#  Math4302 Modern Algebra (Lecture 7)

## Subgroups

### Cyclic group

Last time, let $G$ be a group and $a\in G$. $|\langle a\rangle|=$ smallest positive $n$ such that $a^n=e$.

$\langle a\rangle=\{a^0,a^1,a^2,\cdots,a^{n-1}\}$.


#### Lemma subgroup of cyclic group is cyclic

Every subgroup of a cyclic group is cyclic.

$G=\langle a\rangle$.

<details>
<summary>Proof</summary>

Let $H\leq G$ be a subgroup. 

If $H=\{e\}$, we are done. 

Otherwise, let $m$ be the smallest positive integer such that $a^m\in H$. We claim $H=\langle a^m\rangle$.

- $\langle a^m\rangle\subseteq H$. trivial since $a^m\in H$ and $H$ is a subgroup.
- $H\subseteq\langle a^m\rangle$. Suppose $a^k\in H$, need to show $a^k\in \langle a^m\rangle$
  Divide $k$ by $m$: $k=qm+r$, $0\leq r\leq m-1$, Then $a^k\in H\implies a^{qm+r}\in H$. Also $a^m\in H$, then $(a^m)^q\in H$, so $a^mq\in H$, $a^-mq\in H$, so $a^{k}a^{-mq}\in H$, so $a^r\in H$, so $r$ has to be zero.  
  By our choice of $m$, $k=mq$, so $a^k=a^mq\in \langle a^m\rangle$.

</details>

<details>
<summary>Example</summary>

Every subgroup of $(\mathbb{Z},+)$ is of the form 

like the multiples of $n$: $n\mathbb{Z}=\langle n\rangle$ for some $n\geq 0$.

In particular, if $n,m\geq 1$ are in $\mathbb{Z}$, then the subgroup $\{nr+ms|r,s\in \mathbb{Z}\}\leq \mathbb{Z}$.

is equal to $d\mathbb{Z}$ where $d=\operatorname{gcd}(n,m)$.

</details>

Skip $\operatorname{gcd}$ part, check for Math 4111 notes in this site.


#### Lemma for size of cyclic subgroup

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

<details>
<summary>Proof</summary>

Recall $|H|$ is the smallest power of $a^m$ which is equal to $e$.

Let $d=\operatorname{gcd}(m,n)$, so $m=m_1d$, $n=n_1d$. and $\frac{n}{\operatorname{gcd}(m,n)}=n_1$,

- $(a^m)^{n_1}=a^{mn_1}=a^{m_1dn_1}=a^{m_1n}=(a^n)^{m_1}=e$.
- If $(a^m)^k=e$, the $a^{mk}=e\implies$ $mk$ is a multiple of $n$, 
  - If $a^\ell=e$, divide $\ell$ by $n$, $\ell=nq+r$, $0\leq r\leq n-1$, then $e=a^\ell=a^{nq+r}=a^r$, $r$ has to be zero, so $a^\ell=a^r=e$. $n|\ell$.
- $n_1d|m_1dk$, but by the definition of smallest common divisor, $m_1,n_1$ should not have common divisor other than $1$. So $n_1|m_1k$, $n_1|k\implies k\geq n_1$.

</details>

<details>
<summary>Example Applying the lemma</summary>

Let $G=\langle a \rangle$, $|G|=6$, $H=\langle a^4\rangle$. Then $|H|=\frac{6}{d}=3$ where $d=\operatorname{gcd}(6,4)=2$.

To check this we do enumeration $\langle a^4\rangle=\{e,a^4,a^2\}$.

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Find generator of $\mathbb{Z}_9$:

Using the coprime, we have $g=\{1,2,4,5,7,8\}$.

</details>

Corollary: $\langle a^m\rangle=G\iff |H|=n\iff \frac{n}{d}=n\iff \operatorname{gcd}(m,n)=1$ $m,n$ are coprime.