Normal Subgroup Whose Order is Relatively Prime to Its Index

Problem 621

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 $m$ 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.

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3 Responses

1. Lalitha says:

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 🙂

• Yu says:

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.

• Lalitha says: