Quotient Groups

Cosets of a normal N form G/N, of order |G|/|N|.

The idea

Sometimes you want to stop distinguishing certain elements — to declare a whole subgroup $N$ to be the identity and ask what group is left. Doing that means collapsing each set $aN = \{an : n \in N\}$, the coset of $a$, down to a single point. A coset is a shifted copy of $N$ sitting inside $G$, and the cosets partition $G$ into pieces all of the same size as $N$. The figure shows this partition: the cosets $gH$ of a subgroup $H$ cut $G$ into equal stripes, and with $H = N$ quotienting collapses each stripe to a single element of $G/N$.

For a group to survive the collapse, multiplication has to be well defined on the pieces: the product of the coset holding $a$ with the coset holding $b$ must not depend on which representatives you happened to pick. That is what fails for an arbitrary subgroup and what normality buys — $gN = Ng$ for every $g \in G$, which is automatic when $G$ is abelian.

Definition (Quotient group).

Let $N$ be a normal subgroup of $G$. The quotient group $G/N$ is the set of cosets of $N$ in $G$ with the multiplication $(aN)(bN) = abN.$ Its identity is the coset $N$ itself.

Proposition.

If $G$ is finite and $N$ is a normal subgroup, then $|G/N| = \dfrac{|G|}{|N|}$.

The count is free because the cosets are equal in size and partition $G$. The construction is the group-theoretic version of arithmetic mod $n$, where the multiples of $n$ are declared to be zero, and it is how a complicated group gets replaced by a smaller one that remembers only the features you care about.

Ways to work on it

• Walkthrough. Build G/N and compute its order and structure.
• Practice. Order of a quotient group.
• Hardest. G/Z(G) cyclic forces G abelian.

Not sure where to start? Take the ten-question placement test.