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!

  • $\begingroup$ In a group, identity element exists. How about saying inverse 'b' of a when ab=identity? $\endgroup$ – kayak Jan 7 '17 at 0:26
  • $\begingroup$ A number in $\Bbb Z/n\Bbb Z$ has a multiplicative inverse iff $1$ appears in its row of the multiplication table. $\endgroup$ – Arthur Jan 7 '17 at 0:27
  • $\begingroup$ @mimyo It's important to note that $\mathbb Z/n\mathbb Z$ is not a group, at least not in the sense that you are using (it's a group under addition but under multiplication it's only a semigroup). The question that you linked to is being very careless by referring to these structures as groups. Modular multiplication is not definable as a group operation (one has to use the fact that these are not just elements of a group but also concretely represented by integers). $\endgroup$ – Erick Wong Jan 7 '17 at 3:15

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 $1,3,5,7$.

  • 2
    $\begingroup$ You could also explain that unit is a term used for an element for which the multiplicative inverse exists. For example in $Z/8Z$ the element $a=6$ cannot have an inverse since $ax$ would always be "even" ("even" is well-defined in $Z/8Z$) an so never $1$. $\endgroup$ – Jeppe Stig Nielsen Jan 7 '17 at 0:33
  • 3
    $\begingroup$ The inverse of $x$ is necessarily $1/x$, by definition of "\". However, what that actually means is dependent of context. In $\Bbb Q$, we have $1/2=0.5$. In $\Bbb Z/3\Bbb Z$, we have $1/2=2$. $\endgroup$ – Arthur Jan 7 '17 at 0:37

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.


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 $a$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.