MATH 415 Lecture 14

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Test Review

Problem 3

Subgroup diagram: Arrow points toward supergroup:

Problem 5

Theorem. If $H⊴G$ and $m=(G:H)$ then $a^m\in H$ for all $a\in G$.

Proof. Consider factor group $G/H$. then there exists a $\varphi$ such that $\varphi :G\to G/H$, mapping $g$ to $gH$.

Therefore, for $a\in G$, $\varphi (a)=aH=\overline{a}\in G/H$.

Since $m=(G:H)$, it holds that $g^m=e$ for all $g\in G$.

Note that

$a^m\in H$

if and only if

$\varphi (a^m)=e_{G/H}$

. We can show this as follows:

$\varphi (a^m)=\varphi (a{)}^m$

by the homomorphism property. Then

$\varphi (a{)}^m=(aH{)}^m={\overline{a}}^m=e_{G/H}$

quod erat demonstrandum

Bonus Problem 6

Theorem. A group $G$ that has only a finite number of subgroups must be a finite group.

Proof. Consider the cyclic subgroup generated by

$a$

. Note that every group has at least one cyclic subgroup. Then

$⟨a⟩\simeq \mathbb{\mathbb{Z}}$

or

$⟨a⟩\simeq {\mathbb{\mathbb{Z}}}_n$

.

quod erat demonstrandum

...

Therefore $G$ must be finite.

Section 18: Rings and Fields

Rings

A ring $⟨R,+,\cdot ⟩$ consists of a set $R$ and two binary operations: addition ($+$) and multiplication ($\cdot$) satisfying a bunch of axioms:

1. The additive group of the ring $⟨R,+⟩$ is an abelian group
2. Multiplication is associative: $(a\cdot b)\cdot c=a\cdot (b\cdot c)=a\cdot b\cdot c$
3. Left and right distributive laws: $a\cdot (b+c)=a\cdot b+a\cdot c$ and $(a+b)\cdot c=a\cdot c+b\cdot c$

Examples

• $⟨\mathbb{\mathbb{Z}},+,\cdot ⟩$
• $⟨\mathbb{\mathbb{Q}},+,\cdot ⟩$
• $⟨\mathbb{ℝ},+,\cdot ⟩$
• $⟨\mathbb{ℂ},+,\cdot ⟩$
• $⟨M_n(R),+,\cdot ⟩$ ($n\times n$ matrices with entries from ring $R$)
• $⟨\{f:\mathbb{ℝ}\to \mathbb{ℝ}\},+,\cdot ⟩$
• $⟨n\mathbb{\mathbb{Z}},+,\cdot ⟩$
• $⟨{\mathbb{\mathbb{Z}}}_n,+,\cdot ⟩$ (all operations mod $n$)
• $⟨R\times R,+,\cdot ⟩$ (Direct product of rings with component-wise operations: $(a_1,b_1)+(a_2,b_2)=(a_1+a_2,b_1+b_2)$ and $(a_1,b_1)\cdot (a_2,b_2)=(a_1\cdot a_2,b_1\cdot b_2)$)

Multiplicative Semigroup

The binary structure $⟨R,\cdot ⟩$ is called a semigroup. If $a,b\in R$ and $a\cdot b=b\cdot a$, then $R$ is commutative. Multiplicative identity $1$ may or may not belong to $R$. For example, the ring $2\mathbb{\mathbb{Z}}$ does not have a multiplicative identity.

Note: A semigroup with an identity element still might not be a group: it may not be closed on multiplication, and there could be elements without an inverse.

Theorem 18.8

Let $R$ be a ring, and $a,b\in R$:

1. $0\cdot a=a\cdot 0=0$
2. $a\cdot (-b)=(-a)\cdot b=-(a\cdot b)$
3. $(-a)\cdot (-b)=a\cdot b$.

Proof of 1.

$a\cdot 0+a\cdot 0=a\cdot (0+0)=a\cdot 0$

. Note we have

$a\cdot 0+a\cdot 0=a\cdot 0$

, therefore

$a\cdot 0=0$

. The reverse holds by right distributivity.

quod erat demonstrandum

Ring Homomorphisms

Let $R$ and $R^{\prime }$ be rings. Then $\varphi :R\to R^{\prime }$ is a homomorphism if for all $a,b\in R$,

1. $\varphi (a+b)=\varphi (a)+\varphi (b)$ ($\varphi$ is a homomorphism of abelian groups $⟨R,+⟩$ and $⟨R^{\prime },+⟩$.)
2. $\varphi (a\cdot b)=\varphi (a)\cdot \varphi (b)$ ($\varphi$ is a homomorphism of multiplicative semigroups $⟨R,\cdot ⟩$ and $⟨R^{\prime },\cdot ⟩$

Kernel

Let $\mathrm{K}\mathrm{e}\mathrm{r}\varphi =\{x\in R\mid \varphi (x)=0\}$