Let $g$ be an element in $G$ of order $5$.
Then the subgroup $\langle g \rangle$ generated by $g$ is a cyclic group of order $5$.
That is, $\langle g \rangle=\{e, g, g^2, g^3, g^4\}$, where $e$ is the identity element in $G$.

Note that the order of each non-identity element in $\langle g \rangle$ is $5$.

Also, if $h$ is another element in $G$ of order $5$, then we have either $\langle g \rangle=\langle h \rangle$ or $\langle g \rangle \cap \langle h \rangle = \{e\}$.
This follows from the fact that the intersection $\langle g \rangle \cap \langle h \rangle$ is a subgroup of the order $5$ group $\langle g \rangle$, and thus the order of $\langle g \rangle \cap \langle h \rangle$ is either $5$ or $1$.

On the other hand, if $H$ is a subgroup of $G$ of order $5$, then every non-identity element in $H$ has order $5$.

These observations imply that each subgroup of order $5$ contains exactly $4$ elements of order $5$ and each element of order $5$ appears in exactly one of such subgroups.

As there are $28$ elements of order $5$, there are $28/4=7$ subgroups of order $5$.

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 […]

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 […]

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 […]

Determine the Number of Elements of Order 3 in a Non-Cyclic Group of Order 57
Let $G$ be a group of order $57$. Assume that $G$ is not a cyclic group.
Then determine the number of elements in $G$ of order $3$.
Proof.
Observe the prime factorization $57=3\cdot 19$.
Let $n_{19}$ be the number of Sylow $19$-subgroups of $G$.
By […]

Normal Subgroup Whose Order is Relatively Prime to Its Index
Let $G$ be a finite group and let $N$ be a normal subgroup of $G$.
Suppose that the order $n$ of $N$ is relatively prime to the index $|G:N|=m$.
(a) Prove that $N=\{a\in G \mid a^n=e\}$.
(b) Prove that $N=\{b^m \mid b\in G\}$.
Proof.
Note that as $n$ and […]

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, […]

A Group of Order $20$ is Solvable
Prove that a group of order $20$ is solvable.
Hint.
Show that a group of order $20$ has a unique normal $5$-Sylow subgroup by Sylow's theorem.
See the post summary of Sylow’s Theorem to review Sylow's theorem.
Proof.
Let $G$ be a group of order $20$. The […]

Infinite Cyclic Groups Do Not Have Composition Series
Let $G$ be an infinite cyclic group. Then show that $G$ does not have a composition series.
Proof.
Let $G=\langle a \rangle$ and suppose that $G$ has a composition series
\[G=G_0\rhd G_1 \rhd \cdots G_{m-1} \rhd G_m=\{e\},\]
where $e$ is the identity element of […]

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