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.
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 […]
Torsion Subgroup of an Abelian Group, Quotient is a Torsion-Free Abelian Group
Let $A$ be an abelian group and let $T(A)$ denote the set of elements of $A$ that have finite order.
(a) Prove that $T(A)$ is a subgroup of $A$.
(The subgroup $T(A)$ is called the torsion subgroup of the abelian group $A$ and elements of $T(A)$ are called torsion […]
Pullback Group of Two Group Homomorphisms into a Group
Let $G_1, G_1$, and $H$ be groups. Let $f_1: G_1 \to H$ and $f_2: G_2 \to H$ be group homomorphisms.
Define the subset $M$ of $G_1 \times G_2$ to be
\[M=\{(a_1, a_2) \in G_1\times G_2 \mid f_1(a_1)=f_2(a_2)\}.\]
Prove that $M$ is a subgroup of $G_1 \times G_2$.
[…]
Quotient Group of Abelian Group is Abelian
Let $G$ be an abelian group and let $N$ be a normal subgroup of $G$.
Then prove that the quotient group $G/N$ is also an abelian group.
Proof.
Each element of $G/N$ is a coset $aN$ for some $a\in G$.
Let $aN, bN$ be arbitrary elements of $G/N$, where $a, b\in […]
Abelian Group and Direct Product of Its Subgroups
Let $G$ be a finite abelian group of order $mn$, where $m$ and $n$ are relatively prime positive integers.
Then show that there exists unique subgroups $G_1$ of order $m$ and $G_2$ of order $n$ such that $G\cong G_1 \times G_2$.
Hint.
Consider […]
The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$
Let $p$ be a prime number.
Let $G$ be a non-abelian $p$-group.
Show that the index of the center of $G$ is divisible by $p^2$.
Proof.
Suppose the order of the group $G$ is $p^a$, for some $a \in \Z$.
Let $Z(G)$ be the center of $G$. Since $Z(G)$ is a subgroup of $G$, the order […]
Sylow Subgroups of a Group of Order 33 is Normal Subgroups
Prove that any $p$-Sylow subgroup of a group $G$ of order $33$ is a normal subgroup of $G$.
Hint.
We use Sylow's theorem. Review the basic terminologies and Sylow's theorem.
Recall that if there is only one $p$-Sylow subgroup $P$ of $G$ for a fixed prime $p$, then $P$ […]
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, […]