Use of the symbol $G/N$ for the quotient group

Question

I am reading Dummit and Foote's Abstract Algebra, 3rd edition and have a question about their use of the symbol $G/N$.

On page 76, they define the quotient group as follows:

Definition. Let $\varphi:G\to H$ be a homomorphism with kernel $K$. The quotient group, $G/K$, is the group whose elements are the fibers of $\varphi$ with the following group operation: if $X$ is the fiber above $a$ and $Y$ is the fiber above $b$ then the product of $X$ with $Y$ is defined to be the fiber above the product $ab$.

So at this point, the quotient group $G/K$ is defined only when we know $K$ is the kernel of some homomorphism.

Later in the section on page 82, they prove the following proposition:

Proposition 7. A subgroup $N$ of the group $G$ is normal if and only if it is the kernel of some homomorphism.

In the proof, they say "if $N$ is a subgroup of $G$, let $H=G/N$". At this point, they haven't proved that every subgroup is the kernel of some homomorphism. In fact, that's what they are trying to prove here. My question is why they can use the notation $G/N$ before proving that $N$ is the kernel?

Answer 1

Generally speaking, when $H$ is a subgroup of $G$, $G/H$ denotes the set of left cosets of $G$ modulo $H$: $$G/H=\{gH\mid g\in G\},$$ and similarly $H\backslash G$ is the set of right cosets of $G$ modulo $H$: $$H\backslash G=\{Hg\mid g\in G\}.$$

If one tries to endow these sets with an operation deduced from the group operation on $G$, one shows this is possible if and only if $H$ is a normal subgroup of $G$. In this case, the left cosets and the right cosets of $G$ modulo $H$ are the same set, which is called the quotient group of $G$ by $H$.

Answer 2

Not sure about your book, but generally the notation $G/H$ is used for the set of left cosets of $H$, that is, $G/H = \{gH : g \in G\}$. This makes sense for any subgroup $H$. One can try to define a group operation on this set by $(gH)(g'H) = gg'H$, but this is well-defined if and only if $H$ is a normal subgroup. (And once it is, there is a surjective homomorphism $G \to G/H$ by $g \mapsto gH$, whose kernel is $H$ — this makes it agree with your definition of quotient group.)

Answer 3

Until Theorem 3 he is talking about what a quotient group looks alike which is $G/N$ where $N$ is the kernel of some homomorphism of $G$.

Then in the above of page 80, he is talking about what if a more general subgroup used to define a quotient group just like the quotient group explained up to Theorem 3.

It turns out that the operation is not well defined because operation on cosets such $uN \: vN$ is not necessarily equal with coset $uvN$, unless $g N g^{-1} \in N$. It means that $N$ must be a normal subgroup of $G$, and without it, operations on cosets are not well defined.

Then based on that fact, he shows that a subgroup $N$ must be a kernel of some homomorphism.