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

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?

• Thank you for correction.
– Koda
Commented Aug 7, 2018 at 21:25
• I don't have a copy of Dummit and Foote. But check to see if there isn't an alternate definition of a quotient group of $G/N$ as the cosets of $N$ in $G.$ Commented Aug 7, 2018 at 21:32
• That is what I was hoping to find, but they do not define $G/N$ in any other way.
– Koda
Commented Aug 8, 2018 at 0:33

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 $H$.

• Really? I've never seen $H\setminus G$ used with that meaning. I've only seen "$\setminus$" used to denote exclusion: $H\setminus G=\{h\in H\mid h\notin G\}$. Is the notation you mention commonly used?
– MPW
Commented Aug 7, 2018 at 21:41
• It is Bourbaki's notation. Remark the spacing is not the same as with ‘set minus’. Commented Aug 7, 2018 at 21:45
• I never would have guessed that. Learned something new today. +1, thanks.
– MPW
Commented Aug 7, 2018 at 21:52
• @Bernard Thank you for your reply. I am aware of that definition of $G/H$. What I am hoping to understand here is what the authors mean by $G/N$ in their proof given that the only definition of $G/N$ they give is the one using the kernel of a homomorphism.
– Koda
Commented Aug 8, 2018 at 0:42
• As I don't have this book it's hard to tell precisely. My guess is they have defined this notation for left cosets somewhere before. B.t.w., not every subgroup is the kernel of a homomorphism. Only normal subgroups are. Commented Aug 8, 2018 at 1:26

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.

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.)