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.

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 […]