Comments on: Normal Subgroup Whose Order is Relatively Prime to Its Index

By: Lalitha

Lalitha — Mon, 11 Dec 2017 08:39:40 +0000

Thanks for your explanation

By: Yu

Yu — Sun, 10 Dec 2017 16:00:55 +0000

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.

By: Lalitha

Lalitha — Sun, 10 Dec 2017 12:55:29 +0000

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 🙂