MATH 415 Lecture 11

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Factor Groups

Theorem 14.1. Let $\varphi :G\to G^{\prime }$ be a homomorphism and $H$ be the kernel of $\varphi$. We denote $G/H$ as the set of left cosets (identical to right cosets since $H$ is the kernel).

We call $⟨G/H,*⟩$ the factor group obtained from $G$ and the homomorphism $\varphi$.

we define $*$ as the operation $(aH)*(bH)=(ab)H$

The function $\mu :G/H\to \varphi [G]$ is an isomorphism on the image of $\varphi$.

Similarly, we define $H\setminus G=\{Hg\mid g\in G\}$, and note that $G/H\ne H\setminus G$ in general.

Theorem 14.4

Recall that a subgroup $H⊴G$ is normal if its left and right cosets coincide.

Let $H\le G$. Then left coset multiplication is well-defined by the equation $(aH)*(bH)=(ab)H$ if and only if $H$ is a normal subgroup of $G$.

Proof. Suppose that $*$ does not give a well-defined binary operation; so for $a\in G$, is it true that $aH=Ha$? Let $x\in aH$, $a\in aH$, and $a^{-1}\in a^{-1}H$. By definition $x$ and $a$ are from the same coset.

$\begin{array}{rl}(xH)(a^{-1}H)& =(x{a}^{-1})H\\ (aH)(a^{-1}H)& =eH=H\\ (xH)(a^{-1}H)& =H\\ x{a}^{-1}& =h\in H\\ x& =ha\end{array}$

Therefore, $x\in Ha$, when we have assumed it to be a member of a left coset. Therefore, $aH\subseteq Ha$ and $Ha\subseteq aH$ imply that $aH=Ha$ and thus $H$ is normal.

So if $H$ is normal in $G$, let's take a look at the operation $(aH)(bH)=(ab)H$. Is it true that if $a_1\in aH$ and $b_1\in bH$, then $(a_{1}H)(b_{1}H)=(a_1{b}_1)H=(ab)H$?

Using our definition of

$a_1$

and

$b_1$

, we arrive at the following:

$a_1{b}_1=a{h}_{1}b{h}_2$

. If we look at

$h_{1}b$

, we deduce that

$h_{1}b$

must be equivalent to

$b{h}_3$

for some

$h_3$

since

$H$

is normal and its cosets coincide. Therefore

$a{h}_{1}b{h}_2=ab{h}_3{h}_2$

. Since

$H$

is closed on

$*$

, we can represent

$h_3{h}_2$

as

$h^{\prime }\in H$

, therefore

$(a_1{b}_1)H$

is equivalent to

$(ab)H$

.

quod erat demonstrandum

Corollary. If $H⊴G$, then the cosets of $H$ form a group $G/H$ under the binary operation $(aH)(bH)=(ab)H$.

• Associativity: $((aH)(bH))(cH)=((ab)H)(cH)=((ab)c)H=(a(bc))H=(aH)((bc)H)=(aH)((bH)(cH))$
• Identity: $e=eH=H$, which is the ID element in $G/H$
• Inverse: $(aH{)}^{-1}=a^{-1}H$, so $(aH)(a^{-1}H)=(a{a}^{-1})H=eH=H=id$

Examples

$G=\mathbb{\mathbb{Z}}$, and $n\mathbb{\mathbb{Z}}=H⊴G$

What is the quotient group $G/H$?

$G/H=\mathbb{\mathbb{Z}}/n\mathbb{\mathbb{Z}}$. This group is exactly isomorphic to ${\mathbb{\mathbb{Z}}}_n$ on addition mod $n$.

The Fundamental Homomorphism Theorem

First, we need Theorem 14.9. Let $H⊴G$. Then $\gamma :G\to G/H$ given by $\gamma (x)=xH$ is a homomorphism with the kernel $H$.

Proof. For $x,y\in G$, we need to check that $\gamma (xy)=\gamma (x)\gamma (y)$. By definition $\gamma (xy)=(xy)H=(xH)(yH)$. Observe that the first factor is $\gamma (x)$ and the second is $\gamma (y)$. Therefore it is a homomorphism.

What about kernel? By definition, the kernel is

$\{x\in G\mid \gamma (x)=e_{G/H}\}$

. Now

$\gamma (x)=xH$

. Since

$x\in H$

, we get

$xH=eH$

, so the kernel of

$\gamma$

is

$H$

. Furthermore, we can say that

$\gamma$

is onto.

quod erat demonstrandum

From before, we have $\mu (gH)\simeq \varphi (g)$

Now we can state that $\varphi =\mu \circ \gamma$ since $\varphi :G\to G^{\prime }$ (in particular, $\varphi [G]\subseteq G^{\prime }$), $\mu :G/H\to \varphi [G]$, and $\gamma :G\to G/H$.

Theorem 14.11. [The Fundamental Homomorphism Theorem]. 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 $\varphi (g)=(\mu \circ \gamma )(g)$ for each $g\in G$.

Example

$G={\mathbb{\mathbb{Z}}}_4\times {\mathbb{\mathbb{Z}}}_2$, and $H=\left\{0\right\}\times \mathbb{\mathbb{Z}}=\{(0,x)\mid x\in {\mathbb{\mathbb{Z}}}_2\}$.

Note $\left|H\right|=2$ and $\left|G\right|=8$.

$G/H\simeq {\mathbb{\mathbb{Z}}}_4/\left\{0\right\}\times {\mathbb{\mathbb{Z}}}_2/{\mathbb{\mathbb{Z}}}_2$ for a very specific reason: $G=K\times L$ and $H=K_1\times L_1$ such that $K_1\le K$ and $L_1\le L$, then $G/H=K/K_1\times L/L_1$.

Back to the example, we have ${\mathbb{\mathbb{Z}}}_4/\left\{0\right\}={\mathbb{\mathbb{Z}}}_4$ and ${\mathbb{\mathbb{Z}}}_2/{\mathbb{\mathbb{Z}}}_2=\left\{0\right\}$ because

• $\left\{e\right\}⊴G$ and $G⊴G$
• $G/\left\{e\right\}=G$
• $G/G=\left\{e\right\}$

And now ${\mathbb{\mathbb{Z}}}_2\times \left\{0\right\}={\mathbb{\mathbb{Z}}}_2$ because $G\times \left\{e\right\}=G$.

Theorem 14.13

The following are three equivalent conditions for a subgroup $H$ of a group $G$ to be a normal subgroup of $G$.

1. $gh{g}^{-1}\in H$ for all $g\in G$, $h\in H$.
2. $gH{g}^{-1}=H$ for all $g\in G$.
3. $gH=Hg$ for all $g\in G$.

Note: The transformation $h\to gh{g}^{-1}$ is called the conjugation of $h$ by $g$.

Assume $H⊴G$, so 3 holds. Let $g\in G$, $h\in H$.

$gh{g}^{-1}=h_1$, so $gh=h_{1}g$ which holds by the assumption.

Automorphisms

Let $\mathrm{A}\mathrm{u}\mathrm{t}(G)$ be the group of automorphisms, or isomorphisms of a group with itself, over the composition operation.

• It is closed over composition since if $\varphi :G\to G$ and $\psi :G\to G$ are isomorphisms, then $\varphi \circ \psi :G\to G$ will be an isomorphism as well.
• We already know that composition is associative
• The identity element will be the identity isomorphism (mapping elements in $G$ to themselves)