# Let G be a group, and $N\triangleleft G$. Then, the quotient map $q: G\rightarrow G/N$ given by $q(g) = g\circ N$ is an epimorphism.

My attempt:

1. Showing that the given map is a homomorphism. Let $$a, b ∈ G$$ Then, $$q(a\circ b) = (a\circ b)\circ N = (a\circ N)\circ(b\circ N) = q(a)\circ q(b)$$.
2. Showing surjectivity [I am stuck up here...]
• What's the def of $G/N$ you use? – coffeemath Jun 27 at 4:12
• The definition I am using is: (Quotient Group) G/N is the group of all left cosets of N in G. @coffeemath – Aakash Singh Bais Jun 27 at 4:16
• Then if a left coset is $a \circ N$ the map is clearly surjective, you already checked homomorphism. – coffeemath Jun 27 at 4:29

Pick an element (coset) $$g N \in G/N$$. Then just use $$g \in G$$ (the coset representative) and $$g \mapsto g N$$ to give us surjectivity.

Also, I presume you are using epimorphism and surjection interchangeably in the setting of groups, but just for completeness, in the category of groups, we have that a morphism is a surjection if and only if it is an epimorphism, so the fact that this is a surjection shows that it is an epimorphism in $$\mathsf{Grp}$$.

• I am using: Epimorphism = Surjective homomorphism. Is anything wrong in this? – Aakash Singh Bais Jun 27 at 4:18
• No nothing to worry about because in the category of groups, the two notions are the same. They are not the same in other categories however! – mathphys Jun 27 at 4:19
1. Let $$a,b \in G$$. Then $$q(ab)=abN=(aN)(bN)=q(a)q(b).$$ That is, $$q$$ is a group homomorphism.

(Since $$N$$ is a normal subgroup of $$G$$,$$(aN)(bN)=abN$$ multiplication is well-defined.

Because let $$x\in(aN)(bN)$$. Then $$x=an_1 bn_2$$ for some $$n_1 , n_2 \in N$$

Since $$N$$ is normal, $$gN=Ng$$ for all $$g\in G$$. Then $$x=an_1 bn_2=a(n_1 b)n_2 =a(b {n'}_1)n_2 =ab{n'}_1 n_2 \in abN$$.

Conversely, choosing $$a\in aN$$ and $$b \in bN$$, we obtain the coset $$abN$$.)

1. Let $$\bar g \in G/N$$. Then $$\bar g = gN$$ for some $$g\in G$$. We choose $$g$$. Then $$q(g)=\bar g=gN$$