^2 \pmod{13}&= 9 \pmod{13} \equiv -4 \pmod{13} = -4,\\ 3^3 \pmod{13}& = 3\times(-4) \pmod{13} = -12 \pmod{13} = 1.\end{align}

As you can see the inverse of $-4$ is $9$, but my question is the final set of elements going to be like this:

$\langle 3\rangle = \{3,-4,1\}$ or like this: $\langle 3\rangle = \{3,9,1\}$.

Any help would be grateful.

Answer 1

The group ${\Bbb Z}_{13}^*$ is a subset of the ring ${\Bbb Z}_{13}=\{0,1,\ldots,12\}$ whose elements are the residue classes modulo 13. Here $-4\equiv 9\mod 13$ and so $-4$ and $9$ represent the same element.

Answer 2

Actually the residue classes partition the set of integers. When it comes to $\Bbb Z_{13}$, there are thirteen residue classes. Every integer is in exactly one of these classes. Which representative you use (for each equivalence class) is up to you. You could always pick a negative integer if you like.

Typically you will see $\Bbb Z_{13}=\{0,1,\dots,12\}$. Here it is understood, for instance, that $\bar2=\{2+13k: k\in\Bbb Z\}$. That is, these numbers are all identified. So, again, the elements of $\Bbb Z_n$ are actually equivalence classes. Also, the $"\bar{}"$ is often just skipped.