Equivalence class as a homomorphism from a group to its quotient group

Question

Given a normal subgroup $N$ of group $G$, the quotient group $G/N$is defined by the usual group multiplication $$[a]\star [b]=[a \cdot b]$$ where $\star$ is the group multiplication in $G/N$ and $\cdot$ is the group multiplication in $G$, and $[\cdot]$ denotes the equivalence classes, or the elements of $G/N$ defined by $$[a]=\{g\in G | g\cdot a^{-1} \in N\}$$.

I want to know that if the sense that $[\cdot] : G \rightarrow G/N$ can be viewed as a homomorphism from $G$ to $G/N$ is correct.

Answer 1

Can't you just check this directly? Call your map $\phi: G \to G/N$. Then for any $a, b \in G$, $$ \phi(ab) = abN = aN bN = \phi(a)\phi(b). $$ Or, if you want to stick with your class-box notation, $$ \phi(ab) = [ab] = [a][b] = \phi(a)\phi(b). $$ This follows directly from how we define the operation on cosets.

Answer 2

Yes, it can; it is usually denoted the quotient homomorphism or canonical projection, let me call it $p$. In fact, it already completely determines $H$, as $H\cong \operatorname{ker}(p)$. This can be seen as a reason that instead of viewing normal subgroups and quotients by those as the fundamental construction, one can more abstractly (and in greater generality) use such morphisms as the underlying construction, as for an arbitrary morphism, there is an easy condition that tells us whether it arises as a projection. This is an important approach taken in category theory.

Answer 3

This is also called Natural or Canonical Homomorphism. when you have N as a Normal Subgroup of G then ${G/N}$ is called Factor or Quotient Group.

• The map ${f:g\rightarrow gN}$ is called surjective group homomorphism, called canonical projection, as mentioned by Markus and Michael in his comment. In your case ${g \cdot a^{-1} \in N}$ as mentioned in above mapping.
• The surjective map from ${G}$ onto ${G/N}$, which maps each element to its equivalence class, is called the canonical surjection or the canonical projection map
• To prove homomorphism property you have to remember that product of two right cosets of Normal subgroup is again a right Coset.