What is a multiplicative inverse? From my understanding, the “multiplicative inverse” of a number is what you have to multiply it by to get $1$, i.e. the inverse; in general, the multiplicative inverse of $x$ would be $\frac{1}{x}$. However, I came across a question to do with modular arithmetic and I feel like it has a completely different meaning. 
Given a group $\mathbb {Z}/n\mathbb{Z}$, how do you identify which elements are the multiplicative inverses of a group? I have little to no group theory knowledge, so is there a simple way of understanding this?
Question I came across, (i) I understood, but for (ii) I'm not sure how to get the units of each group.
Thank you! 
 A: The inverse of $x$ is not necessarily $1/x$; it depends on the space you are talking about. The inverse of an element $a$ is defined to be the element $b$ such that $ab=1$, where $1$ is the multiplicative identity element.
Consider $Z/5Z$ with multiplication, which can be thought of as $\{0,1,2,3,4\}$ with the usual multiplication, but take the result mod 5. So, for example, $2*3=1$. Therefore, the inverse of 2 in $Z/5Z$ is $3$.
The units of $Z/nZ$ are precisely those $a$ with $a$ coprime to $n$. So, for example, for $Z/8Z$, the units are $1,3,5,7$.
A: The only difference is in the way multiplication is defined. In $\mathbf Z/n\mathbf Z$, which is a commutative ring (but  not a group for multiplication), an element $a$ has an inverse mod $m$ if there exists an element $a'$ such that $aa'\equiv 1\mod n$.
This means there is a Bézout's relation between $a$ and $n$: $aa'-kn=1$. It happens if and only if $a$ and $n$ are coprime. In particular, if $n$ is a prime number, any $a\not\equiv 0\mod n$ is a unit. In other words, if $n$ is prime,  $\mathbf Z/n\mathbf Z$ is a field (and conversely).
For instance, in $\mathbf Z/15\mathbf Z$, $8$ is a unit, and its inverse is  $2$, since $2\cdot =16\equiv 1\mod15$.
The coefficients of a Bézout's relation, henve the inverse of $a$ mod $n$ are found with the extended Euclidean algorithm.
A: And just to complement other answers, Euler's theorem is also a very useful tool in determining inverses. If $a,n$ coprime then $a^{\varphi(n)} \equiv 1 \pmod{n}$ or $a\cdot a^{\varphi(n)-1} \equiv 1 \pmod{n}$. So $a^{\varphi(n)-1} \pmod{n}$ (or the remainder of $a^{\varphi(n)-1}$ divided by $n$) is the multiplicative inverse of $a$. 
