# Math4302 Modern Algebra (Lecture 6)

## Subgroups

### Dihedral group

The dihedral group $D_n$ is the group of all rotations and reflections about the center of the regular polygon of $n$ sides.

$|S_n|=n!, |D_n|=2n$

### Cyclic group

$G=\langle a\rangle=\{1,a,a^2,\cdots\}$ for some $a\in G$

<details>
<summary>Example of cyclic group</summary>

$(\mathbb{Z}_n,+)$ is cyclic and $\mathbb{Z}_n=\langle 1\rangle=\{0,1,2,\cdots,n-1\}$

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$(\mathbb{Z},+)$ is cyclic and $\mathbb{Z}=\langle 1\rangle=\langle -1 \rangle$

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$S_3$ is not cyclic

$\langle e\rangle=\{e\}$
$\langle (1,2)\rangle=\{e,(1,2)\}$
$\langle (1,3)\rangle=\{e,(1,3)\}$
$\langle (2,3)\rangle=\{e,(2,3)\}$
$\langle (1,2,3)\rangle=\{e,(1,2,3),(1,3,2)\}$
$\langle (1,3,2)\rangle=\{e,(1,3,2),(1,2,3)\}$

</details>

#### Every cyclic group is abelian

Every cyclic group is abelian

<details>
<summary>Proof</summary>

Let $G=\langle a\rangle$ be a cyclic group, then $\forall g_1,g_2\in G$ we have $g_1g_2=g_2g_1$ since $g_1g_2=a^k_1a^k_2=a^{k_1+k_2}$ and $g_2g_1=a^k_2a^k_1=a^{k_1+k_2}$

</details>

#### Definition for order of element

Let $G$ be a group, then the order of $g\in G$ is defined to be the size of the smallest subgroup containing $g$.

If $|\langle g\rangle|$ is infinite, then we say that $g$ has infinite order.

<details>
<summary>Example of order of element</summary>

$5$ in $(\mathbb{Z},+)$ has infinite order.

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$5$ in $(\mathbb{Z}_{10},+)$ has order $2$.

$\langle 5\rangle=\{0,5\}$.

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$5$ in $(\mathbb{Z}_{6},+)$ has order $6$.

$\langle 5\rangle=\{0,5,4,3,2,1\}$.

</details>

#### Lemma for order of element

Let $G$ be a group, then $a\in G$ has order $n$ if $n$ is the smallest positive integer such that $a^n=e$.

<details>
<summary>Proof</summary>

There are 2 cases:

Case 1:

There is no positive $n$ such that $a^n=e$.

Then $a^i\neq a^j$ if $i\neq j, i,j\in \mathbb{N}$.

Reason: if $a^i=a^j$, then $a^{i-j}=e$.

Then the order of group is infinite.

Case 2:

There is a positive $n$ such that $a^n=e$.

Let $n$ be the smallest such positive integer. Then we claim $\langle a^n\rangle=\{e,a^1,a^2,\cdots,a^{n-1}\}$.

We claim they are all distinct.

Suppose not, then we can have $a^i=a^j$ for $i\neq j$, $0\leq i,j\leq n-1$.

Then $a^{i-j}=e$ but $i-j\leq n-1$. Therefore $n$ is not the smallest positive integer such that $a^n=e$.

</details>

#### Theorem for cyclic group up to isomorphism

Suppose $G$ is a cyclic group,

- If $|G|=n$, then $|G|\simeq \mathbb{Z}_n^+$
- If $|G|=\infty$, then $|G|\simeq \mathbb{Z}$.

<details>
<summary>Proof</summary>

Case 1:

If $|G|=\infty$, then we can map $G$ to $(\mathbb{Z},+)$, where $G=\langle a\rangle$. $\phi(n)=a^n$. This gives a bijection between $G$ and $(\mathbb{Z},+)$.

where $\phi(n+m)=a^{n+m}=a^n a^m=\phi(n)\phi(m)$.

Case 2:

If $|G|=n$, then we can map $G$ to $(\mathbb{Z}_n,+)$, where $G=\langle a\rangle$. $\phi(n)=a^n$. This gives a bijection between $G$ and $(\mathbb{Z}_n,+)$.

where $\phi(n+m)=\phi(r)=a^{n+m}=a^n a^m=\phi(n)\phi(m)$.
</details>

<details>
<summary>Example</summary>

Let $H=\langle (12)(345)\rangle\subseteq S_5$. Then $H\simeq \mathbb{Z}_6^+$.

Let $\tau=(12)(345)$

All the elements of $H$ are:

- $\tau^0=(12)(345)$
- $\tau^1=(453)$
- $\tau^2=(12)(534)$
- $\tau^3=(345)$
- $\tau^4=(12)(453)$
- $\tau^5=(534)$

</details>

#### GCD and order

If $G=\langle a\rangle$, then $H=\langle a^k\rangle$, $|H|=\frac{n}{d}$ where $d=\operatorname{gcd}(n,k)$.