reading:abstract_algebra

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revision Previous revision
reading:abstract_algebra [2023-05-12 06:21] – [Chapter 3] asdfreading:abstract_algebra [2023-05-29 00:12] (current) – [Chapter 3] asdf
Line 3: Line 3:
  
 ===== Notes ===== ===== Notes =====
-==== Chapter 3 ====+==== Quotient Groups and Homomorphisms ====
 I've always had a tenuous grasp on cosets and quotient groups, but this book is helping me make sense of them. Unlike previous texts which use cosets to define quotient groups, this one uses the structure of quotient groups to describe cosets. Its initial focus is on how group homomorphisms give rise to quotient groups. That is, the fibers of a homomorphism $\varphi:G\to H$ with kernel $K$ naturally form the group $G/K$. It reduces cosets to objects (in this case subsets of a group) that may or may not possess a group structure, then lays out the conditions under which those objects do form a group (the cosets must be of a normal subgroup). Similarly, normal subgroups are not defined in terms of cosets, but rather in terms of quotient groups.  I've always had a tenuous grasp on cosets and quotient groups, but this book is helping me make sense of them. Unlike previous texts which use cosets to define quotient groups, this one uses the structure of quotient groups to describe cosets. Its initial focus is on how group homomorphisms give rise to quotient groups. That is, the fibers of a homomorphism $\varphi:G\to H$ with kernel $K$ naturally form the group $G/K$. It reduces cosets to objects (in this case subsets of a group) that may or may not possess a group structure, then lays out the conditions under which those objects do form a group (the cosets must be of a normal subgroup). Similarly, normal subgroups are not defined in terms of cosets, but rather in terms of quotient groups. 
  
Line 39: Line 39:
 Let $a\in (\mathbb{Z}/n\mathbb{Z})^\times$. Note that $\gcd(n,a)=1$ and $|(\mathbb{Z}/n\mathbb{Z})^\times|=\varphi(n)$ by definition of $(\mathbb{Z}/n\mathbb{Z})^\times$. By corollary 9, $a^{|(\mathbb{Z}/n\mathbb{Z})^\times|}=1$. By analogy (be more specific), $a^{|\varphi(n)|}\equiv 1\mod n$. $\square$ Let $a\in (\mathbb{Z}/n\mathbb{Z})^\times$. Note that $\gcd(n,a)=1$ and $|(\mathbb{Z}/n\mathbb{Z})^\times|=\varphi(n)$ by definition of $(\mathbb{Z}/n\mathbb{Z})^\times$. By corollary 9, $a^{|(\mathbb{Z}/n\mathbb{Z})^\times|}=1$. By analogy (be more specific), $a^{|\varphi(n)|}\equiv 1\mod n$. $\square$
 ++++ ++++
 +
 +Corollary 17 closely resembles some important results from linear algebra (see [[Linear Algebra Done Right]], 3.16 and 3.22). This makes sense, as linear transformations are definitionally group homomorphisms. 
 +
 +<WRAP center round box 60%>
 +**Proof of Corollary 17(1)**
 +
 +($\implies$) Suppose $\varphi$ is injective; we must show that $\ker\varphi = \{1\}$. $\varphi(1)=1\implies 1\in\ker\varphi\implies\{1\}\subseteq\ker\varphi$. Let $g\in \ker\varphi$. $\varphi(g)=1=\varphi(1)$; since $\varphi$ is injective, we must have $g=1$ and thus $\ker\varphi=\{1\}$.
 +
 +($\impliedby$) Now suppose $\ker\varphi=\{1\}$; we must show that $\varphi$ is injective. Let $a,b\in G$ such that $\varphi(a)=\varphi(b)$. Then $\varphi(a)*\varphi(b)^{-1}=1\iff\varphi(a*b^{-1})=1\implies a*b^{-1}\in\ker\varphi$. By hypothesis, it must be the case that $a*b^{-1}=1$; that is, $a=b$. Hence $\varphi$ is injective, as desired.
 +</WRAP>
 +
 +<WRAP center round todo 80%>
 +Revisit normalizers, centralizers, and stabilizers (§2.2, p. 49) and take notes on this page. They have appeared more often than expected when I skimmed past them.
 +</WRAP>
  
 {{tag>math algebra}} {{tag>math algebra}}
  • reading/abstract_algebra.1683872501.txt.gz
  • Last modified: 2023-05-12 06:21
  • by asdf