MATH 415 Lecture 14

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Test Review

Problem 3

Subgroup diagram: Arrow points toward supergroup:

Problem 5

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

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

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

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 $\phi (a^{m})=e_{G/H}$. We can show this as follows: $\phi (a^{m})=\phi (a)^{m}$ by the homomorphism property. Then $\phi (a)^{m}=(aH)^{m}={\bar {a}}^{m}=e_{G/H}$

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 $\left\langle a\right\rangle \simeq \mathbb {Z}$ or $\left\langle a\right\rangle \simeq \mathbb {Z} _{n}$.

quod erat demonstrandum

...

Therefore $G$ must be finite.

Section 18: Rings and Fields

Rings

A ring $\left\langle R,+,\cdot \right\rangle$ 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 $\left\langle R,+\right\rangle$ is an abelian group
2. Multiplication is associative: $\left(a\cdot b\right)\cdot c=a\cdot \left(b\cdot c\right)=a\cdot b\cdot c$
3. Left and right distributive laws: $a\cdot \left(b+c\right)=a\cdot b+a\cdot c$ and $\left(a+b\right)\cdot c=a\cdot c+b\cdot c$

Examples

• $\left\langle \mathbb {Z} ,+,\cdot \right\rangle$
• $\left\langle \mathbb {Q} ,+,\cdot \right\rangle$
• $\left\langle \mathbb {R} ,+,\cdot \right\rangle$
• $\left\langle \mathbb {C} ,+,\cdot \right\rangle$
• $\left\langle M_{n}(R),+,\cdot \right\rangle$ ($n\times n$ matrices with entries from ring $R$)
• $\left\langle \left\{f:\mathbb {R} \to \mathbb {R} \right\},+,\cdot \right\rangle$
• $\left\langle n\mathbb {Z} ,+,\cdot \right\rangle$
• $\left\langle \mathbb {Z} _{n},+,\cdot \right\rangle$ (all operations mod $n$)
• $\left\langle R\times R,+,\cdot \right\rangle$ (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 $\left\langle R,\cdot \right\rangle$ 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 $2\mathbb {Z}$ 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 $R'$ be rings. Then $\phi :R\to R'$ is a homomorphism if for all $a,b\in R$,

1. $\phi (a+b)=\phi (a)+\phi (b)$ ($\phi$ is a homomorphism of abelian groups $\left\langle R,+\right\rangle$ and $\left\langle R',+\right\rangle$.)
2. $\phi (a\cdot b)=\phi (a)\cdot \phi (b)$ ($\phi$ is a homomorphism of multiplicative semigroups $\left\langle R,\cdot \right\rangle$ and $\left\langle R',\cdot \right\rangle$

Kernel

Let $\mathrm {Ker} \ \phi =\left\{x\in R~\mid ~\phi (x)=0\right\}$