36b3b8f371
Some checks failed
Sync from Gitea (main→main, keep workflow) / mirror (push) Has been cancelled
95 lines
2.7 KiB
Markdown
95 lines
2.7 KiB
Markdown
# Math4302 Modern Algebra (Lecture 16)
|
|
|
|
## Group
|
|
|
|
### Normal subgroup
|
|
|
|
Suppose $H\leq G$, then the following are equivalent:
|
|
|
|
1. $aH=Ha$ for all $a\in G$
|
|
2. $aHa^{-1}= H$ for all $a\in G$
|
|
3. $aha^{-1}\subseteq H$ for all $a\in G$
|
|
4. the operation $(aH)(bH)=abH$ is well defined for all $a,b\in G$, on the set of left coset of $H\leq G$
|
|
|
|
Then $H\trianglelefteq G$
|
|
|
|
If $H\trianglelefteq G$, then the set of left coset of $H\leq G$ is a group under the operation $(aH)(bH)=abH$
|
|
|
|
$G/H$ is factor (or quotient) group of $G$ by $H$
|
|
|
|
identity $H$=$eH$
|
|
|
|
<details>
|
|
<summary>Example</summary>
|
|
|
|
If $|H|=\frac{1}{2}|G|$, then $H$ is a normal subgroup of $G$, then $G/H$ is isomorphic to $\mathbb{Z}_2$
|
|
|
|
---
|
|
|
|
Let $\phi:G\to G'$ be a homomorphism, then $\ker(\phi)\trianglelefteq G$
|
|
|
|
$\mathbb{Z}/5\mathbb{Z}\trianglelefteq \mathbb{Z}$
|
|
|
|
And $\mathbb{Z}/5\mathbb{Z}$ is isomorphic to $\mathbb{Z}_5$
|
|
|
|
---
|
|
|
|
$G/G$ is isomorphic to trivial group
|
|
|
|
---
|
|
|
|
$G/\{e\}$ is isomorphic to $G$
|
|
|
|
---
|
|
|
|
$\mathbb{R}/\mathbb{Z}$ is isomorphic to $S^1$
|
|
|
|
---
|
|
|
|
$\mathbb{Z}_3\times\mathbb{Z}_6/\langle (1,1)\rangle$ is isomorphic to $\mathbb{Z}_3$
|
|
|
|
$\langle (1,1)\rangle=\{(1,1),(2,2),(0,3),(1,4),(2,5),(0,0)\}$
|
|
|
|
</details>
|
|
|
|
Recall
|
|
|
|
1. The lagrange theorem, if $G$ is finite and $H\leq G$, then $|H| | |G|$.
|
|
2. If $G$ is finite, abelian, $d||G|$, then $G$ has a subgroup of order $d$.
|
|
|
|
We will show that 2. is not true if $G$ is not abelian. (consider $A_4$ with order 12, have no subgroup of order 6)
|
|
|
|
<details>
|
|
<summary>Proof</summary>
|
|
|
|
Suppose $H\leq A_4$, and $|H|=6$. Then $A_4/H$ is normal in $A_4$, (since $|A_4/H|=2$), and $A_4/H$ is isomorphic to $\mathbb{Z}_2$.
|
|
|
|
In other words, every element in $A_4/H$ has either order 1 or 2.
|
|
|
|
So for any $\sigma\in A_4$, $(\sigma H)(\sigma H)=\sigma^2 H$. Therefore $\sigma^2=H$.
|
|
|
|
But $\sigma=(1,3)(1,2)\in A_4$ and $\sigma^2=(1,3,2)\in H$.
|
|
|
|
Similarly, $(1,3,2)(1,2,4)\dots(1,4,3)$ are all even permutations, making $|H|\geq 8$, that is a contradiction.
|
|
|
|
</details>
|
|
|
|
#### Fundamental homomorphism theorem (first isomorphism theorem)
|
|
|
|
If $\phi:G\to G'$ is a homomorphism, then the function $f:G/\ker(\phi)\to \phi(G)$, ($\phi(G)\subseteq G'$) given by $f(a\ker(\phi))=\phi(a)$, $\forall a\in G$, is an well-defined isomorphism.
|
|
|
|
<details>
|
|
<summary>Proof</summary>
|
|
|
|
First we will prove the well definedness and injectivity of $f$.
|
|
|
|
We need to check the map will not map the same coset represented in different ways to different elements.
|
|
|
|
Suppose $a\ker(\phi)=a'\ker(\phi)$, then $a^{-1}b\in \ker(\phi)$, this implies $\phi(a^{-1}b)=e'$ so $\phi(a)=\phi(b)$.
|
|
|
|
Reverse the direction to prove the converse.
|
|
|
|
Other properties are trivial.
|
|
|
|
</details>
|