# Math4302 Modern Algebra (Lecture 23)

## Group

### Group acting on a set

#### Theorem for the orbit of a set with prime power group

Suppose $X$ is a $G$-set, and $|G|=p^n$ where $p$ is prime, then $|X_G|\equiv |X|\mod p$.

Where $X_G=\{x\in X|g\cdot x=x\text{ for all }g\in G\}=\{x\in X|\text{orbit of }x\text{ is trivial}\}$

#### Corollary: Cauchy's theorem

If $p$, where $p$ is a prime, divides $|G|$, then $G$ has a subgroup of order $p$. (equivalently, $g$ has an element of order $p$)

> This does not hold when $p$ is not prime.
>
> Consider $A_4$ with order $12$, and $A_4$ has no subgroup of order $6$.

#### Corollary: Center of prime power group is non-trivial

If $|G|=p^m$, then $Z(G)$ is non-trivial. ($Z(G)\neq \{e\}$)

<details>
<summary>Proof</summary>

Let $G$ act on $G$ via conjugation, then $g\cdot h=ghg^{-1}$. This makes $G$ to a $G$-set.

Apply the theorem, the set of elements with trivial orbit is; Let $X=G$, then $X_G=\{h\in G|g\cdot h=h\text{ for all }g\in G\}=\{h\in G|ghg^{-1}=h\text{ for all }g\in G\}=Z(G)$.

Therefore $|Z(G)|\equiv |G|\mod p$.

So $p$ divides $|Z(G)|$, so $|Z(G)|\neq 1$, therefore $Z(G)$ is non-trivial.
</details>

#### Proposition: Prime square group is abelian

If $|G|=p^2$, where $p$ is a prime, then $G$ is abelian.

<details>
<summary>Proof</summary>

Since $Z(G)$ is a subgroup of $G$, $|Z(G)|$ divides $p^2$ so $|Z(G)|=1, p$ or $p^2$.

By corollary center of prime power group is non-trivial, $Z(G)\neq 1$.

If $|Z(G)|=p$. If $|Z(G)|=p$, then consider the group $G/Z(G)$ (Note that $Z(G)\trianglelefteq G$). We have $|G/Z(G)|=p$ so $G/Z(G)$ is cyclic (by problem 13.39), therefore $G$ is abelian.

If $|Z(G)|=p^2$, then $G$ is abelian.

</details>

### Classification of small order

Let $G$ be a group

- $|G|=1$
  - $G=\{e\}$
- $|G|=2$
  - $G\simeq\mathbb{Z}_2$ (prime order)
- $|G|=3$
  - $G\simeq\mathbb{Z}_3$ (prime order)
- $|G|=4$
  - $G\simeq\mathbb{Z}_2\times \mathbb{Z}_2$
  - $G\simeq\mathbb{Z}_4$
- $|G|=5$
  - $G\simeq\mathbb{Z}_5$ (prime order)
- $|G|=6$
  - $G\simeq S_3$
  - $G\simeq\mathbb{Z}_3\times \mathbb{Z}_2\simeq \mathbb{Z}_6$
<details>
<summary>Proof</summary>

$|G|$ has an element of order $2$, namely $b$, and an element of order $3$, namely $a$.

So $e,a,a^2,b,ba,ba^2$ are distinct.

Therefore, there are only two possibilities for value of $ab$. ($a,a^2$ are inverse of each other, $b$ is inverse of itself.)

If $ab=ba$, then $G$ is abelian, then $G\simeq \mathbb{Z}_2\times \mathbb{Z}_3$.

If $ab=ba^2$, then $G\simeq S_3$.
</details>

- $|G|=7$
  - $G\simeq\mathbb{Z}_7$ (prime order)
- $|G|=8$
  - $G\simeq\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$
  - $G\simeq\mathbb{Z}_4\times \mathbb{Z}_2$
  - $G\simeq\mathbb{Z}_8$
  - $G\simeq D_4$
  - $G\simeq$ quaternion group $\{e,i,j,k,-1,-i,-j,-k\}$ where $i^2=j^2=k^2=-1$, $(-1)^2=1$. $ij=l$, $jk=i$, $ki=j$, $ji=-k$, $kj=-i$, $ik=-j$.
- $|G|=9$
  - $G\simeq\mathbb{Z}_3\times \mathbb{Z}_3$
  - $G\simeq\mathbb{Z}_9$ (apply the corollary, $9=3^2$, these are all the possible cases)
- $|G|=10$
  - $G\simeq\mathbb{Z}_5\times \mathbb{Z}_2\simeq \mathbb{Z}_{10}$
  - $G\simeq D_5$
- $|G|=11$
  - $G\simeq\mathbb{Z}_11$ (prime order)
- $|G|=12$
  - $G\simeq\mathbb{Z}_3\times \mathbb{Z}_4$
  - $G\simeq\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_3$
  - $A_4$
  - $D_6\simeq S_3\times \mathbb{Z}_2$
  - ??? One more
- $|G|=13$
  - $G\simeq\mathbb{Z}_{13}$ (prime order)
- $|G|=14$
  - $G\simeq\mathbb{Z}_2\times \mathbb{Z}_7$
  - $G\simeq D_7$

#### Lemma for group of order $2p$ where $p$ is prime

If $p$ is prime, $p\neq 2$, and $|G|=2p$, then $G$ is either abelian $\simeq \mathbb{Z}_2\times \mathbb{Z}_p$ or $G\simeq D_p$

<details>
<summary>Proof</summary>

We know $G$ has an element of order 2, namely $b$, and an element of order $p$, namely $a$.

So $e,a,a^2,\dots ,a^{p-1},ba,ba^2,\dots,ba^{p-1}$ are distinct elements of $G$.

Consider $ab$, if $ab=ba$, then $G$ is abelian, then $G\simeq \mathbb{Z}_2\times \mathbb{Z}_p$.

If $ab=ba^{p-1}$, then $G\simeq D_p$.

$ab$ cannot be inverse of other elements, if $ab=ba^t$, where $2\leq t\leq p-2$, then $bab=a^t$, then $(bab)^t=a^{t^2}$, then $ba^tb=a^{t^2}$, therefore $a=a^{t^2}$, then $a^{t^2-1}=e$, so $p|(t^2-1)$, therefore $p|t-1$ or $p|t+1$.

This is not possible since $2\leq t\leq p-2$.

</details>