# Math4302 Modern Algebra (Lecture 8)

## Subgroups

### Cyclic group

#### Subgroup of cyclic group is cyclic

Every subgroup of a cyclic group is cyclic.

#### Order of subgroup of cyclic group

If $a\in G$ and $|\langle a\rangle|$ be the smallest positive $n$ such that $a^n=e$, then $\langle a\rangle=\{e,a,a^2,\cdots,a^{n-1}\}$ and $a^{m_1}=a^{m_2}\iff m_1=m_2\mod n$. ($n$ divides $m_1-m_2$)

#### Size of subgroup of cyclic group

Let $G=\langle a\rangle$ and $H=\langle a^m\rangle$. Then $|H|=\frac{|G|}{d}$ where $d=\operatorname{gcd}(|G|,|H|)$. In particular, $\langle a^m\rangle=G\iff \operatorname{gcd}(n,m)=1$.

#### GCD decides the size of subgroup

Suppose $G=\langle a\rangle$, $|G|=n$.

Then $\langle a^{m_1}\rangle=\langle a^{m_2}\rangle\iff \operatorname{gcd}(n,m_2)=\operatorname{gcd}(n,m_1)$.

<details>
<summary>Proof</summary>

$\implies$:

$\langle a^{m_1}\rangle=\langle a^{m_2}\rangle\implies \operatorname{gcd}(n,m_1)=\operatorname{gcd}(n,m_2)$

$\impliedby$:

Suppose $d=\operatorname{gcd}(n,m_1)=\operatorname{gcd}(n,m_2)$.

Enough to show $a^{m_1}\in \langle a^{m_2}\rangle$. (then we conclude $\langle a^{m_1}\rangle=\langle a^{m_2}\rangle$ and by symmetry $\langle a^{m_2}\rangle=\langle a^{m_1}\rangle$.)

Equivalent to show that $a^{m_1}=(a^{m_2})^k$ for some integer $k$. That is $n$ divides $m_1-km_2$ for some $k\in \mathbb{Z}$.

From last lecture, we know that $d$ can be written as $d=nr+m_2 s$ for some $r,s\in \mathbb{Z}$.

Multiply by $\frac{m_1}{d}$, (since $d$ divides $m_1$, this is an integer).

So $m_1=nr\frac{m_1}{d}+m_2s\frac{m_1}{d}$.

Therefore $n$ divides $m_1-(\frac{m_1}{d}s)m_2$, so $k=\frac{m_1}{d}s$. works.

</details>

#### Corollaries for subgroup of cyclic group

Let $G=\langle a\rangle$ be a cyclic group of finite order.

1. If $H\leq G$, then $|H|$ is a divisor of $|G|$. (More generally true for finite groups.)
2. For any $d$ divides $|G|$, there is exactly one subgroup of $G$ of order $d$. $\langle a^m\rangle$ where $m=\frac{|G|}{d}$.

<details>
<summary>Examples</summary>

$(\mathbb{Z}_18,+)$.

The subgroup with size $6$ is $\langle 3\rangle=\{0,3,6,9,12,15\}=\langle 15\rangle$.

Note that $\operatorname{gcd}(18,3)=3=\operatorname{gcd}(18,15)$.

$\langle 6\rangle=\{0,6,12\}$.

$\langle 9\rangle=\{0,9\}$.

$\langle 2\rangle=\{0,2,4,6,8,10,12,14,16\}$ (generators are $2,4,8,10,14,16$ since they have gcd $2$ with $18$).

</details>

### Non-cyclic groups

Let $G$ be a group and $a,b\in G$, then we use $\langle a,b\rangle$ to mean the subgroup of $G$ generated by combination of $a$ and $b$.

$$
\langle a,b\rangle\coloneqq \{e,a,b,ab,ba,a^{-1},b^{-1},(ab)^{-1},(ba)^{-1},\ldots\}
$$

This is a subgroup of $G$ since it is closed and $e=a^0$.

#### Klein 4 group

Klein 4 group is abelian but not cyclic.

|*|e|a|b|c|
|--|---|---|---|---|
|e|e|a|b|c|
|a|a|e|c|b|
|b|b|c|e|a|
|c|c|b|a|e|

The subgroups are

$\langle e\rangle=\{e\}$

$\langle a\rangle=\{e,a\}$

$\langle b\rangle=\{e,b\}$

$\langle c\rangle=\{e,c\}$

Therefore $G$ is **not cyclic** and **not isomorphic** to $\mathbb{Z}_4$.

Here $G=\langle a,b\rangle=\{e,a,b,ab=c\}$.

More generally, if we have $a_i\in G$, where $i\in I$, then $\langle a_i,i\in I\rangle=$ all possible combinations of $a_i$ with their inverses. Is a subgroup of $G$.

Another way to describe is that $\langle a_i,i\in I\rangle=\bigcap_{H\leq G, a_i\in H,i\in I}H$.

#### Definition of finitely generated group

If $G$ is a group and if there is a finite set $a_1,\ldots, a_n\in G$ such that $G=\langle a_1,\ldots, a_n\rangle$, then $G$ is called finitely generated.

<details>
<summary>Examples</summary>

Any finite group is finitely generated.

---

$(\mathbb{Q},+)$ is not finitely generated.

Suppose for the contrary, there is a finite set $\frac{a_1}{b_1},\ldots,\frac{a_n}{b_n}\in \mathbb{Q}$ such that 

$$
\mathbb{Q}=\langle \frac{a_1}{b_1},\ldots,\frac{a_n}{b_n}\rangle=\{t_1\frac{a_1}{b_1},\ldots,t_n\frac{a_n}{b_n}|t_1,t_2,\ldots,t_n\in \mathbb{Z}\}
$$.

Pick prime $p$ such that $p>|b_1|,\ldots,|b_n|$. Then $\frac{1}{p}\in \mathbb{Q}$.

$$
\frac{1}{p}=t_1\frac{a_1}{b_1}+t_2\frac{a_2}{b_2}+\cdots+t_n\frac{a_n}{b_n}=\frac{A}{b_1b_2\cdots b_n}
$$

This implies that $pA=b_1b_2\cdots b_n$.

Since $p$ is prime, $p|b_i$ for some $i$.

However, by our construction, $p>|b_i|$ and cannot divide $b_i$.

Contradiction.

</details>
