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.

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