# Math4302 Modern Algebra (Lecture 5)

## Groups

### Subgroups

A subset $H\subseteq G$ is a subgroup of $G$ if

- $e\in H$
- $\forall a,b\in H, a b\in H$
- $a\in H\implies a^{-1}\in H$

_$H$ with $*$ is a group_

We denote as $H\leq G$.

<details>
<summary>Example</summary>

For an arbitrary group $(G,*)$,

$(\{e\},*)$ and $(G,*)$ are always subgroups.

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$(\mathbb{Z},+)$ is a subgroup of $(\mathbb{R},+)$.

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Non-example:

$(\mathbb{Z}_+,+)$ is not a subgroup of $(\mathbb{Z},+)$.

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Subgroup of $\mathbb{Z}_4$:

$(\{0,1,2,3\},+)$ (if $1\in H$, $3\in H$)

$(\{0,2\},+)$

$(\{0\},+)$

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Subgroup of $\mathbb{Z}_5$:

$(\{0,1,2,3,4\},+)$

$(\{0\},+)$

_Cyclic group with prime order has only two subgroups_

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Let $D_n$ denote the group of symmetries of a regular $n$-gon. (keep adjacent points pairs).

$$
D_n=\{\sigma\in S_n\mid i,j\text{ are adjacent } \iff \sigma(i),\sigma(j)\text{ are adjacent }\}
$$

$$
\begin{pmatrix}
1&2&3&4\\
2&3&1&4
\end{pmatrix}\notin D_4
$$

$D_4$ has order $8$ and $S_4$ has order $24$.

$|D_n|=2n$. ($n$ option to rotation, $n$ option to reflection. For $\sigma(1)$ we have $n$ option, $\sigma(2)$ has 2 option where the remaining only has 1 option.)

Since $1-4$ is not adjacent in such permutation.

$D_n\leq S_n$ ($S_n$ is the symmetric group of $n$ elements).

</details>

#### Lemma of subgroups

If $H\subseteq G$ is a non-empty subset of a group $G$.

then ($H$ is a subgroup of $G$) if and only if ($a,b\in H\implies ab^-1\in H$).

<details>
<summary>Proof</summary>

If $H$ is subgroup, then $e\in H$, so $H$ is non-empty and if $a,b\in H$, then $b^{-1}\in H$, so $ab^{-1}\in H$.

---

If $H$ has the given property, then $H$ is non-empty and if $a,b\in H$, then $ab^-1\in H$, so

- There is some $a,a\in H$, $aa^{-1}\in H$, so $e\in H$.
- If $b\in H$, then $e\in H$, so $eb^{-1}\in H$, so $b^{-1}\in H$.
- If $b,c\in H$, then $c^{-1}$, so $bc^{-1}^{-1}\in H$, so $bc\in H$.

</details>

#### Cyclic group

$G$ is cyclic if $G$ is a subgroup generated by $a\in G$. (may be infinite)

$\mathbb{Z}_n\leq D_n\leq S_n$.

Cyclic group is always abelian.