Relatively Prime Subgroup Orders

The product formula becomes especially simple when two subgroups have relatively prime orders. If

|H|

and

|K|

have no common divisor other than

1

, then the intersection

H\cap K

must have only one element. Since every subgroup contains the identity element, this means

H\cap K=\{e\}

. This lecture shows how Lagrange's theorem controls intersections through divisibility and then uses that control to count products of subgroups.

Let

(G,\circ)

be a group and let

H,K

be finite subgroups of

G

. The subgroups

H

and

K

are said to have

\textbf{relatively prime orders}

\begin{equation*} \gcd(|H|,|K|)=1. \end{equation*}

The product formula becomes especially simple when two subgroups have relatively prime orders. If

|H|

and

|K|

have no common divisor other than

1

, then the intersection

H\cap K

must have only one element. Since every subgroup contains the identity element, this means

H\cap K=\{e\}

. This lecture shows how Lagrange's theorem controls intersections through divisibility and then uses that control to count products of subgroups.

Let

(G,\circ)

be a group and let

H,K

be finite subgroups of

G

. The subgroups

H

and

K

are said to have

\textbf{relatively prime orders}

\begin{equation*} \gcd(|H|,|K|)=1. \end{equation*}

Let

(G,\circ)

be a group and let

H,K

be finite subgroups of

G

. If

\begin{equation*} \gcd(|H|,|K|)=1, \end{equation*}

then

\begin{equation*} H\cap K=\{e\}. \end{equation*}

Given that

(G,\circ)

is a group,

H,K

are finite subgroups of

G

, and

\gcd(|H|,|K|)=1

. To prove that

\begin{equation*} H\cap K=\{e\}. \end{equation*}

Since

H\cap K

is a subgroup of

H

, Lagrange's theorem gives

\begin{equation*} |H\cap K|\mid |H|. \end{equation*}

Since

H\cap K

is a subgroup of

K

, Lagrange's theorem gives

\begin{equation*} |H\cap K|\mid |K|. \end{equation*}

Therefore

|H\cap K|

is a common divisor of

|H|

and

|K|

. Since

\gcd(|H|,|K|)=1

, we get

\begin{equation*} |H\cap K|=1. \end{equation*}

Since

e\in H\cap K

, the subgroup

H\cap K

has exactly one element. Therefore

\begin{equation*} H\cap K=\{e\}. \end{equation*}

Hence,

\begin{equation*} H\cap K=\{e\}. \end{equation*}

\square

We observe why relatively prime subgroup orders force a trivial intersection. Choose the orders of

H

and

K

, and compare their positive divisors side by side. The order of

H\cap K

must appear in both divisor lists, because the intersection is a subgroup of both

H

and

K

. When the only common divisor is

1

, the identity is the only possible element of the intersection.

The proof is a divisibility argument. Once the only common divisor is

1

, Lagrange's theorem leaves no room for a larger intersection subgroup.

Let

(G,\circ)

be a group and let

H,K

be finite subgroups of

G

. If

\gcd(|H|,|K|)=1

, then

\begin{equation*} |HK|=|H||K|. \end{equation*}

Given that

(G,\circ)

is a group,

H,K

are finite subgroups of

G

, and

\gcd(|H|,|K|)=1

. To prove that

\begin{equation*} |HK|=|H||K|. \end{equation*}

By the previous theorem,

\begin{equation*} H\cap K=\{e\}. \end{equation*}

Therefore

\begin{equation*} |H\cap K|=1. \end{equation*}

By the product formula for finite subgroups,

\begin{align*} |HK| &=\frac{|H||K|}{|H\cap K|}\\ &=\frac{|H||K|}{1}\\ &=|H||K|. \end{align*}

Hence,

\begin{equation*} |HK|=|H||K|. \end{equation*}

\square

Let

(G,\circ)

be a finite group and let

H,K

be subgroups of

G

. If

\gcd(|H|,|K|)=1

, then

\begin{equation*} |H||K|\leq |G|. \end{equation*}

Given that

(G,\circ)

is a finite group,

H,K

are subgroups of

G

, and

\gcd(|H|,|K|)=1

. To prove that

\begin{equation*} |H||K|\leq |G|. \end{equation*}

By the previous corollary,

\begin{equation*} |HK|=|H||K|. \end{equation*}

Since

HK\subseteq G

, we get

\begin{equation*} |HK|\leq |G|. \end{equation*}

Therefore

\begin{equation*} |H||K|\leq |G|. \end{equation*}

Hence,

\begin{equation*} |H||K|\leq |G|. \end{equation*}

\square

Let

G=\mathbb{Z}_{12}

under addition modulo

12

. Let

\begin{equation*} H=\{0,3,6,9\} \end{equation*}

and

\begin{equation*} K=\{0,4,8\}. \end{equation*}

Then

\begin{equation*} |H|=4 \end{equation*}

and

\begin{equation*} |K|=3. \end{equation*}

Since

\gcd(4,3)=1

, we get

\begin{equation*} H\cap K=\{0\}. \end{equation*}

Also,

\begin{equation*} |H+K|=|H||K|=12. \end{equation*}

Let

(G,\circ)

be a group and let

H,K

be finite subgroups of

G

. Suppose

|H|=15

and

|K|=8

. Prove that

H\cap K=\{e\}

Let

(G,\circ)

be a group and let

H,K

be finite subgroups of

G

. Since

\begin{equation*} \gcd(15,8)=1, \end{equation*}

we get

\begin{equation*} \gcd(|H|,|K|)=1. \end{equation*}

By the theorem on relatively prime subgroup orders,

\begin{equation*} H\cap K=\{e\}. \end{equation*}

We compute the greatest common divisor of

|H|

and

|K|

and use it to decide what the theorem guarantees. When the gcd is

1

, the calculator reports the trivial intersection and computes

|HK|=|H||K|

. If an ambient group order is entered, the same calculation checks whether the inequality

|H||K|\leq |G|

is possible. Notice that when the gcd is not

1

, the theorem simply does not apply; it does not prove that the intersection is nontrivial.

1. Let

H,K

be finite subgroups of a group

G

with

|H|=9

and

|K|=10

. Find

H\cap K

. 2. Let

H,K

be finite subgroups of a group

G

with

|H|=7

and

|K|=12

. Find

|HK|

. 3. Let

G

be a finite group with subgroups

H,K

such that

|H|=5

and

|K|=9

. What inequality involving

|G|

follows?

1. Since

\gcd(9,10)=1

, we get

\begin{equation*} H\cap K=\{e\}. \end{equation*}

2. Since

\gcd(7,12)=1

, we get

\begin{equation*} |HK|=|H||K|=7\cdot 12=84. \end{equation*}

3. Since

\gcd(5,9)=1

, we have

\begin{equation*} |H||K|\leq |G|. \end{equation*}

Therefore

\begin{equation*} 45\leq |G|. \end{equation*}

The order of the intersection divides both subgroup orders, so it must divide their greatest common divisor.

No. Trivial intersection can occur even when the subgroup orders are not relatively prime.

No. It counts the set

HK

; it does not automatically prove closure.

BMLabs Mathematics

BMLabs Mathematics

Relatively Prime Subgroup Orders

Relatively Prime Subgroup Orders

OVERVIEW

RELATIVELY PRIME SUBGROUP ORDERS

EXPLORE COMMON DIVISORS

COPRIME SUBGROUP INTERSECTION EXPLORER

WHAT THE DIVISORS REVEAL

SUBGROUPS OF ORDERS FOUR AND THREE

TRIVIAL INTERSECTION FROM COPRIME ORDERS

SOLUTION

TEST COPRIME PRODUCTS

COPRIME PRODUCT CALCULATOR

EXERCISES

ANSWERS

Coordinates and Locus

Invariants under Orthogonal Transformation