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By definition, the Carmichael function maps a positive integer $n$ to the smallest positive integer $t$ such that $a^t\equiv1\pmod n$ for all integers $a$ with $\gcd(a,n)=1$. It is denoted as $\lambda(n)$.

The Wikipedia page on Carmichael function states that $\lambda(n)=\max\{\operatorname{ord}_n(a):\gcd(a,n)=1\}$. My question is: why is this true? In other words, why is it the case that there always exists an integer $x$ coprime to $n$ with $\operatorname{ord}_n(x)=\lambda(n)$?

Bill Dubuque
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Iming Cheng
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    $\lambda(n)$ is , per definition, the largest order that occurs. This immediately implies the existence of the desired $x$ – Peter Aug 05 '21 at 08:50
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    Is it not possible that $\lambda(n)=lcm(ord_n(a),ord_n(b))$ for some integers $a$ and $b$ but there doesn't exist an integer $x$ with $ord_n(x)=\lambda(n)$? – Iming Cheng Aug 05 '21 at 08:54
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    Can't happen in these groups. Makes a good exercise to prove it. Maybe look at https://math.stackexchange.com/questions/4003768/carmichael-lambda-function-proof – Gerry Myerson Aug 05 '21 at 10:29
  • The element $x$ such that $\mathrm{ord}_n(x)=\lambda(n)$ is called the primitive $\lambda$-root modulo $n$. There are at least $\phi(\phi(n))$ primitive $\lambda$-roots modulo $n$ where $\phi$ is the Euler totient function. – Sungjin Kim Aug 05 '21 at 20:06

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Let $S$ be the set of elements coprime to $n$.

The smallest number $\lambda(n)$ such that $a^{\lambda(n)} \equiv 1 \bmod n$ for all $a\in S$ is the smallest number that is a multiple of all of the orders of the elements of $S$.

In other words we want to show that there is an element whose order is the lcm of all the orders of $S$.

We can do this by showing this lemma:

If there is an element $s$ with order $a$ and an element $t$ with order $b$ then there is an element with order $\operatorname{lcm}(a,b)$.

Proof: First assume $a$ and $b$ are coprime, then $st$ has the desired order, because if $(st)^l \equiv 1 \bmod n$ then $s^l \equiv t^{-l}$ but if $t^{-l} \not \equiv 1 \bmod n$ then the order of $t^{-l}$ is a divisor of $b$ which cannot be the order of $s^l$ because the order of $s^l$ is a divisor of $a$, which is coprime to $b$. Therefore if $(st)^l \equiv 1 $ we must have $s^l \equiv 1$ and $t^{-l} \equiv 1$ which implies $l$ is a multiple of $a$ and $b$.

If $a$ and $b$ are not coprime then let $a = \prod\limits_{i=1}^ s p_i^{a_i}$ and $b = \prod\limits_{i=1}^s p_i^{b_i}$, and let $A$ be the set of indices with $a_i \geq b_i$.

Define $a'$ as $\prod\limits_{i\in A} p_i^{a_i}$ and $b'=\prod\limits_{i \not \in A} p_i^{b_i}$.

And now take $s'=s^{a/a'}$ and $t'=t^{b/b'}$and apply the previous result with $s',t',a'$ and $b'$.

Asinomás
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