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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!

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  • $\begingroup$ 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
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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 $1,3,5,7$.

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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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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 $a$.

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