MATH 415 Lecture 11

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Factor Groups

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

We call $\left\langle G/H,*\right\rangle$ the factor group obtained from $G$ and the homomorphism $\phi$.

we define $*$ as the operation $(a\,H)*(b\,H)=(ab)\,H$

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

Similarly, we define $H\setminus G=\left\{H\,g~\mid ~g\in G\right\}$, and note that $G/H\neq H\setminus G$ in general.

Theorem 14.4

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

Let $H\leq G$. Then left coset multiplication is well-defined by the equation $(a\,H)*(b\,H)=(ab)\,H$ 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.

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

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 $a\,h_{1}\,b\,h_{2}=a\,b\,h_{3}\,h_{2}$. Since $H$ is closed on $*$, we can represent $h_{3}\,h_{2}$ as $h'\in H$, therefore $(a_{1}\,b_{1})H$ is equivalent to $(a\,b)H$.

quod erat demonstrandum

Corollary. If $H\trianglelefteq G$, 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

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

What is the quotient group $G/H$?

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

The Fundamental Homomorphism Theorem

First, we need Theorem 14.9. Let $H\trianglelefteq G$. 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 $\gamma (x\,y)=\gamma (x)\,\gamma (y)$. By definition $\gamma (x\,y)=(x\,y)H=(xH)(yH)$. 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 $\left\{x\in G~\mid ~\gamma (x)=e_{G/H}\right\}$. 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 $\mu (gH)\simeq \phi (g)$

Now we can state that $\phi =\mu \circ \gamma$ since $\phi :G\to G'$ (in particular, $\phi [G]\subseteq G'$), $\mu :G/H\to \phi [G]$, and $\gamma :G\to G/H$.

Theorem 14.11. [The Fundamental Homomorphism Theorem]. Let $\phi :G\to G'$ be a group homomorphism with kernel $H$. Then $\phi [G]$ is a group, and $\mu :G/H\to \phi [G]$ given by $\mu (gH)=\phi (g)$ is an isomorphism. If $\gamma :G\to G/H$ is the homomorphism given by $\gamma (g)=gH$, then $\phi (g)=(\mu \circ \gamma )(g)$ for each $g\in G$.

Example

$G=\mathbb {Z} _{4}\times \mathbb {Z} _{2}$, and $H=\left\{0\right\}\times \mathbb {Z} =\left\{\left(0,x\right)~\mid ~x\in \mathbb {Z} _{2}\right\}$.

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

$G/H\simeq \mathbb {Z} _{4}/\left\{0\right\}\times \mathbb {Z} _{2}/\mathbb {Z} _{2}$ for a very specific reason: $G=K\times L$ and $H=K_{1}\times L_{1}$ such that $K_{1}\leq K$ and $L_{1}\leq L$, then $G/H=K/K_{1}\times L/L_{1}$.

Back to the example, we have $\mathbb {Z} _{4}/\left\{0\right\}=\mathbb {Z} _{4}$ and $\mathbb {Z} _{2}/\mathbb {Z} _{2}=\left\{0\right\}$ because

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

And now $\mathbb {Z} _{2}\times \left\{0\right\}=\mathbb {Z} _{2}$ 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. $g\,h\,g^{-1}\in H$ for all $g\in G$, $h\in H$.
2. $g\,H\,g^{-1}=H$ for all $g\in G$.
3. $gH=Hg$ for all $g\in G$.

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

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

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

Automorphisms

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

• It is closed over composition since if $\phi :G\to G$ and $\psi :G\to G$ are isomorphisms, then $\phi \circ \psi :G\to G$ 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)