Math 120: Groups and Rings (Fall 2025)

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Instructor: Daniel Bump (office 383U)
Course Assistant: Hongrui Chen (office 380H)

This class will meet MWF from 10:30-11:20 in 380X

The text will be Dummit and Foote Abstract Algebra, Third edition. We will cover group theory (through the Sylow theorems), and beginning ring theory. Although groups are more "basic" algebraic objects, rings are also pervasive and useful even in thinking about groups. I will talk about rings from an early stage, and I recommend that you read Section 7.1 in the book early.

Office hours

Daniel Bump: MWF 12:00-1 in 383U or by appointment.
Hongrui Chen: Tu-Th 4-5:30 PM in 380H.

There will be one evening Midterm on October 22, 2025. Additionally there will be a final and a WIM project

Grading Plan

You will have four scores: Homework, Midterm 1, WIM Project, Final. These will be given the following weights:

Homework 15 %

Midterm 30 %

WIM Project 15 %

Final 40 %

The grades will be curved against the resulting course score.

Caveat: this scheme could be modified in some minor way. For example I might take several weights including the above and take their maximum.

Midterm

The midterm will take place in 380X from 7-9 PM in 380X. It is closed book. It will cover Chapters 1-3 in Dummit and Foote with the following exception. From Section 3.4, you need to know what a simple group is, but otherwise I will be skipping this section.

Final

The final will take place on Monday, December 8 in 380X from 3:30-6:30 PM in 380X.

The First Isomorphism Theorem

On Page 97, Dummit and Foote don't prove the First Isomorphism Theorem. They leave the proof as an Exercise, vaguely suggesting that they've already proved it in Section 1. But this is an important basic fact so here is a proof.

Theorem. Let $\varphi:G\to H$ be a surjective homomorphism. Let $K$ be the kernel of $\varphi$. Then $K$ is a normal subgroup and $H\cong G/K$.

Note: Dummit and Foote do not assume that $\varphi$ is surjective, but this formulation is equivalent to what they prove. (Why?) We will make use of the fact that the kernel $K$ is a normal subgroup. (Proposition 7 on page 82.)

Proof: Define a map $\theta:G/K\to H$ by $\theta(gK)=\varphi(g)$ for the coset $gK\in G/K$. Since there may be more than one choice for the coset representative $g$, we must check that this is well-defined. This means that if $gK=hK$ we must prove that $\varphi(g)=\varphi(h)$. Indeed, $gK=hK$ implies that $h^{-1}gK=K$ so $h^{-1}g\in K$. By definition of the kernel, this means that $\varphi(h^{-1}g)=1_H$ so $\varphi(h)^{-1}\varphi(g)=1_H$ or $\varphi(h)=\varphi(g)$, proving that $\theta$ is well-defined.

Now we want to prove that $\theta$ is an isomorphism. First, we check that it is a homomorphism. If $g,h\in G$ then $gK\cdot hK=gh K$ (Proposition 5) so $\theta(gK\cdot hK)=\theta(ghK)=\varphi(gh)=\varphi(g)\varphi(h)$ which equals $\theta(gK)\theta(hK)$, proving that $\varphi$ is a group homomorphism.

Now we check that $\theta$ is an isomorphism. It is surjection because we are assuming $\varphi$ is surjective, so every element of $H$ equals $\varphi(g)=\theta(gK)$ for some $g\in G$. To show that it is injective we just reverse the argument that it is well defined: if $\theta(gK)=\theta(hK)$ then $\varphi(g)=\varphi(h)$ so $g^{-1}h$ is in the kernel $K$ of $\varphi$. Therefore $g^{-1}hK=K$ so $gK=hK$. Thus $\theta$ is a bijective homomorphism, that is, an isomorphism.

WIM Project

In preparation for the WIM assignment, please read the following essay about writing proofs.

Writing Mathematics.
Here's the tex file for this document: tex file.
normal.tex and normal.pdf. The "Writing" document refers to this sample text as a latex model.

Here is the WIM project. It is due Thursday, November 13.