# 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. ($Z(G)\neq \{e\}$)
Proof
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.
#### Proposition: Prime square group is abelian
If $|G|=p^2$, where $p$ is a prime, then $G$ is abelian.
Proof
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.
### 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)
- $|G|=6$
- $G\simeq S_3$
- $G\simeq\mathbb{Z}_3\times \mathbb{Z}_2\simeq \mathbb{Z}_6$
Proof
$|G|$ 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 $G\simeq S_3$.
- $|G|=7$
- $G\simeq\mathbb{Z}_7$ (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 $G\simeq D_p$
Proof
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 $2\leq t\leq p-2$.