Note that as $n$ and $m$ are relatively prime integers, there exits $s, t\in \Z$ such that
\[sn+tm=1. \tag{*}\]
Also, note that as the order of the group $G/N$ is $|G/N|=|G:N|=m$, we have
\[g^mN=(gN)^m=N\]
for any $g \in G$ by Lagrange’ theorem, and thus
\[g^m\in N \tag{**}.\]

(a) Prove that $N=\{a\in G \mid a^n=e\}$.

Suppose $a\in \{a\in G \mid a^n=e\}$. Then we have $a^n=e$.
It follows that
\begin{align*}
a \stackrel{(*)}{=} a^{sn+tm}= a^{sn}a^{tm}=a^{tm}=(a^t)^m\in N
\end{align*}
by (**).
This proves that $\{a\in G \mid a^n=e\} \subset N$.

On the other hand, if $a\in N$, then we have $a^n=e$ as $n$ is the order of the group $N$.
Hence $N\subset \{a\in G \mid a^n=e\}$.

Putting together these inclusions yields that $N=\{a\in G \mid a^n=e\}$ as required.

(b) Prove that $N=\{b^m \mid b\in G\}$.

Let $b^m \in \{b^m \mid b\in G\}$. Then by (**), we know that $b^m\in N$.
Thus, we have $\{b^m \mid b\in G\}\subset N$.

On the other hand, let $a\in N$. Then we have $a^n=e$ as $n=|N|$.
Hence it follows that
\begin{align*}
a \stackrel{(*)}{=} a^{sn+tm}= a^{sn}a^{tm}=a^{tm}=b^m,
\end{align*}
where we put $b:=a^t$.
This implies that $a\in \{b^m \mid b\in G\}$, and hence we have $N \subset \{b^m \mid b\in G\}$.

So we see that $N=\{b^m \mid b\in G\}$ by these two inclusions.

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, […]

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 […]

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 […]

Subgroup Containing All $p$-Sylow Subgroups of a Group
Suppose that $G$ is a finite group of order $p^an$, where $p$ is a prime number and $p$ does not divide $n$.
Let $N$ be a normal subgroup of $G$ such that the index $|G: N|$ is relatively prime to $p$.
Then show that $N$ contains all $p$-Sylow subgroups of […]

Subgroup of Finite Index Contains a Normal Subgroup of Finite Index
Let $G$ be a group and let $H$ be a subgroup of finite index. Then show that there exists a normal subgroup $N$ of $G$ such that $N$ is of finite index in $G$ and $N\subset H$.
Proof.
The group $G$ acts on the set of left cosets $G/H$ by left multiplication.
Hence […]

Any Subgroup of Index 2 in a Finite Group is Normal
Show that any subgroup of index $2$ in a group is a normal subgroup.
Hint.
Left (right) cosets partition the group into disjoint sets.
Consider both left and right cosets.
Proof.
Let $H$ be a subgroup of index $2$ in a group $G$.
Let $e \in G$ be the identity […]

A Subgroup of the Smallest Prime Divisor Index of a Group is Normal
Let $G$ be a finite group of order $n$ and suppose that $p$ is the smallest prime number dividing $n$.
Then prove that any subgroup of index $p$ is a normal subgroup of $G$.
Hint.
Consider the action of the group $G$ on the left cosets $G/H$ by left […]

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 […]

I have a doubt, can someone help me!!!
What is the meaning of the below step & how did it come ?
g^m N=(gN)m=N
and how a^sn disappered in the below step ?
a=(∗)asn+tm=asnatm=atm=(at)m∈N

Recall that if the order of a group $G$ is $n$, then for any element $g$ in $G$, we have $g^n=e$, where $e$ is the identity element in $G$.

Now, for your first question, the group is $G/N$ and its order is $m$. Note that $gN$ is an element in $G/N$. Thus, $(gN)^m$ is the identity element in $G/N$, which is $N$. So we have $(gN)^m=N$.

For the second question, $a^{sn}=(a^n)^s=e^s=e$. Thus $a^{sn}=e$ and that’s why it dissapeared.

I have a doubt, can someone help me!!!

What is the meaning of the below step & how did it come ?

g^m N=(gN)m=N

and how a^sn disappered in the below step ?

a=(∗)asn+tm=asnatm=atm=(at)m∈N

Thanks 🙂

Dear Lalitha,

Recall that if the order of a group $G$ is $n$, then for any element $g$ in $G$, we have $g^n=e$, where $e$ is the identity element in $G$.

Now, for your first question, the group is $G/N$ and its order is $m$. Note that $gN$ is an element in $G/N$. Thus, $(gN)^m$ is the identity element in $G/N$, which is $N$. So we have $(gN)^m=N$.

For the second question, $a^{sn}=(a^n)^s=e^s=e$. Thus $a^{sn}=e$ and that’s why it dissapeared.

Thanks for your explanation