Section 14 Factor Groups

14.1 Definition Factor Group

Let $H$ be a normal subgroup of $G$ We first divide $G$ into several partitions or formally cosets (written in the form $gH$ with $g\in G$, presenting $\{g\cdot h|h\in H\}$). Each partition has “size” $|H|$, and the set of all cosets forms a new group denoted as $G/H$. This group $G/H$ has

• Identity: $H$
• Inverse:$(gH{)}^{-1}=g^{-1}H$
• Operation: $(gH)\cdot (kH)=(gk)H$

From the view of homomorphism

Let $\varphi :G\to G^{\prime }$ Let $H=Ker(\varphi )$, $a\in G$ The set of all cosets of H is denoted by $G/H$ (read as G over H or G modulo H)

• There exists a one-to-one relationship between $G/H$ and the elements of the group $\varphi [G]$ . 13.15 Theorem
• $G/H$ and $\varphi [G]$ have same amount of elements. Thus there is a onto relationship 13.15 Theorem
• A homomorphism exist: Definition of Homomorphism $\mu :G/H\to \varphi [G]$ if $\mu (xH)=\varphi (x)$ and $\mu (yH)=\varphi (y)$ and $\mu (zH)=\varphi (x)\varphi (y)$ Then $(xH)(yH)=zH$ $G/H$ is isomorphic to $\varphi [G]$

Definition

$G/H$ is called as a factor group (quotient group) or more precisely factor group of G modulo H

Tip

H can also be any normal subgroup subgroup of $G$, otherwise we cannot form a factor group

14.2 Example

The factor Group $\mathbb{Z}/n\mathbb{Z}$ is isomorphic to ${\mathbb{Z}}_n$

Taking the example from 3Z Let $\mu :\mathbb{Z}/3\mathbb{Z}\to {\mathbb{Z}}_3$

• $\mu (3\mathbb{Z})=0$
• $\mu (3\mathbb{Z}+1)=1$
• $\mu (3\mathbb{Z}+2)=2$

Computation in factor group

find any element from two cosets. calculate using the operation in $G$, and the result lies in the coset that is the product of the original two cosets.

NOTE

congruent modulo H: Elements in the same coset of H

Let H be a subgroup of G $(aH)(bH)=(ab)H$ This is well defined(independent from the choice of $h\in H$) if and only if H is a normal subgroup of G

$(a{h}_1)(b{h}_2)=a(h_{1}b)h_2=a(b{h}_3)h_2=ab(h_3{h}_2)$

Tip

• $a(h_{1}b)h_2=a(b{h}_3)h_2$ because left coset is the same as the right coset (see normal group)
• $h_3{h}_2\in H$ because H is a group

The Fundamental Homomorphism Theorem

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=145&selection=166,1,264,1&color=note|p.140]]

Let $\varphi :G\to G^{{}^{\prime }}$ be a group homomorphism with kernel $H$ . Then $\varphi [G]$ is a group, and $\mu :G/H\to \varphi [G]$ given by $\mu (gH)=\varphi (g)$ is an isomorphism.

If $\gamma :G\to G/H$ is the homomorphism given by $\gamma (g)=gH$ , then $\varphi (g)=\mu \gamma (g)$ for each $g\in G$.

• $\gamma (x)=xH$
• $\mu (xH)=\varphi (x)$
• $\varphi (x)=\mu \gamma (x)$

Let H be a normal subgroup of G. Then $\gamma (x)=xH$ is a homomorphism with kernel H.

• Proof 14.9🔽

• $\mu$ is an isomorphism of G/H with $\varphi [G]$ 14.1 Definition Factor Group

• $\gamma$ is an homomorphism of G with G/H

14.12 Example

Classify the group $({\mathbb{Z}}_4\times {\mathbb{Z}}_2)/(\{0\}\times {\mathbb{Z}}_2)$

Warning

${\mathbb{Z}}_4\times {\mathbb{Z}}_2$ is not isomorphic to ${\mathbb{Z}}_8$ See Fundamental Theorem of Finitely Generated Abelian Groups

• Let G be ${\mathbb{Z}}_4\times {\mathbb{Z}}_2$, and H be $\{0\}\times {\mathbb{Z}}_2$
• Find a homomorphism $\varphi$ from $G$ to $\varphi [G]$ (Here to ${\mathbb{Z}}_4$) that has the Kernel $(\{0\}\times {\mathbb{Z}}_2)$.
• $\varphi :{\mathbb{Z}}_4\times {\mathbb{Z}}_2\to {\mathbb{Z}}_4$
• $\varphi (x,y)=x$
• According to The Fundamental Homomorphism Theorem ${\mathbb{Z}}_4$ is isomorphic to $({\mathbb{Z}}_4\times {\mathbb{Z}}_2)/(\{0\}\times {\mathbb{Z}}_2)$

A general construction method of classification

• Step 1: Write $G$ in its decomposition as a direct sum of cyclic factors:

$$G\cong {\mathbb{Z}}_{n_1}\oplus {\mathbb{Z}}_{n_2}\oplus \cdots \oplus {\mathbb{Z}}_{n_k},n_1\mid n_2\mid \cdots \mid n_k.$$

• Step 2: Describe how $H$ sits inside this decomposition, i.e., which factors or subcycles of factors it occupies.

• Step 3: Construct a projection (or combination of projections)

$$\varphi :G⟶K$$

by “discarding” the factors corresponding to $H$ and keeping the remaining factors, ensuring

$$\ker \varphi =H.$$

• Step 4: By the First Isomorphism Theorem, conclude

$$G/H\cong \varphi (G).$$

• Step 5: If needed, adjust $\varphi (G)$ into standard form as a direct sum of cyclic groups:

$$\varphi (G)\cong {\mathbb{Z}}_{d_1}\oplus {\mathbb{Z}}_{d_2}\oplus \cdots \oplus {\mathbb{Z}}_{d_r},$$

where $d_1\mid d_2\mid \cdots \mid d_r$.

14.14 Example

NOTE

Every subgroup H of an abelian group G is normal

14.15 Definition: Automorphism

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=146&selection=590,1,678,1&color=note|p.141]]

• An isomorphism $\phi :G\to G$ of a group $G$ with itself is an automorphism of $G$.
• The automorphism $i_g:G\to G$, where $i_g(x)=gx{g}^{-1}$ for all $x\in G$, is the inner automorphism of G by g.
• Performing $i_g$ on x is called conjugation of x by g.
• if $i_g[H]=H$ for all $g\in G$, H is a normal subgroup

• Definition of conjugate subgroup: Given a group $G$ and a subgroup $H\subseteq G$, for any $g\in G$ define

$$i_g[H]=gH{g}^{-1}=\{gh{g}^{-1}:h\in H\}.$$

The set $gH{g}^{-1}$ is a subgroup of $G$, called a conjugate subgroup of $H$.

• Isomorphism Property: The map $H⟶gH{g}^{-1},h\mapsto gh{g}^{-1}$ is a group isomorphism.

NOTE

Hence every conjugate subgroup is isomorphic to $H$.

See how Automorphism is defined in Category Theory: Definition: Automorphism

Ways to Prove $H$ is a Normal Subgroup - Summary

1. $gh{g}^{-1}\in H$ for all $g\in G$ and $h\in H$ .
2. $gH{g}^{-1}=H$ for all $g\in G$.
3. $i_g[H]=H$ for all $g\in G$
4. $gH=Hg$ for all $g\in G$. See Ways to Prove H is a Normal Subgroup