In my abstract algebra course, the instructor is calling $G/N$ (the set of left Cosets of $N$ in $G$) $G\: mod\: N$. This has not yet been explained. Why is this the case?

My immediate suspicion is some isomorphism between $G/N$ and the integers modulo $n$.

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    $\begingroup$ It is common in math to start with one object (be it a group, ring, topological space, etc...)(call it $X$) and then consider a subset of that object (call it $Y$), and some logical condition which relates elements of $X$ based on which elements are in $Y$ and create a new object based on this relation denoted $X/Y$ and called "$X$ mod $Y$." The integers modulo $n$ are just an instance of this. Such an object is called a quotient of $X$. $\endgroup$ – Alex S Jul 6 '15 at 21:58
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    $\begingroup$ The integers modulo $n$ is simply the quotient of the group $\mathbf Z$ by the subgroup $n\mathbf Z$. $\endgroup$ – Bernard Jul 6 '15 at 22:00
  • $\begingroup$ @Alex: Is it not "sine qua non" than your logical condition be an equivalence relation on $X$? Is it not that an equivalence relation on $X$ induces canonically una partition (your $X/Y$)? $\endgroup$ – Piquito Jul 6 '15 at 22:09
  • $\begingroup$ @Ataulfo Yes, it is necessary that the relation be an equivalence relation, and $X/Y$ consists of equivalence classes resulting from this relation. $\endgroup$ – Alex S Jul 6 '15 at 22:16
  • $\begingroup$ Man, I hate this terminology. It's like calling $6/3$ both "6 mod 3" and "6 divided by 3". $\endgroup$ – user2357112 Jul 7 '15 at 2:42

Your intuition is a bit off but not entirely. This quotient group essentially creates a group in which all of $N$ is now the identity (I'm sure you know we can only do this if $N\triangleleft G$). So, for example, let's consider the integers along with the normal subgroup $6\mathbb{Z}$. Then we can define a quotient group


We can add these together in ways such as $$(3+6\mathbb{Z})+(4+6\mathbb{Z})=1+6\mathbb{Z}$$

If you've been dealing with the groups of integers under modular addition you'll notice this quotient group we've defined does precisely the same that we would expect of $\mathbb{Z}_6$. And, indeed, the two groups are isomorphic.

One important use of this idea as well is called the First Isomorphism Theorem. Let's say we've set up a homomorphism that's not injective (i.e. a nontrivial kernel). Well, the kernel is always a normal subgroup, so we can "mod out" by the kernel, essentially condensing our entire kernel into one element. We then have something that's injective. This is stated more formally as:

Let $\phi:G\rightarrow H$ be a group homomorphism. Then,


This is extremely useful in proving certain isomorphisms.


A little more generally, the use of the word "modulo" is connected to equivalence relations. Given a set $X$ and an equivalence relation $R$ defined on $X$, the set of equivalence classes is frequently denoted $X/R$ and called "X mod R".

In your example, the subgroup $N$ can be used to define an equivalence relation on $G$: Declare elements $g, h \in N$ to be equivalent if $g^{-1}h \in N$. The equivalence classes are precisely the left cosets of $N$ in $G$. There's thus a very nice affinity between the general notation $X/R$ and the group-specific notation $G/N$.


$$A \text{ is the same as } B \text{ modulo } C.$$

A Wikipedia disambiguation page says this informally means $A$ is the same as $B$ except for differences explained by $C$. But it can mean that the difference between $A$ and $B$ is in $C$, and things in $C$ are treated as equivalent to zero. That's what it means in quotient groups: the "difference" between $A$ and $B$ is $AB^{-1}$ or $BA^{-1}$, and it is in the group $C$. That's what it means in modular arithmetic: when one says that $(53\equiv98)\bmod 5$ it means the difference between $53$ and $98$ is in the set of all multiplies of $5$, which is a subgroup of the group of integers with addition.


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