Intersections of Subgroups

After learning how to test a single subset for being a subgroup, we now study how subgroups interact with one another. The focus keyword for this lecture is intersection of subgroups. If two subgroups lie inside the same group, their intersection consists of the elements common to both. The important fact is that this common part is again a subgroup. This result is used frequently when constructing smaller subgroups from known ones.

Let

(G,\circ)

be a group and let

H

and

K

be two subgroups of

G

. Then

H\cap K

is a subgroup of

G

Let

(G,\circ)

be a group and let

H

and

K

be two subgroups of

G

. Then

H\cap K

is a subgroup of

G

Given that

(G,\circ)

is a group and

H

and

K

are two subgroups of

G

. To prove that

H\cap K

is a subgroup of

G

. Since

H

and

K

are subgroups of

G

, the identity element

e

G

belongs to both

H

and

K

. Therefore

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

Thus

H\cap K

is non-empty. Let

a,b\in H\cap K

. Then

\begin{equation*} a,b\in H \end{equation*}

and

\begin{equation*} a,b\in K. \end{equation*}

Since

H

is a subgroup of

G

, the one-step subgroup criterion gives

\begin{equation*} a\circ b^{-1}\in H. \end{equation*}

Since

K

is a subgroup of

G

, the one-step subgroup criterion gives

\begin{equation*} a\circ b^{-1}\in K. \end{equation*}

Therefore

\begin{equation*} a\circ b^{-1}\in H\cap K. \end{equation*}

By the one-step subgroup criterion,

H\cap K

is a subgroup of

G

. Hence,

H\cap K\leq G

\square

Let

(G,\circ)

be a group and let

\{H_i:i\in I\}

be a non-empty family of subgroups of

G

. Then

\begin{equation*} \bigcap_{i\in I}H_i \end{equation*}

is a subgroup of

G

Given that

(G,\circ)

is a group and

\{H_i:i\in I\}

is a non-empty family of subgroups of

G

. To prove that

\bigcap_{i\in I}H_i

is a subgroup of

G

. Since each

H_i

is a subgroup of

G

, we get

\begin{equation*} e\in H_i\quad \forall\,i\in I. \end{equation*}

Therefore

\begin{equation*} e\in \bigcap_{i\in I}H_i. \end{equation*}

Thus

\bigcap_{i\in I}H_i

is non-empty. Let

a,b\in \bigcap_{i\in I}H_i

. Then

\begin{equation*} a,b\in H_i\quad \forall\,i\in I. \end{equation*}

Since each

H_i

is a subgroup of

G

, we get

\begin{equation*} a\circ b^{-1}\in H_i\quad \forall\,i\in I. \end{equation*}

Therefore

\begin{equation*} a\circ b^{-1}\in \bigcap_{i\in I}H_i. \end{equation*}

By the one-step subgroup criterion,

\bigcap_{i\in I}H_i

is a subgroup of

G

. Hence,

\bigcap_{i\in I}H_i\leq G

\square

An intersection keeps only the elements that satisfy all membership conditions at once. Since each subgroup contains the identity, the intersection is never empty. Since each subgroup is closed under

a\circ b^{-1}

, the intersection is also closed under

a\circ b^{-1}

. This is why intersections are safe operations for subgroups. Students should contrast this with unions, where combining elements from different subgroups may destroy closure.

(\mathbb{Z},+)

, consider

2\mathbb{Z}

and

3\mathbb{Z}

. Their intersection is

\begin{equation*} 2\mathbb{Z}\cap 3\mathbb{Z}=6\mathbb{Z}. \end{equation*}

Indeed, an integer divisible by both

2

and

3

is divisible by

6

, and every multiple of

6

is divisible by both

2

and

3

. Since

6\mathbb{Z}

is a subgroup of

\mathbb{Z}

, this example agrees with the intersection theorem.

We observe intersections of subgroups of

\mathbb{Z}

. Enter two positive integers

m

and

n

to compare

m\mathbb{Z}

and

n\mathbb{Z}

. Their intersection is the set of all integers divisible by both

m

and

n

, so it equals

\operatorname{lcm}(m,n)\mathbb{Z}

. The preview lists nearby multiples so the common pattern is visible.

The listed elements are not chosen from either subgroup separately; they are exactly the elements satisfying both divisibility conditions at once.

Prove that

4\mathbb{Z}\cap 6\mathbb{Z}=12\mathbb{Z}

and hence it is a subgroup of

(\mathbb{Z},+)

Let

m\in 4\mathbb{Z}\cap 6\mathbb{Z}

. Then

m

is divisible by

4

and

6

. Therefore

m

is divisible by the least common multiple of

4

and

6

, which is

12

. Thus

\begin{equation*} m\in 12\mathbb{Z}. \end{equation*}

Hence

\begin{equation*} 4\mathbb{Z}\cap 6\mathbb{Z}\subseteq 12\mathbb{Z}. \end{equation*}

Conversely, let

m\in 12\mathbb{Z}

. Then

m=12k

for some

k\in\mathbb{Z}

. Now

\begin{align*} m &=4(3k), \end{align*}

m\in 4\mathbb{Z}

. Also

\begin{align*} m &=6(2k), \end{align*}

m\in 6\mathbb{Z}

. Therefore

m\in 4\mathbb{Z}\cap 6\mathbb{Z}

. Thus

\begin{equation*} 12\mathbb{Z}\subseteq 4\mathbb{Z}\cap 6\mathbb{Z}. \end{equation*}

Therefore

\begin{equation*} 4\mathbb{Z}\cap 6\mathbb{Z}=12\mathbb{Z}. \end{equation*}

Since

12\mathbb{Z}

is a subgroup of

(\mathbb{Z},+)

, the intersection is a subgroup.

\square

This calculator returns the generator of

m\mathbb{Z}\cap n\mathbb{Z}

. It uses the least common multiple because the intersection must contain exactly the integers divisible by both

m

and

n

. The result is automatically a subgroup of

(\mathbb{Z},+)

1. Find

2\mathbb{Z}\cap 5\mathbb{Z}

(\mathbb{Z},+)

. 2. Prove that

m\mathbb{Z}\cap n\mathbb{Z}=\operatorname{lcm}(m,n)\mathbb{Z}

for positive integers

m,n

. 3. Let

H,K,L\leq G

. Prove that

H\cap K\cap L\leq G

2\mathbb{Z}\cap 5\mathbb{Z}=10\mathbb{Z}

. 2. An integer belongs to both

m\mathbb{Z}

and

n\mathbb{Z}

if and only if it is divisible by both

m

and

n

, equivalently by

\operatorname{lcm}(m,n)

. 3. Since the identity belongs to all three subgroups and

a\circ b^{-1}

remains in each subgroup whenever

a,b

lie in all three, the one-step criterion gives

H\cap K\cap L\leq G

Every subgroup contains the identity element of the original group.

Yes. Any non-empty family of subgroups has a subgroup as its intersection.

No. Intersection keeps common elements, while product forms elements of the form

h\circ k

BMLabs Mathematics

BMLabs Mathematics

Intersections of Subgroups

Intersections of Subgroups

COMMON ELEMENTS OF SUBGROUPS

INTERSECTION OF TWO SUBGROUPS

PROOF OF INTERSECTION OF TWO SUBGROUPS

INTERSECTION OF A FAMILY OF SUBGROUPS

PROOF OF INTERSECTION OF A FAMILY OF SUBGROUPS

WHY INTERSECTIONS BEHAVE WELL

INTERSECTION IN INTEGERS

EXPLORE INTEGER SUBGROUP INTERSECTIONS

INTEGER INTERSECTION EXPLORER

INTERSECTION KEEPS COMMON STRUCTURE

INTERSECTION OF TWO INTEGER SUBGROUPS

SOLUTION OF INTERSECTION OF TWO INTEGER SUBGROUPS

COMPUTE AN INTERSECTION OF INTEGER SUBGROUPS

INTEGER SUBGROUP INTERSECTION CALCULATOR

EXERCISES ON INTERSECTIONS OF SUBGROUPS

ANSWERS ON INTERSECTIONS OF SUBGROUPS

FREQUENTLY ASKED QUESTIONS

CONTINUE TO UNIONS OF SUBGROUPS

Coordinates and Locus

Invariants under Orthogonal Transformation