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Assume that an integer $a$ has a multiplicative inverse modulo an integer $n$. Prove that this inverse is unique modulo $n$.

I was given a hint that proving this Lemma: \begin{align} n \mid ab \ \wedge \ \operatorname{gcd}\left(n,a\right) = 1 \qquad \Longrightarrow \qquad n \mid b \end{align} should help me in finding the answer.

Here are my steps in trying to solve the problem: \begin{align} \operatorname{gcd}\left(n,a\right) = 1 & \qquad \Longrightarrow \qquad sn + ta = 1 \qquad \Longrightarrow \qquad sn = 1 - ta \\ & \qquad \Longrightarrow \qquad 1 \equiv ta \mod n \qquad \Longrightarrow \qquad ta \equiv 1 \mod n . \end{align} I know that having the GCD of m and a equal to 1 proves there is a multiplicative inverse mod n, but I'm not sure on how to prove $n \mid b$ with and how it helps prove the multiplicative inverse is unique.

Bill Dubuque
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JohnnyBoi
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3 Answers3

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If $\,b,\bar b$ are inverses of $\,a\,$ then $\ b\equiv b(a\bar b)\equiv (ba)\bar b\equiv \bar b\, $ by Congruence Product Rule. We used only commutativity & associativity of products so the proof works in any commutative ring or monoid.

Note $ $ The proof slickly cancels $\,\color{#c00}a\,$ from $\,\color{#c00}ab\equiv 1\equiv \color{#c00}a\bar b\,$ by scaling by $\,\color{#c00}a^{-1},\,$ yielding $\,b\equiv \bar b.\,$ Hence we see that the uniqueness of inverses is simply a special case of the uniquitous general fact that invertible elements are cancellable, via scaling by the inverse.

Proofs like this can often be discovered by a general method of "overlapping" equations to find a term that both equations apply to (above the overalapped term is $\,ba\bar b\,$ which can be reduced by both rules $\,ab\to 1,\ a\bar b\to 1).\, $ Below is another example with further explanation.

Below we explain a general way to discover a simpler proof (vs. pulling it out of a hat like magic). The key idea is very simple: we can discover consequences of identities (axioms) by "overlapping" them: $ $looking for a "unified" term that they both apply to. Let's try that to prove the uniqueness of additive identity elements. Suppose that $\,0\,$ and $\,0'$ are both additive identities. This means that

$$\begin{eqnarray} x = && \color{#0a0}0 + \color{#c00}x\\ && \color{#0a0}y + \color{#c00}{0'} = y\\ \hline \Rightarrow\ \ 0' = && 0 + 0' = 0\end{eqnarray}$$

where we chose the values of the specialization $\,\color{#0a0}{y=0},\ \color{#c00}{x = 0'}\,$ in order unify $\,0+x\,$ with $\,y+0', \,$ yielding the "unified" term $\,0+0'\,$ that both axioms apply to. Applying both axioms to the unified term we can rewrite it in two different ways, deducing the new consequence $\,0' = 0.\,$ Presto!

This is a very widely applicable method of deriving consequences of axioms, i.e. by "unifying" or "overlapping" terms of both so that both axioms apply, yielding a rewriting of the term in two different ways (e.g. another example). In fact in some cases it can be used to algorithmically derive all of the consequences of the axioms, so yielding algorithms for deciding equality, e.g. see the Knuth-Bendix equational completion algorithm and the Grobner basis algorithm, and see George Bergman's classic paper The Diamond Lemma in Ring Theory.

Bill Dubuque
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For the first part note as $(n,a) = 1$, then there exist $x,y \in \mathbb{Z}$, s.t. $nx + ay = 1$. Multiply both sides by $b$ and you will get $nxb + (ab)y = b$. The LHS is obviously divisible by $n$, so then the RHS must be too. Hence $n \mid b$.

This thing proves that the inverse exist. To note that it's unique modulo $n$ assume that $aa_1 \equiv 1 \pmod n$ and $aa_2 \equiv 1 \pmod n$. Then we have that there exist integer $x,y$ s.t. $aa_1 + nx = 1$ and $aa_2 + ny = 1$. Subtracting the two equations we have that $a(a_1 - a_2) = n(y-x) \implies a(a_1 - a_2) \equiv 0 \pmod n \implies a_1 \equiv a_2 \pmod n$. Hence the multiplicative inverse is unique in $\mathbb{Z}_n$

Stefan4024
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  • The argument is incomplete, since you seem to be applying the first paragraph in your final inference (which should be mentioned explicitly), but to apply that you need to first prove $a$ is coprime to $n$ when $a$ is invertible. This follows simply from the listed Bezout equation, but that needs to be explicitly stated in order to be rigorous. Note that we are given that $a$ is invertible, so you don't need to prove that. The logic is not correct. However, none of this is needed since the hinted method is rather roundabout - see my answer. – Bill Dubuque Apr 23 '21 at 20:30
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What you have shown already is a technique that, using the Extended Euclidean Algorithm will give you the inverse (if it exists). Note also that you can determine if an element $a \mod n$ has an inverse by checking that they have no common divisors (that is, $\gcd(a,n)=1$).

Now to answer your question. Suppose that you have two multiplicative inverses $b,c$ such that $ab\equiv 1 \equiv ac \mod n$. (Then you also have $ba\equiv 1 \equiv ca \mod n$) Consider

$b \equiv b\cdot 1 \equiv bac \equiv 1\cdot c \equiv c \mod n$.

That shows that modulo $n$, there exists a unique inverse. This technique is very useful and you can also use it to show that: there exists exactly one $0$-vector in every vector space, group, ring... and every invertible element in a group, ring... has a unique inverse.

Note: For instance $2\cdot 3 \equiv 1 \mod 5$, but also $2\cdot 8 \equiv 1 \mod 5$ (Why is this not a contradiction to the above?)

JSchoone
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