how to find the elements of additive Group - $\mathbb{Z_7^+}$

Question

I am given this additive Group G=$\mathbb{Z_7^+}$

I tried to find all its elements and I did:

$$gcd(1,7) = 1 \\ gcd(2,7) = 1 \\ gcd(3,7) = 1 \\ gcd(4,7) = 1 \\ gcd(5,7) = 1 \\ gcd(6,7) = 1 \\ gcd(7,7) = 7 \\ $$ Then I took all which gives $1$. so $1,2,3,4,5,6$ are all elements of this Group $G$. Am I right?

is it the only and good way of finding elements of given Group? regardless whether multiplicative or additive?

thanks a lot

Answer 1

The gcd is irrelevant when you are looking for elements of $\mathbb{Z}_7$. The elements of this group are in reality subsets of $\mathbb{Z}$ : You say that two numbers $n$ and $m$ in $\mathbb{Z}$ are equivalent iff $7\mid (n-m)$. Then equivalence class of $n$ is $$ [n] = \{m \in \mathbb{Z} : 7\mid (n-m)\} $$ This is a subset of $\mathbb{Z}$. There are exactly 7 such subsets, because for any $n \in \mathbb{Z}$, $n$ is equivalent to its remainder upon division by 7. (For instance, $10$ is related to $3$, and so $$ [10] = [3] $$ The elements of $\mathbb{Z}_7$ are these equivalence classes $$ \{[0], [1], \ldots, [6]\} $$

Answer 2

You actually take the modulus (ie $n \pmod 7$ ) for the additive method. The multiplication group here consists of the 6 relatively primes to 7, which form a closed group to multiplication.

What happens in this sort of group is that if $m$ shares a common factor with $n$ it acts like a partial zero, destroying some information. An example is in $\mathbb{Z}_{10}$, where one can easily construct $x$ from $3x$, but not from $2x$.