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 $a=e$ is the identity element, this is trivial. So we assume that $a \neq e$.

Note that $a^2=a\cdot a\in H$, $a^3=a^2\cdot a\in H$, and repeating this we see that $a^n\in H$ for any positive integer $n$.
Since $G$ is finite, not all of $a^n$ can be different.
Thus there exists positive integers $m, n$ such that $a^m=a^n$ and $m>n$.

Note that we actually have $m>n+1$.
For if $m=n+1$, then we have $a^{n+1}=a^n$ and this implies that $a=e$.
This contradicts out choice of $a$. Thus we have $m>n+1$, or equivalently we have
\[m-n-1>0.\]

Since we have
\[a^{m-n}=e,\]
multiplying by $a^{-1}$ we obtain
\[a^{-1}=a^{m-n-1}.\]

Since $m-n-1>0$, the element $a^{m-n-1}\in H$, hence the inverse $a^{-1}\in H$.
Therefore, $H$ is closed under the group operation and inverse, thus $H$ is a subgroup of $G$.

Remark.

In fact, the group $G$ itself can be an infinite group.
We just need that $H$ is a finite subset satisfying the closure property:for any $a,b \in H$, $ab\in H$.

The proof of this generalization is identical to the proof given above.

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Define the subset $M$ of $G_1 \times G_2$ to be
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