Unions of Subgroups

Intersections of subgroups are always subgroups, but unions behave very differently. The focus keyword for this lecture is union of subgroups. If

H

and

K

are subgroups of a group

G

, then

H\cup K

contains elements from both subgroups. The difficulty is that multiplying one element from

H

and one element from

K

may produce an element outside the union. In this lesson, we prove the exact condition under which a union of two subgroups is again a subgroup.

Let

(G,\circ)

be a group and let

H

and

K

be two subgroups of

G

. Then

H\cup K

is a subgroup of

G

if and only if

\begin{equation*} H\subseteq K \end{equation*}

\begin{equation*} K\subseteq H. \end{equation*}

Intersections of subgroups are always subgroups, but unions behave very differently. The focus keyword for this lecture is union of subgroups. If

H

and

K

are subgroups of a group

G

, then

H\cup K

contains elements from both subgroups. The difficulty is that multiplying one element from

H

and one element from

K

may produce an element outside the union. In this lesson, we prove the exact condition under which a union of two subgroups is again a subgroup.

Let

(G,\circ)

be a group and let

H

and

K

be two subgroups of

G

. Then

H\cup K

is a subgroup of

G

if and only if

\begin{equation*} H\subseteq K \end{equation*}

\begin{equation*} K\subseteq H. \end{equation*}

Given that

(G,\circ)

is a group and

H

and

K

are two subgroups of

G

. To prove that

H\cup K

is a subgroup of

G

if and only if

H\subseteq K

K\subseteq H

. [1] Let

H\cup K

be a subgroup of

G

. To prove that

H\subseteq K

K\subseteq H

. If possible let

H\nsubseteq K

and

K\nsubseteq H

. Then there exist

h\in H

and

k\in K

such that

\begin{equation*} h\notin K \end{equation*}

and

\begin{equation*} k\notin H. \end{equation*}

Since

h,k\in H\cup K

and

H\cup K

is a subgroup of

G

, we get

\begin{equation*} h\circ k\in H\cup K. \end{equation*}

Therefore

h\circ k\in H

h\circ k\in K

. If

h\circ k\in H

, then

\begin{align*} h\circ k\in H &\implies h^{-1}\circ(h\circ k)\in H\\ &\implies k\in H, \end{align*}

which contradicts

k\notin H

. If

h\circ k\in K

, then

\begin{align*} h\circ k\in K &\implies (h\circ k)\circ k^{-1}\in K\\ &\implies h\in K, \end{align*}

which contradicts

h\notin K

. A contradiction. Hence,

H\subseteq K

K\subseteq H

. [2] Let

H\subseteq K

K\subseteq H

. If

H\subseteq K

, then

\begin{equation*} H\cup K=K. \end{equation*}

Since

K

is a subgroup of

G

H\cup K

is a subgroup of

G

. If

K\subseteq H

, then

\begin{equation*} H\cup K=H. \end{equation*}

Since

H

is a subgroup of

G

H\cup K

is a subgroup of

G

. Hence,

H\cup K

is a subgroup of

G

if and only if

H\subseteq K

K\subseteq H

\square

The union contains all elements of

H

and all elements of

K

, but it does not automatically contain products mixing the two subgroups. If neither subgroup contains the other, we can choose

h\in H\setminus K

and

k\in K\setminus H

. Closure of the union would force

h\circ k

to lie in

H

K

, and either case gives a contradiction. This is the precise reason unions are not as stable as intersections.

Let

H=2\mathbb{Z}

and

K=4\mathbb{Z}

(\mathbb{Z},+)

. Since

4\mathbb{Z}\subseteq 2\mathbb{Z}

, we get

\begin{equation*} H\cup K=2\mathbb{Z}. \end{equation*}

Therefore

H\cup K

is a subgroup of

(\mathbb{Z},+)

Let

H=2\mathbb{Z}

and

K=3\mathbb{Z}

(\mathbb{Z},+)

. Neither

H\subseteq K

nor

K\subseteq H

. Also

2\in H

and

3\in K

, but

\begin{equation*} 2+3=5. \end{equation*}

Since

5\notin 2\mathbb{Z}

and

5\notin 3\mathbb{Z}

, we get

5\notin H\cup K

. Therefore

H\cup K

is not closed under addition. Hence,

H\cup K

is not a subgroup of

(\mathbb{Z},+)

We observe the union condition with subgroups

m\mathbb{Z}

and

n\mathbb{Z}

\mathbb{Z}

. One subgroup contains the other exactly when one of the divisibility conditions is stronger than the other. If neither contains the other, mixed sums can leave the union. The preview compares the containment relation and gives a typical mixed-sum warning when the union is not a subgroup.

A union can pass all tests inside

H

and inside

K

separately, yet fail when one element comes from each subgroup. This is the key difference from intersections.

Determine whether

6\mathbb{Z}\cup 3\mathbb{Z}

is a subgroup of

(\mathbb{Z},+)

Let

H=6\mathbb{Z}

and

K=3\mathbb{Z}

. Given that

H

and

K

are subgroups of

(\mathbb{Z},+)

. To determine whether

H\cup K

is a subgroup of

\mathbb{Z}

. Since every multiple of

6

is a multiple of

3

, we get

\begin{equation*} 6\mathbb{Z}\subseteq 3\mathbb{Z}. \end{equation*}

Therefore

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

Since

3\mathbb{Z}

is a subgroup of

(\mathbb{Z},+)

, we get that

6\mathbb{Z}\cup 3\mathbb{Z}

is a subgroup of

(\mathbb{Z},+)

\square

Determine whether

4\mathbb{Z}\cup 6\mathbb{Z}

is a subgroup of

(\mathbb{Z},+)

Let

H=4\mathbb{Z}

and

K=6\mathbb{Z}

. Given that

H

and

K

are subgroups of

(\mathbb{Z},+)

. To determine whether

H\cup K

is a subgroup of

\mathbb{Z}

. We have

4\in H

and

6\in K

. But

\begin{align*} 4+6 &=10. \end{align*}

Since

10

is not divisible by

4

and

10

is not divisible by

6

, we get

\begin{equation*} 10\notin H\cup K. \end{equation*}

Therefore

H\cup K

is not closed under addition. Hence,

4\mathbb{Z}\cup 6\mathbb{Z}

is not a subgroup of

(\mathbb{Z},+)

\square

This calculator decides whether

m\mathbb{Z}\cup n\mathbb{Z}

is a subgroup of

\mathbb{Z}

. The union is a subgroup exactly when

m\mathbb{Z}\subseteq n\mathbb{Z}

n\mathbb{Z}\subseteq m\mathbb{Z}

. For integer subgroups, this is checked by divisibility of the generators.

1. Decide whether

2\mathbb{Z}\cup 6\mathbb{Z}

is a subgroup of

(\mathbb{Z},+)

. 2. Decide whether

2\mathbb{Z}\cup 5\mathbb{Z}

is a subgroup of

(\mathbb{Z},+)

. 3. Let

H,K\leq G

. Prove that if

H\subseteq K

, then

H\cup K\leq G

1. Yes, because

6\mathbb{Z}\subseteq 2\mathbb{Z}

, so the union is

2\mathbb{Z}

. 2. No, because neither subgroup contains the other. Also

2+5=7

is in neither

2\mathbb{Z}

nor

5\mathbb{Z}

. 3. If

H\subseteq K

, then

H\cup K=K

, and

K

is a subgroup of

G

No. It is a subgroup exactly when one subgroup is contained in the other.

Intersections preserve all subgroup conditions, while unions may fail closure for mixed products.

It can be, but the two-subgroup containment theorem does not directly describe all three-subgroup cases.

BMLabs Mathematics

BMLabs Mathematics

Unions of Subgroups

Unions of Subgroups

UNIONS ARE MORE DELICATE

UNION OF TWO SUBGROUPS

PROOF OF UNION OF TWO SUBGROUPS

WHY THE CONTAINMENT CONDITION IS NECESSARY

A UNION THAT IS A SUBGROUP

A UNION THAT IS NOT A SUBGROUP

EXPLORE UNIONS OF INTEGER SUBGROUPS

INTEGER SUBGROUP UNION EXPLORER

WHY MIXED ELEMENTS MATTER

UNION OF INTEGER SUBGROUPS

SOLUTION OF UNION OF INTEGER SUBGROUPS

UNION FAILS WITHOUT CONTAINMENT

SOLUTION OF UNION FAILS WITHOUT CONTAINMENT

TEST THE UNION CONDITION

UNION OF INTEGER SUBGROUPS CALCULATOR

EXERCISES ON UNIONS OF SUBGROUPS

ANSWERS ON UNIONS OF SUBGROUPS

FREQUENTLY ASKED QUESTIONS

CONTINUE TO PRODUCTS OF SUBGROUPS

Coordinates and Locus

Invariants under Orthogonal Transformation