Consider subgroups
\[G_1=\{ x\in G \mid x^m=1 \}\]
and
\[G_2=\{ x\in G \mid x^n=1 \}.\]

Proof.

We first show that the existence of such subgroups of $G$.
Let
\[G_1=\{ x\in G \mid x^m=1 \}\]
and
\[G_2=\{ x\in G \mid x^n=1 \}.\]
We claim that $G=G_1 \times G_2$.
To show this, we prove the following conditions.

(a) $G_1$ and $G_2$ are normal in $G$,
(b) $G_1\cap G_2=e$, where $e$ is the identity element of $G$, and
(c) $G=G_1G_2$

Conditions (a) and (b) imply that $G_1G_2 \cong G_1 \times G_2$ and condition (c) concludes that $G\cong G_1 \times G_2$.

Condition (a) is clear since $G$ is an abelian group.
To show condition (b), take $x \in G_1 \cap G_2$, thus $x^m=x^n=1$.
Since $m$ and $n$ are relatively prime, there exist integers $a, b$ such that $am+bn=1$.

Then we have
\[x=x^1=x^{am+bn}=x^{am}x^{bn}=(x^m)^a (x^n)^b=e^a e^b=e.\]
Therefore we proved $G_1 \cap G_2=e$, hence condition (b) holds as well and we conclude that $G_1G_2=G_1\times G_2$.

Next we prove condition (c).
The inclusion $G_1G_2 \subset G$ is clear. For any $x\in G$, we can write
\[ x=x^{am+bn}=x^{bn}x^{am}.\]
Then $x^{bn}$ is in $G_1$ since
\[(x^{bn})^m=(x^{mn})^b=e^b=e\]
where the second equality follows since the order of $G$ is $mn$.

Similarly, $x^{an}$ is in $G_2$ since
\[(x^{an})^m= (x^{mn})^a=e^a=e.\]
Hence $x=x^{bn}x^{am}\in G_1 G_2$, and we have $G=G_1 G_2$.
We checked all the conditions and hence $G \cong G_1 \times G_2$. From this it follows that the order of $G_1$ is $m$ and the order of $G_2$ is $n$.

Now we show the uniqueness of such subgroups.
If $G_1’$ is a subgroup of $G$ of order $m$, then by the definition of $G_1$, we have
\[G_1’\subset G_1.\]
Since both groups are of order $m$, they must be equal. Thus $G_1$ is the unique subgroup of order $m$. Similarly for $G_2$. This completes the proof.

Fundamental Theorem of Finitely Generated Abelian Groups and its application
In this post, we study the Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem.
Problem.
Let $G$ be a finite abelian group of order $n$.
If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic […]

Non-Abelian Group of Order $pq$ and its Sylow Subgroups
Let $G$ be a non-abelian group of order $pq$, where $p, q$ are prime numbers satisfying $q \equiv 1 \pmod p$.
Prove that a $q$-Sylow subgroup of $G$ is normal and the number of $p$-Sylow subgroups are $q$.
Hint.
Use Sylow's theorem. To review Sylow's theorem, check […]

Group of Order $pq$ Has a Normal Sylow Subgroup and Solvable
Let $p, q$ be prime numbers such that $p>q$.
If a group $G$ has order $pq$, then show the followings.
(a) The group $G$ has a normal Sylow $p$-subgroup.
(b) The group $G$ is solvable.
Definition/Hint
For (a), apply Sylow's theorem. To review Sylow's theorem, […]

Surjective Group Homomorphism to $\Z$ and Direct Product of Abelian Groups
Let $G$ be an abelian group and let $f: G\to \Z$ be a surjective group homomorphism.
Prove that we have an isomorphism of groups:
\[G \cong \ker(f)\times \Z.\]
Proof.
Since $f:G\to \Z$ is surjective, there exists an element $a\in G$ such […]

Group of Order 18 is Solvable
Let $G$ be a finite group of order $18$.
Show that the group $G$ is solvable.
Definition
Recall that a group $G$ is said to be solvable if $G$ has a subnormal series
\[\{e\}=G_0 \triangleleft G_1 \triangleleft G_2 \triangleleft \cdots \triangleleft G_n=G\]
such […]

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 […]

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 […]

Normal Subgroups, Isomorphic Quotients, But Not Isomorphic
Let $G$ be a group. Suppose that $H_1, H_2, N_1, N_2$ are all normal subgroup of $G$, $H_1 \lhd N_2$, and $H_2 \lhd N_2$.
Suppose also that $N_1/H_1$ is isomorphic to $N_2/H_2$. Then prove or disprove that $N_1$ is isomorphic to $N_2$.
Proof.
We give a […]