From 30f7615b5857713a7e00a1f6af9a9619e813c00c Mon Sep 17 00:00:00 2001
From: jxu <7989982+jxu@users.noreply.github.com>
Date: Sat, 14 Oct 2023 21:42:47 -0400
Subject: [PATCH 1/2] Phi: add multiplicative group info

---
 src/algebra/phi-function.md | 4 ++++
 1 file changed, 4 insertions(+)

diff --git a/src/algebra/phi-function.md b/src/algebra/phi-function.md
index ac92ea4c1..44040e68e 100644
--- a/src/algebra/phi-function.md
+++ b/src/algebra/phi-function.md
@@ -143,6 +143,10 @@ $$a^n \equiv a^{n \bmod \phi(m)} \pmod m$$
 
 This allows computing $x^n \bmod m$ for very big $n$, especially if $n$ is the result of another computation, as it allows to compute $n$ under a modulo.
 
+In group theory, $\phi(n)$ is the [order of the multiplicative group mod n](https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n) $(\mathbb Z / n\mathbb Z)^\times$, that is the group of multiplicative units (elements with inverses). The elements with multiplicative inverses are precisely those coprime to $n$.
+
+By Lagrange's Theorem, the multiplicative order of any $a$ must divide $\phi(n)$. If the multiplicative order of $a$ is $\phi(n)$, the largest possible, then $a$ is a [primitive root](primitive-root.md) and the group is cyclic by definition. 
+
 ## Generalization
 
 There is a less known version of the last equivalence, that allows computing $x^n \bmod m$ efficiently for not coprime $x$ and $m$.

From da008c43ed272a0056294cae08e7347d866044ce Mon Sep 17 00:00:00 2001
From: jxu <7989982+jxu@users.noreply.github.com>
Date: Sun, 15 Oct 2023 17:37:48 -0400
Subject: [PATCH 2/2] Phi: clarify info on groups

---
 src/algebra/phi-function.md | 5 +++--
 1 file changed, 3 insertions(+), 2 deletions(-)

diff --git a/src/algebra/phi-function.md b/src/algebra/phi-function.md
index 44040e68e..14a04be95 100644
--- a/src/algebra/phi-function.md
+++ b/src/algebra/phi-function.md
@@ -143,9 +143,10 @@ $$a^n \equiv a^{n \bmod \phi(m)} \pmod m$$
 
 This allows computing $x^n \bmod m$ for very big $n$, especially if $n$ is the result of another computation, as it allows to compute $n$ under a modulo.
 
-In group theory, $\phi(n)$ is the [order of the multiplicative group mod n](https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n) $(\mathbb Z / n\mathbb Z)^\times$, that is the group of multiplicative units (elements with inverses). The elements with multiplicative inverses are precisely those coprime to $n$.
+### Group Theory
+$\phi(n)$ is the [order of the multiplicative group mod n](https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n) $(\mathbb Z / n\mathbb Z)^\times$, that is the group of units (elements with multiplicative inverses). The elements with multiplicative inverses are precisely those coprime to $n$.
 
-By Lagrange's Theorem, the multiplicative order of any $a$ must divide $\phi(n)$. If the multiplicative order of $a$ is $\phi(n)$, the largest possible, then $a$ is a [primitive root](primitive-root.md) and the group is cyclic by definition. 
+The [multiplicative order](https://en.wikipedia.org/wiki/Multiplicative_order) of an element $a$ mod $n$, denoted $\operatorname{ord}_n(a)$, is the smallest $k>0$ such that $a^k \equiv 1 \pmod m$. $\operatorname{ord}_n(a)$ is the size of the subgroup generated by $a$, so by Lagrange's Theorem, the multiplicative order of any $a$ must divide $\phi(n)$. If the multiplicative order of $a$ is $\phi(n)$, the largest possible, then $a$ is a [primitive root](primitive-root.md) and the group is cyclic by definition. 
 
 ## Generalization