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 […]

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 […]

Use Lagrange’s Theorem to Prove Fermat’s Little Theorem
Use Lagrange's Theorem in the multiplicative group $(\Zmod{p})^{\times}$ to prove Fermat's Little Theorem: if $p$ is a prime number then $a^p \equiv a \pmod p$ for all $a \in \Z$.
Before the proof, let us recall Lagrange's Theorem.
Lagrange's Theorem
If $G$ is a […]

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