MATH 415 Lecture 9

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First test will be Oct. 8

Generating Sets

Given a group $G$ and $X\subset G$ is a generating set for $G$ if the minimum subgroup of $G$ containing $X$ is $G$

if $H\leq G$ and $X\subset H$, then $H=G$.

For all $g\in G$, we can represent $g$ as a product of elements of $X$ to power $\pm 1$:

$g=x_{1}^{\epsilon _{1}}\,\dots \,x_{n}^{\epsilon _{n}}$ for $x\in X$ and $\epsilon _{i}\in \left\{-1,1\right\}$

We denote $G=\left\langle X\right\rangle =\left\{x_{1}^{\epsilon _{1}}\,\dots \,x_{n}^{\epsilon _{n}}~\mid ~x\in X,\ \epsilon _{i}\in \left\{-1,1\right\}\right\}$ as the subgroup generated by the generating set $X$.

$G$ is finitely generated if there is a finite generating set for $G$.

Examples

• $\mathbb {Z}$ is finitely generated by $X=\left\{1\right\}$ or by $X=\left\{2,3\right\}$.
• $\mathbb {Z} \oplus \mathbb {Z}$ is also finitely generated: $X=\left\{(0,1),(1,0)\right\}$
• $\left\langle \mathbb {Q} ,+\right\rangle$ is not finitely generated: If $\mathbb {Q}$ is finitely generated, then $|X|<\infty$ and $X\subset \mathbb {Q}$, then $\mathbb {Q} =\left\langle X\right\rangle$ must be cyclic, which it is not. Contradiction.

Fundamental Theorem of Finitely Generated Abelian Groups

Theorem 11.12

The fundamental theorem for finitely generated abelian groups

Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups of the form

$\mathbb {Z} _{p_{1}^{r_{1}}}\times \mathbb {Z} _{p_{2}^{r_{2}}}\times \dots \times \mathbb {Z} _{p_{n}^{r_{n}}}\times \mathbb {Z} \times \dots \times \mathbb {Z} \,\!$

Where $p_{i}$ are prime numbers, and $r_{i}\in \mathbb {N}$

This representation is unique up to the rearrangement of factors.

Recall that all cyclic groups are isomorphic to $\mathbb {Z}$ if infinite and $\mathbb {Z} _{n}$ if finite and of order $n$.

Also recall that $\mathbb {Z} _{m}\times \mathbb {Z} _{n}\simeq \mathbb {Z} _{m\,n}$ if $m$ and $n$ are coprime. All primes are coprime to other primes

The (number of)^? infinite factors $\beta (G)$ is an invariant of $G$ and is called the Betti number.

The (number of)^? prime groups is also an invariant.

Example

Let $G$ be an abelian group with $\left|G\right|=360=2^{3}\cdot 3^{2}\cdot 5$. What is $G$?

1. $\mathbb {Z} _{2}\times \mathbb {Z} _{2}\times \mathbb {Z} _{2}\times \mathbb {Z} _{3}\times \mathbb {Z} _{3}\times \mathbb {Z} _{5}$
2. $\mathbb {Z} _{2}\times \mathbb {Z} _{4}\times \mathbb {Z} _{3}\times \mathbb {Z} _{3}\times \mathbb {Z} _{5}$
3. $\mathbb {Z} _{2}\times \mathbb {Z} _{2}\times \mathbb {Z} _{2}\times \mathbb {Z} _{9}\times \mathbb {Z} _{5}$
4. $\mathbb {Z} _{2}\times \mathbb {Z} _{4}\times \mathbb {Z} _{9}\times \mathbb {Z} _{5}$
5. $\mathbb {Z} _{8}\times \mathbb {Z} _{3}\times \mathbb {Z} _{3}\times \mathbb {Z} _{5}$
6. $\mathbb {Z} _{8}\times \mathbb {Z} _{9}\times \mathbb {Z} _{5}$

This is an exhaustive list (by theorem 11.12) of all abelian groups that contain 360 elements. Any other groups can be decomposed into above: $\mathbb {Z} _{360}\simeq \mathbb {Z} _{40}\times \mathbb {Z} _{9}\simeq \mathbb {Z} _{8}\times \mathbb {Z} _{9}\times \mathbb {Z} _{5}$.

Decomposable Groups

A group $G$ is decomposable if $G\simeq H\times K$ with $\left|H\right|\neq 1$ and $\left|K\right|\neq 1$.

Theorem 11.15

The finite indecomposible abelian groups are exactly the cyclic groups whose order is a power of a prime: $\mathbb {Z} _{p}^{r}$.

Proof. Suppose $G$ is indecomposible. Apply the fundamental theorem to arrive at the conclusion that $G\simeq \mathbb {Z} _{p_{i}^{r_{i}}}$.

Conversely, let $G=\mathbb {Z} _{p^{r}}$. Assume $G=\mathbb {Z} _{p^{r}}=\mathbb {Z} _{p_{i}}\times \mathbb {Z} _{p_{j}}$ such that $r=i+j$ and $i>j$. We know that $\mathbb {Z} _{p^{r}}$ is cyclic, thus it has a generator of order $p^{r}$. If we take an element $g=(a,b)\in \mathbb {Z} _{p^{i}}\times \mathbb {Z} _{p^{j}}$, then the order of $g$ is the least common multiple of the order of $a\in \mathbb {Z} _{p^{i}}$ and the order of $b\in \mathbb {Z} _{p^{j}}$. Thus Failed to parse (unknown function "\lcm"): {\displaystyle \mathrm{order}(g) \le \lcm{(p^i, p^j)}} Since we have the order of an element is at most $p^{i}$ for the subgroup $\mathbb {Z} _{p^{i}}$, this contradicts the fact that the same element in $G=\mathbb {Z} _{p^{r}}$ must have an order $p^{r}$ because $p^{i}<p^{r}$.

quod erat demonstrandum

Theorem 11.16

If $m\mid \left|G\right|$, where $G$ is a finite abelian group, then there is a subgroup $H\leq G$ with $\left|H\right|=m$.

Homomorphisms

A map $\phi :G\to G'$ is called a homomorphism if $\phi (x\,y)=\phi (x)\,\phi (y)$ for all $x,y\in G$.

1. If $\phi$ is injective ($x\neq y$ implies $\phi (x)\neq \phi (y)$), then it is called a monomorphism
2. If $\phi$ is surjective (for all $y\in G'$ there exists $x\in G$ such that $\phi (x)=y$), then it is called an epimorphism

The trivial homomorphism $\phi (x)=e$, where $e$ is the identity element in $G'$, is simple but not interesting.

Theorem

If $\phi :G\to G'$ is surjective and $G$ is abelian, then $G'$ is also abelian.

Proof. Choose $y,z\in G'$. Is $y\,z=z\,y$?

Because $\phi$ is onto, we know that there exist $u,v\in G$ such that $\phi (u)=y$ and $\phi (v)=z$. Since $G$ is abelian, we know $u\,v=v\,u$. By applying our homomorphism $\phi$, we get

Failed to parse (syntax error): {\displaystyle \phi(u\,v) &= \phi(u) \, \phi(v) = y \, z \\ \phi(v\,u) &= \phi(v) \, \phi(u) = z \, y }

Thus $G'$ is abelian by the above result.

quod erat demonstrandum

Examples

Symmentric Group of Permutations

$S_{n}$ is a symmetric group. Then $\phi :S_{n}\to \mathbb {Z} _{2}$ with

$\phi (\sigma )={\begin{cases}0&\sigma \ {\mbox{is an even permutation}}\\1&{\mbox{otherwise}}\end{cases}}$

$\phi$ is a homomorphism.

Note that $S_{n}$ is not commutative, but $\mathbb {Z} _{2}$ is commutative.

Evaluation Homomorphism

Define the set $F=\left\{f:\mathbb {R} \to \mathbb {R} \right\}$ as a function over real numbers. We claim $\left\langle F,+\right\rangle$ is an abelian group.

Indeed, $f(x)+g(x)=g(x)+f(x)$.

We can represent each function as a graph on cartesian plane. Define $\phi _{c}:F\to \mathbb {R}$, where $\mathbb {R}$ is the additive group of real numbers, as

$\phi _{c}(f)=f(c)\in \mathbb {R}$

Thus $\phi _{c}(f+g)=(f+g)(c)=f(c)+g(c)$, so $\phi$ is a homomorphism

Example: Vector Addition

Let $\mathbb {R} ^{n}$ be the additive group of column vectors. then ${\vec {x}}=\left\langle x_{1},\ldots ,x_{n}\right\rangle \in \mathbb {R} ^{n}$. We define $\phi _{A}:\mathbb {R} ^{n}\to \mathbb {R} ^{m}$ to be

$\phi _{A}({\vec {x}})=A\,{\vec {x}}$

Where $A$ is a $m\times n$ matrix.

Then $\phi _{A}({\vec {x}}+{\vec {y}})=A({\vec {x}}+{\vec {y}})=A\,{\vec {x}}+A\,{\vec {y}}=\phi _{A}({\vec {x}})+\phi _{A}({\vec {y}})$. Therefore $\phi$ is a homomorphism.

Example: Determinants

Recall that $\det {A\,B}=\det {(A)}\,\det {(B)}$. Then the determinant is a homomorphism:

$\det :M_{n}^{*}(\mathbb {R} )\to \mathbb {R} ^{*}$, where $\mathbb {R} ^{*}$ is the multiplicative group over nonzero real numbers.

Note that $M_{n}^{*}(\mathbb {R} )$ is not commutative, but $\mathbb {R} ^{*}$ is commutative.