Elements of multiplicative group equal generators of additive group

Statement
Let $$n$$ be a positive integer, and consider the group of integers modulo n. Then, an element in this is a generator for the group of integers modulo $$n$$ if and only if it is an element of the multiplicative group modulo $$n$$.

Note that these are also the same as the elements in $$\{0,1,2,\dots,n-1\}$$ that are relatively prime to $$n$$, and the number of such elements is $$\varphi(n)$$.

Generator of additive group implies element of multiplicative group
If $$x$$ is a generator of $$\mathbb{Z}/n\mathbb{Z}$$, then some integer multiple of $$x$$ must be equal to the element $$1 \in \mathbb{Z}/n\mathbb{Z}$$. Thus, there exists $$m$$ such that $$mx = 1$$ in $$\mathbb{Z}/n\mathbb{Z}$$. Viewing $$m$$ as a congruence class modulo $$n$$, we see that $$x$$ is invertible modulo $$n$$, and hence is in the multiplicative group.

Element of multiplicative group implies generator of additive group
If $$x$$ is a element of the multiplicative group modulo $$n$$, there exists an integer $$m$$ such that $$mx = 1 \mod n$$. Thus, the cyclic subgroup containing $$x$$ must also contain $$1$$. But any subgroup containing $$1$$ must equal the whole group $$\mathbb{Z}/n\mathbb{Z}$$, so $$x$$ generates the whole group.