Why is $\langle xN\rangle$ necessarily a subgroup of $G/N$?

Question

I'm having some trouble understanding this answer to If a maximal subgroup is normal, it has prime index.

Notation: We denote the normal subgroup by $N$ instead.

By the Correspondence Theorem, there exists a bijection from the set of all subgroups $H$ such that $N\subseteq H\subseteq G$ onto the set of all subgroups of $G/N$. Since the only such subgroups are $H=N$ and $H=G$, $G/N$ has only two subgroups, namely $N/N$ and $G/N$.

Let $xN$ be a nontrivial element in $G/N$. $\langle xN\rangle$ is a nontrivial subgroup of $G/N$, thus $\langle xN\rangle=G/N$. This means $G/N$ is cyclic. If $|G/N|$ is infinite, then $G/N\cong\mathbb{Z}$ which is a contradiction as $\mathbb{Z}$ has infinite subgroups of the form $n\mathbb{Z}$. Therefore $[G:N]=|G/N|$ is finite.

Question:

I understand that $xN$ is a non-trivial element in $G/N$. However, what proof do we have that $(xN)^2 = x^2N$, $(xN)^3 = x^3N$ and other higher order terms exist in the group $G/N$? And without proving that first, I don't think it makes sense to say that $\langle xN\rangle$ is a subgroup of $G$.

I'm not sure I get your problem. If $N$ is normal then $G/N$ is equipped with the quotient group structure given by $gN\cdot hN:=ghN$. And so $(xN)^m=x^mN$ by the definition (plus some trivial induction). — freakish, Apr 14 '20 at 07:26

score 2 · Accepted Answer · answered Apr 14 '20 at 07:29

2

For any element $g\in G$, $gN$ is some coset of $G$, and thereby is some element of $G/N$, which is by definition the group of all cosets of the normal subgroup $N$ in $G$.

answered Apr 14 '20 at 07:29

Parcly Taxel

105,904

Okay, but if $g \in G$ we don't know if $g^2 \in G$ too (?). So how can we say that $g^2N \in G/N$? – Apr 14 '20 at 07:32
@S.D. Remember the group axioms. – Parcly Taxel Apr 14 '20 at 07:32
Are you talking about the closure axiom? But it only says that if two distinct elements of $G$ i.e. $a$ and $b$ are in $G$ then $ab$ is also in $G$. Or do $a$ and $b$ need not necessarily be distinct in this context? – Apr 14 '20 at 07:34
1

@S.D. They don't have to be distinct. – Parcly Taxel Apr 14 '20 at 07:35
I didn't know that. Thanks! – Apr 14 '20 at 07:36

score 1 · Answer 2 · answered Apr 14 '20 at 07:33

1

For any $x\in G$, we have $xN\in G/N$. But once you have your hands on an element of a group, you can always consider the cyclic subgroup it generates.

answered Apr 14 '20 at 07:33

Why is $\langle xN\rangle$ necessarily a subgroup of $G/N$?

2 Answers2