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...]
  • $\begingroup$ What's the def of $G/N$ you use? $\endgroup$ – coffeemath Jun 27 at 4:12
  • $\begingroup$ The definition I am using is: (Quotient Group) G/N is the group of all left cosets of N in G. @coffeemath $\endgroup$ – Aakash Singh Bais Jun 27 at 4:16
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    $\begingroup$ Then if a left coset is $a \circ N$ the map is clearly surjective, you already checked homomorphism. $\endgroup$ – 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 $\mathsf{Grp}$.

  • $\begingroup$ I am using: Epimorphism = Surjective homomorphism. Is anything wrong in this? $\endgroup$ – Aakash Singh Bais Jun 27 at 4:18
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    $\begingroup$ 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! $\endgroup$ – 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 $q(g)=\bar g=gN$

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