Generators of the group of invertible elements of the ring $\mathbb{Z}_{14}$—are they multiplicative or additive? When I was asked to find the generators of the group of invertible elements of the ring $\mathbb{Z}_{14}$, which are denoted as $\phi(14)$, I did not realize whether the generators are multiplicative or additive. Which is it? 
If both are equally valid, then why did the author omit mention of the type of generators?
 A: For any ring $R$, it is standard to consider the set of units
$$R^\times=\{a\in R:\textsf{there exists some }\,b\in R\,\textsf{ with }\,a\cdot b=b\cdot a=1\}$$
as a group under the multiplication operation of $R$. (Here's the relevant Wikipedia page.)
(In fact, if $R$ is not the trivial ring, then $R^\times$ is not closed under the addition operation of $R$, so it certainly will not be a group under that operation.)
You're told to look at the group of units of $R=\mathbb{Z}_{14}$, namely the set 
$$(\mathbb{Z}_{14})^\times=\{\overline{1},\overline{3},\overline{5},\overline{9},\overline{11},\overline{13}\}$$
under the multiplication operation of $\mathbb{Z}_{14}$, and you are looking for the elements $x\in(\mathbb{Z}_{14})^\times$ that generate the entire group, i.e., the elements with the property that
$$(\mathbb{Z}_{14})^\times=\{x^n:n\in\mathbb{Z}\}$$
(Here's the relevant Wikipedia page.)
A: Invertible elements of a ring form a group under multiplication, but never under addition (e.g., 0 does not belong to that set). 
So this should be clear.
