# Union of Two Subgroups is Not a Group

## Problem 625

Let $G$ be a group and let $H_1, H_2$ be subgroups of $G$ such that $H_1 \not \subset H_2$ and $H_2 \not \subset H_1$.

(a) Prove that the union $H_1 \cup H_2$ is never a subgroup in $G$.

(b) Prove that a group cannot be written as the union of two proper subgroups.

## Proof.

### Prove that the union $H_1 \cup H_2$ is never a subgroup in $G$.

Seeking a contradiction, let us assume that the union $H_1 \cup H_2$ is a subgroup of $G$.
Since $H_1 \not \subset H_2$, there exists an element $a\in H_1$ such that $a\notin H_2$.
Similarly, as $H_2 \not \subset H_1$, there exists an element $b\in H_2$ such that $b\notin H_1$.

As we are assuming $H_1 \cup H_2$ is a group, we have $ab\in H_1 \cup H_2$.
It follows that either $ab \in H_1$ or $ab \in H_2$.

If $ab \in H_1$, then we have
$b=a^{-1}(ab) \in H_1$ as both $a^{-1}$ and $ab$ are elements in the subgroup $H_1$.
This contradicts our choice of the element $b$.

Similarly, if $ab \in H_2$, we have
$a=(ab)b^{-1} \in H_2,$ which contradicts the choice of $a$.

In either case, we reached a contradiction.
Thus, we conclude that the union $H_1 \cup H_2$ is not a subgroup of $G$.

### (b) Prove that a group cannot be written as the union of two proper subgroups.

This is a special case of part (a).

If a group $G$ is a union of two proper subgroup $H_1$ and $H_2$, then we must have $H_1 \not \subset H_2$ and $H_2 \not \subset H_1$, otherwise $G=H_1$ or $G=H_2$ and this is impossible as $H_1, H_2$ are proper subgroups.
Then $G=H_1\cup H_2$ is a subgroup of $G$, which is prohibited by part (a).

Thus, any group cannot be a union of proper subgroups.

### More from my site

• The Union of Two Subspaces is Not a Subspace in a Vector Space Let $U$ and $V$ be subspaces of the vector space $\R^n$. If neither $U$ nor $V$ is a subset of the other, then prove that the union $U \cup V$ is not a subspace of $\R^n$.   Proof. Since $U$ is not contained in $V$, there exists a vector $\mathbf{u}\in U$ but […]
• Union of Subspaces is a Subspace if and only if One is Included in Another Let $W_1, W_2$ be subspaces of a vector space $V$. Then prove that $W_1 \cup W_2$ is a subspace of $V$ if and only if $W_1 \subset W_2$ or $W_2 \subset W_1$.     Proof. If $W_1 \cup W_2$ is a subspace, then $W_1 \subset W_2$ or $W_2 \subset […] • Finite Group and Subgroup Criteria Let$G$be a finite group and let$H$be a subset of$G$such that for any$a,b \in H$,$ab\in H$. Then show that$H$is a subgroup of$G$. Proof. Let$a \in H$. To show that$H$is a subgroup of$G$, it suffices to show that the inverse$a^{-1}$is in$H$. If […] • Elements of Finite Order of an Abelian Group form a Subgroup Let$G$be an abelian group and let$H$be the subset of$G$consisting of all elements of$G$of finite order. That is, $H=\{ a\in G \mid \text{the order of a is finite}\}.$ Prove that$H$is a subgroup of$G$. Proof. Note that the identity element$e$of […] • Abelian Normal Subgroup, Intersection, and Product of Groups Let$G$be a group and let$A$be an abelian subgroup of$G$with$A \triangleleft G$. (That is,$A$is a normal subgroup of$G$.) If$B$is any subgroup of$G$, then show that $A \cap B \triangleleft AB.$ Proof. First of all, since$A \triangleleft G$, the […] • A Simple Abelian Group if and only if the Order is a Prime Number Let$G$be a group. (Do not assume that$G$is a finite group.) Prove that$G$is a simple abelian group if and only if the order of$G$is a prime number. Definition. A group$G$is called simple if$G$is a nontrivial group and the only normal subgroups of$G$is […] • The Number of Elements Satisfying$g^5=e$in a Finite Group is Odd Let$G$be a finite group. Let$S$be the set of elements$g$such that$g^5=e$, where$e$is the identity element in the group$G$. Prove that the number of elements in$S$is odd. Proof. Let$g\neq e$be an element in the group$G$such that$g^5=e$. As […] • Image of a Normal Subgroup Under a Surjective Homomorphism is a Normal Subgroup Let$f: H \to G$be a surjective group homomorphism from a group$H$to a group$G$. Let$N$be a normal subgroup of$H$. Show that the image$f(N)$is normal in$G$. Proof. To show that$f(N)$is normal, we show that$gf(N)g^{-1}=f(N)$for any$g \in […]

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

##### Normal Subgroup Whose Order is Relatively Prime to Its Index

Let $G$ be a finite group and let $N$ be a normal subgroup of $G$. Suppose that the order $n$...

Close