MATH 415 Lecture 4

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Subgroups

$H$ is a subgroup of $G$ (denoted $H\leq G$ ) if:

$H$ is closed with respect to $*$
$e\in H$
$a^{-1}\in H$ for all $a\in H$

We say $H$ is a proper subgroup of $G$ (denoted $H<G$ ) if $H\neq G$ .

Cyclic Subgroups

We call $\left\langle a\right\rangle$ a cyclic subgroup of a group $\left\langle G,*\right\rangle$ generated by $a$ , where $a$ is the generator.

We call a group $\left\langle G,*\right\rangle$ cyclic if there is an element $a\in G$ such that $G=\left\langle a\right\rangle$ .

If we take an arbitrary group $\left\langle G,\cdot \right\rangle$ and an element $a\in G$ ,

We can compute a subset $H$ containing identity element $a^{0}$ , $a$ , $a*a=a^{2}$ , $a*a*a=a^{3}$ , and so on up to $a^{n}$ . Then invert all of these to find $a^{-n}$ :

H=\left\{a^{n}~\mid ~n\in \mathbb {Z} \right\}\,\!

This is a subgroup generated by $a$ , with notation $H=\left\langle a\right\rangle$ .

Theorem. $H=\left\langle a\right\rangle$ is the smallest subgroup of $G$ containing $a$ .

Examples on Integers

It's worth noting that $\mathbb {Z} =\left\langle 1\right\rangle =\left\langle -1\right\rangle$ .

All subgroups of $\left\langle \mathbb {Z} ,+\right\rangle$ are of the form $n\,\mathbb {Z}$ . When $n\neq \pm 1$ , then $n\,\mathbb {Z} <\mathbb {Z}$ .

Consider $\left\langle \mathbb {Z} _{12},+\right\rangle$ :

$\left\{0,1,2,3,4,5,6,7,8,9,10,11\right\}$ , with $0$ as identity.

Consider subgroup $\left\langle 3\right\rangle =H=\left\{0,3,6,9\right\}$ , a subgroup of $\mathbb {Z} _{12}$

We call $\left\langle 3\right\rangle$ a subgroup generated by 3.

Examples on Complex Numbers

Recall the Roots of Unity: $U_{n}$ is a cyclic subgroup generated by $\zeta$ .

Theorem 6.1. Every cyclic group is Abelian.

Proof. For $g,h\in G=\left\langle a\right\rangle$ , we have $g=a^{n}$ and $h=a^{m}$ . $a^{m}\,a^{n}=a^{n}\,a^{n}=a^{m+n}$ and is thus commutative.

quod erat demonstrandum

Division Algorithm for Integers

For all $m\in \mathbb {Z} ^{+}$ and for all $n\in \mathbb {Z}$ , there exist $q\in \mathbb {Z}$ and $r\in \mathbb {Z}$ such that

n=m\,q+r\quad \quad 0\leq r<m\,\!

Where $q$ is the quotient and $r$ is the remainder.

Theorem 6.6

A subgroup of a cyclic group is cyclic.

Proof. Let $G=\left\langle a\right\rangle$ be a cyclic subgroup, and let $H\leq G$ .

If $H=\left\{e\right\}$ (trivially cyclic), we're done.

Otherwise, for some element $e\neq c\in H$ , we know $c=a^{m}$ for some smallest positive integer $m$ .

The claim is that $H=\left\langle a^{m}\right\rangle$ .

Let's pick an element $b\in H$ . We know that $b=a^{n}$ for some $n$ . We can write $n$ via the division algorithm: $n=m\,q+r$ . Therefore $b=a^{n}=a^{m\,q+r}=(a^{m})^{q}\cdot a^{r}$ . Since $r<m$ , we must assume $r=0$ since $m$ was defined to be smallest positive integer.

From this, we get "All subgroups of $\left\langle \mathbb {Z} ,+\right\rangle$ are of the form $n\,\mathbb {Z}$ " as a corollary.

Greatest common divisor also comes as a result of this.

Greatest Common Divisor

Exercise 45 in the book (for homework) asks us to prove $H=\left\{n\,r+m\,s~\mid ~n,m\in \mathbb {Z} \right\}$ is a subgroup of $\mathbb {Z}$ . For this we know $H$ is cyclic (because $\mathbb {Z}$ is cyclic), and its generator must be of the form $H=\left\langle d\right\rangle$ for generator $d=\mathrm {gcd} (r,s)$ . Therefore $d=n\,r+m\,s$ for some $r,s\in \mathbb {Z}$ .

Coprimality

We say that two integers $r$ and $s$ are coprime (or relatively prime) if $\mathrm {gcd} (r,s)=1$ . Thus $1=m\,r+n\,s$ for some $m,n\in \mathbb {Z}$ .

If

r

divides a product of integer factors, it must divide at least one of the factors.

Theorem 6.10

Classification of Cyclic Groups

Let $G=\left\langle a\right\rangle$ be cyclic.

If $|G|=\infty$ , then $G\simeq \mathbb {Z}$ .
If $|G|=n<\infty$ , then $G\simeq \mathbb {Z}$

Proof. For all $m\in \mathbb {Z} ^{+}$ , $a^{m}\neq e$ . We claim that $a^{m}\neq a^{n}$ if $m\neq n$ because $a^{m}=a^{n}$ implies $a^{m}\cdot (a^{n})^{-1}=a^{m-n}=e$ . But for $m>n$ , this is not true.

Now let $\phi :G\to \mathbb {Z}$ be defined as $\phi (a^{m})=m$ . $\phi$ is one-to-one because $a^{m}\neq a^{n}\iff m\neq n$ . $\phi$ is also onto. Finally, $\phi (a^{m}\cdot a^{n})=\phi (a^{m})+\phi (a^{n})$ .

Similarly, a group $G=\{e,a,a^{2},\ldots ,a^{n-1}\}$ corresponds to $\mathbb {Z} =\{0,1,2,\ldots ,n-1\}$ .

quod erat demonstrandum

Consequently, $G\simeq \mathbb {Z} _{n}\simeq U_{n}$ . This is why these are called cyclic groups

Structure of Cyclic Subgroups

We already know that $\mathbb {Z}$ has only subgroups $n\,\mathbb {Z}$ . In particular, if $m\,\mathbb {Z} <n\,\mathbb {Z}$ , then $n\mid m$ .

Theorem 6.14

Let $|G|=n$ , so $G\simeq \mathbb {Z} _{n}$ . Furthermore, let $G=\left\langle a\right\rangle$ . Let $b=a^{s}\in G$ . And finally, let $H=\left\langle b\right\rangle \leq G$ .