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 $H\le G$ 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 $\pm 1$:

$g=x_1^{{ϵ}_1}\dots x_n^{{ϵ}_n}$ for $x\in X$ and ${ϵ}_i\in \{-1,1\}$

We denote $G=⟨X⟩=\{x_1^{{ϵ}_1}\dots x_n^{{ϵ}_n}\mid x\in X,{ϵ}_i\in \{-1,1\}\}$ as the subgroup generated by the generating set $X$.

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

Examples

• $\mathbb{\mathbb{Z}}$ is finitely generated by $X=\left\{1\right\}$ or by $X=\{2,3\}$.
• $\mathbb{\mathbb{Z}}\oplus \mathbb{\mathbb{Z}}$ is also finitely generated: $X=\{(0,1),(1,0)\}$
• $⟨\mathbb{\mathbb{Q}},+⟩$ is not finitely generated: If $\mathbb{\mathbb{Q}}$ is finitely generated, then $|X|<\mathrm{\infty }$ and $X\subset \mathbb{\mathbb{Q}}$, then $\mathbb{\mathbb{Q}}=⟨X⟩$ 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

${\mathbb{\mathbb{Z}}}_{p_1^{r_1}}\times {\mathbb{\mathbb{Z}}}_{p_2^{r_2}}\times \dots \times {\mathbb{\mathbb{Z}}}_{p_n^{r_n}}\times \mathbb{\mathbb{Z}}\times \dots \times \mathbb{\mathbb{Z}}$

Where $p_i$ are prime numbers, and $r_i\in \mathbb{\mathbb{N}}$

This representation is unique up to the rearrangement of factors.

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

Also recall that ${\mathbb{\mathbb{Z}}}_m\times {\mathbb{\mathbb{Z}}}_n\simeq {\mathbb{\mathbb{Z}}}_{mn}$ 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. ${\mathbb{\mathbb{Z}}}_2\times {\mathbb{\mathbb{Z}}}_2\times {\mathbb{\mathbb{Z}}}_2\times {\mathbb{\mathbb{Z}}}_3\times {\mathbb{\mathbb{Z}}}_3\times {\mathbb{\mathbb{Z}}}_5$
2. ${\mathbb{\mathbb{Z}}}_2\times {\mathbb{\mathbb{Z}}}_4\times {\mathbb{\mathbb{Z}}}_3\times {\mathbb{\mathbb{Z}}}_3\times {\mathbb{\mathbb{Z}}}_5$
3. ${\mathbb{\mathbb{Z}}}_2\times {\mathbb{\mathbb{Z}}}_2\times {\mathbb{\mathbb{Z}}}_2\times {\mathbb{\mathbb{Z}}}_9\times {\mathbb{\mathbb{Z}}}_5$
4. ${\mathbb{\mathbb{Z}}}_2\times {\mathbb{\mathbb{Z}}}_4\times {\mathbb{\mathbb{Z}}}_9\times {\mathbb{\mathbb{Z}}}_5$
5. ${\mathbb{\mathbb{Z}}}_8\times {\mathbb{\mathbb{Z}}}_3\times {\mathbb{\mathbb{Z}}}_3\times {\mathbb{\mathbb{Z}}}_5$
6. ${\mathbb{\mathbb{Z}}}_8\times {\mathbb{\mathbb{Z}}}_9\times {\mathbb{\mathbb{Z}}}_5$

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: ${\mathbb{\mathbb{Z}}}_{360}\simeq {\mathbb{\mathbb{Z}}}_{40}\times {\mathbb{\mathbb{Z}}}_9\simeq {\mathbb{\mathbb{Z}}}_8\times {\mathbb{\mathbb{Z}}}_9\times {\mathbb{\mathbb{Z}}}_5$.

Decomposable Groups

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

Theorem 11.15

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

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

Conversely, let

$G={\mathbb{\mathbb{Z}}}_{p^r}$

. Assume

$G={\mathbb{\mathbb{Z}}}_{p^r}={\mathbb{\mathbb{Z}}}_{p_i}\times {\mathbb{\mathbb{Z}}}_{p_j}$

such that

$r=i+j$

and

$i>j$

. We know that

${\mathbb{\mathbb{Z}}}_{p^r}$

is cyclic, thus it has a generator of order

$p^r$

. If we take an element

$g=(a,b)\in {\mathbb{\mathbb{Z}}}_{p^i}\times {\mathbb{\mathbb{Z}}}_{p^j}$

, then the order of

$g$

is the least common multiple of the order of

$a\in {\mathbb{\mathbb{Z}}}_{p^i}$

and the order of

$b\in {\mathbb{\mathbb{Z}}}_{p^j}$

. 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

${\mathbb{\mathbb{Z}}}_{p^i}$

, this contradicts the fact that the same element in

$G={\mathbb{\mathbb{Z}}}_{p^r}$

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 $H\le G$ with $\left|H\right|=m$.

Homomorphisms

A map $\varphi :G\to G^{\prime }$ is called a homomorphism if $\varphi (xy)=\varphi (x)\varphi (y)$ for all $x,y\in G$.

1. If $\varphi$ is injective ($x\ne y$ implies $\varphi (x)\ne \varphi (y)$), then it is called a monomorphism
2. If $\varphi$ is surjective (for all $y\in G^{\prime }$ there exists $x\in G$ such that $\varphi (x)=y$), then it is called an epimorphism

The trivial homomorphism $\varphi (x)=e$, where $e$ is the identity element in $G^{\prime }$, is simple but not interesting.

Theorem

If $\varphi :G\to G^{\prime }$ is surjective and $G$ is abelian, then $G^{\prime }$ is also abelian.

Proof. Choose $y,z\in G^{\prime }$. Is $yz=zy$?

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

$G^{\prime }$

is abelian by the above result.

quod erat demonstrandum

Examples

Symmentric Group of Permutations

$S_n$ is a symmetric group. Then $\varphi :S_n\to {\mathbb{\mathbb{Z}}}_2$ with

$\varphi (\sigma )=\{\begin{array}{ll}0& \sigma \text{is an even permutation}\\ 1& \text{otherwise}\end{array}$

$\varphi$ is a homomorphism.

Note that $S_n$ is not commutative, but ${\mathbb{\mathbb{Z}}}_2$ is commutative.

Evaluation Homomorphism

Define the set $F=\{f:\mathbb{ℝ}\to \mathbb{ℝ}\}$ as a function over real numbers. We claim $⟨F,+⟩$ 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 ${\varphi }_c:F\to \mathbb{ℝ}$, where $\mathbb{ℝ}$ is the additive group of real numbers, as

${\varphi }_c(f)=f(c)\in \mathbb{ℝ}$

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

Example: Vector Addition

Let ${\mathbb{ℝ}}^n$ be the additive group of column vectors. then $\overrightarrow{x}=⟨x_1,\dots ,x_n⟩\in {\mathbb{ℝ}}^n$. We define ${\varphi }_A:{\mathbb{ℝ}}^n\to {\mathbb{ℝ}}^m$ to be

${\varphi }_A(\overrightarrow{x})=A\overrightarrow{x}$

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

Then ${\varphi }_A(\overrightarrow{x}+\overrightarrow{y})=A(\overrightarrow{x}+\overrightarrow{y})=A\overrightarrow{x}+A\overrightarrow{y}={\varphi }_A(\overrightarrow{x})+{\varphi }_A(\overrightarrow{y})$. Therefore $\varphi$ is a homomorphism.

Example: Determinants

Recall that $\det AB=\det (A)\det (B)$. 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.