factor group

$(\tilde{a}\circ \tilde{b})\circ N=(a\circ b)\circ NN⊴(G,\circ )$ … generalizing modulo arithmetic …

Different representations of the same coset/element in the factor group.

For the abelian group $(\mathbb{Z},+)$ and $m\ge 1$, its $m$ cosets area normal subgroup $m\mathbb{Z}⊴\mathbb{Z}$.
Looking at the concrete case of ${\mathbb{Z}}_4$, we have the following congruence classes:
$0+4\mathbb{Z}=\bar{0}=4\mathbb{Z}$ (note: $\bar{0}$ is the neutral element of the factor group)
$1+4\mathbb{Z}=\bar{1}$
$2+4\mathbb{Z}=\bar{2}$
$3+4\mathbb{Z}=\bar{3}$
We can do arithmetic with these classes, e.g.: $(1+4\mathbb{Z})+(2+4\mathbb{Z})=3+4\mathbb{Z}$
This is the same as the addition in ${\mathbb{Z}}_4$, adding the elements and taking the remainder modulo 4.
Same as above, but a different representation: $(17+4\mathbb{Z})+(38+4\mathbb{Z})=55+4\mathbb{Z}$
The reason we can do this is because $4\mathbb{Z}⊴\mathbb{Z}$.

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For any group $(G,\circ )$ with a normal subgroup $N⊴G$ and the coses $a\circ N,a\in G$ (left and right coset equiv).
Now (like above) we define the same operation on the cosets as on the group, i.e. $(a\circ N)\circ (b\circ N)=(a\circ b)\circ N$
We can now take this concept of modulo arithmetic and generalize it it to any group with a normal subgroup.
$\tilde{a}\in a\circ N=\tilde{a}\circ N$, $\tilde{b}\in b\circ N=\tilde{b}\circ N$
$(\tilde{a}\circ N)\circ (\tilde{b}\circ N)=(a\circ b)\circ N$

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Proof: $(\tilde{a}\circ \tilde{b})\circ N=(a\circ b)\circ N$
$\exists n_1,n_2,n_3,n_4\in N:$
$\tilde{a}=a\circ n_1=n_2\circ a$ ($n_1,n_2$ are representatives of the same coset)
$\tilde{b}=b\circ n_3=n_4\circ b$
$\tilde{a}\circ \tilde{b}=a\circ n_1\circ b\circ n_3=\underset{u\in N\circ (a\circ b)=(a\circ b)\circ N}{\underbrace{n_2\circ a\circ b}}\circ n_3=\dots$
$(⟹\exists n_5\in N:u=a\circ b=n_5)$
$\cdots =a\circ b\circ n_5\circ n_3\in (a\circ b)\circ N$ (since $n_5\circ n_3\in N$ and $N$ being a subgroup and hence closed).
$⟹(\tilde{a}\circ \tilde{b})\circ N=(a\circ b)\circ N$ $✓$

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Quotient group (or factor group) $(G/N,\circ )$

Let $G$ be a group and $N$ a normal subgroup of $G$ (written $N⊴G$).
The set $G/N:=\{a\circ N\mid a\in G\}$ is the set of all (left=right) cosets of $N$ in $G$ (read “G modulo N” or “G over N”).
The factor group is $(G/N,\circ )$ with the operation:

$$(a\circ N)\circ (b\circ N):=(a\circ b)\circ N$$

The factor group does not need to be commutative, and isn’t if $G$ isn’t.

Group properties of the factor group

closure: $(a\circ N)\circ (b\circ N)=(a\circ b)\circ N$ is always in $G/N$.
associativity: Inherited from $G$ since we’re just working with representatives.
neutral element: $e\circ N=N$ since $(e\circ N)\circ (a\circ N)=a\circ N$ for all cosets.
inverse element: $a^{-1}\circ N$ is inverse to $a\circ N$ since $(a\circ N)\circ (a^{-1}\circ N)=N$.

Example factor group

Consider the integers under addition $G=(\mathbb{Z},+)$ and the normal subgroup $N=n\mathbb{Z}$ of multiples of $n$.
The quotient group $(\mathbb{Z}/n\mathbb{Z},+)$ aka $({\mathbb{Z}}_n,+)$ consists of the following cosets:

$$\begin{array}{rl}\mathbb{Z}/n\mathbb{Z}& =\{0+n\mathbb{Z},1+n\mathbb{Z},\dots ,(n-1)+n\mathbb{Z}\}\\ & =\{\bar{0},\bar{1},\dots ,\overline{n-1}\}\\ & ={\mathbb{Z}}_n\end{array}$$

Note: This quotient group is isomorphic to the cyclic group of order $n$, generated by the congruence class $\bar{1}$: $C_n=⟨\bar{1}⟩$

For instance, in $(\mathbb{Z}/3\mathbb{Z},+)$ : $(1+3\mathbb{Z})+(2+3\mathbb{Z})=3+3\mathbb{Z}=0+3\mathbb{Z}$ $⟺$ arithmetic in ${\mathbb{Z}}_3$.

Interpretation

The quotient group identifies elements that differ by elements of $N$. Each element of $G/N$ represents an entire coset of $N$, effectively treating elements that differ by elements of $N$ as equivalent.

For finite groups, $|G/N|=[G:N]=|G|/|N|$ by Lagrange's Theorem.

The kernel of a homomorphism $f:G\to H$ determines a quotient group $G/\ker (f)$ that is isomorphic to the image of $f$:

First Isomorphism Theorem

For any homomorphism $\varphi :G\to H$:

$$G/\ker (\varphi )\cong \text{im}(\varphi )$$

The set of all cosets of the kernel in the group $G$ is isomorphic to the image of $\varphi$.

This shows that after “collapsing/factoring out” elements that map to the identity (the kernel), the factor group structure exactly matches the image.

Proof $\psi :G/\ker (\varphi )\to \varphi (G)$ is an isomorphism We need to show that

1. $\psi$ is well-defined: $\psi (a\ker (\varphi ))=\varphi (a)$ (can’t have elements from the same coset $G$ point to different elements in $H$)
2. $\psi$ is surjective
3. $\psi$ is injective
4. $\psi$ is a homomorphism

Ad 1: We need to show that if $a\circ \ker \varphi =b\circ \ker \varphi$, then $\psi (a\circ \ker \varphi )=\varphi (a)$ and $\psi (b\circ \ker \varphi )=\varphi (b)$ are the same.
This would mean that there is an $x\in a\circ \ker \varphi ⟹\exists k_1,k_2\in \ker \varphi :x=a\circ k_1=b\circ k_2$
$\varphi (a)=\varphi (a)\star e_H=\varphi (a)\star \varphi (k_1)=\varphi (a\circ k_1)=\varphi (b\circ k_2)=\varphi (b)\star \varphi (k_2)=\varphi (b)\star e_H=\varphi (b)\boxed{}$
Note: $k_1,k_2$ when mapped become the identity in $H$ as they are in the kernel.

Ad 2: Surjectivity is clear as $\psi$ maps all cosets to elements in the image: $x\in \varphi (G):\exists a\in G:\varphi (a)=x⟹\psi (a\circ \ker \varphi )=\varphi (a)=x$
Note: $\psi (a\circ \ker \varphi )=\varphi (a)$ is the definition of $\psi$: $\psi$ maps cosets to elements in the image!

Ad 3: For injectivity we want to show that: $\psi (a\circ \ker \varphi )=\psi (b\circ \ker \varphi )⟺\varphi (a)=\varphi (b)$
In words: The $\psi$-image of a coset is the same as the $\varphi$-image of the representative of that coset (representative = any element in the coset).
$⟹\varphi (a\circ b^{-1})=\varphi (a)\star \varphi (b{)}^{-1}=e_H⟹a\circ b^{-1}\in \ker \varphi ⟹a\in b\circ \ker \varphi ⟹a\circ \ker \varphi =b\circ \ker \varphi$
Note: $a\circ b^{-1}=x\in \ker \varphi ⟹a=x\circ b\in \ker \varphi \circ b⟹a\in \ker \varphi \circ b$ (and we can wrhite $b\circ$ – from the left – since the kernel is normal)
→ If the images are the same the preimages are the same.

Ad 4: $\psi (\underset{(a\circ b)\circ \ker \varphi }{\underbrace{(a\circ \ker \varphi )\circ (b\circ \ker \varphi )}})=\varphi (a\circ b)=\varphi (a)\star \varphi (b)=\psi (a\circ \ker \varphi )\star \psi (b\circ \ker \varphi )\boxed{}$

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