Sem 3 - Algebraic structures

Theory Recap

$(G,\ast )$ group if:

• Associative
• has identity element
• all elements have a symmetric $(R,+,\cdot )$ ring if:
• $(R,+)$ Abelian group
• $(R^{\ast },\cdot )$ semigroup
• Distributivity holds

Distributivity

\begin{align} \forall x,y,z\in R: &x(y+z)=xy+xz\\ &(y+z)x=yz+zx \end {align} $$

$(H,+)$ subgroup of $(G,+)$ if

• H stable part of $(G,+)$
• $(H,+)$ Group

Characterization theorem of a subgroup

$H\ne \varnothing$

$$\begin{array}{ll}\forall x,y\in H:& \in H\\ \forall x\in H:& -x\in H\end{array}\}⟹\forall x,y\in H:x-y\in H$$

$f:(G_1,\ast )\to (G_2,\circ )$ gr. homomorphism $\forall x,y\in G_1:f(x\ast y)=f(x)\circ f(y)$

f group isomorphism if

• f - homomorphism
• f - bijective (inj+surj) ($\exists f^{-1}$)

Exercises

Exercise 1.

$M\ne \varnothing$ $S_M=\{f:M\to M|f\text{ bijective}\}$ Show that $(S_M,\circ$) is a group Associativity: $\forall f,g,x\in S_M;f\circ (g\circ h)=(f\circ g)\circ h$ Let $x\in M⟹(f\circ (g\circ h))(x)=f\circ (g(h(x))=f(g(h(x)))$ $((f\circ g)\circ h)(x)=f(g(x))\circ h=f(g(h(x)))⟹\text{ assoc}$

Identitiy element: Let $e\in S_m$ be the identity element $\forall f\in S_{M}f\circ e=f\text{ and }e\circ f=f$ $e=1_{S_M};e(x)=x$ $(f\circ e)(x)=f(e(x))=f(x)$ $(e\circ$

Exercise 2.

$M\ne \varnothing$ $(R,+,\cdot )$ ring $R^M=\{f|f:M\to R\}$

\forall f,g\in R^M:&(f+g)(x)=f(x)+g(x)\\ \\ &(f\cdot g)(x)=f(x)\cdot g(x) \end{align}$$ Show that $(R^M,+,\cdot)$ is a ring. If R is comm or has id, does $R^M$ have the same property Associativity $$\begin {gathered} \forall f,g,h\in R^M:(f+g)+h=f+(g+h) \\ \end {gathered}

\begin{rcases} ((f+g)+h)(x)=(f+g)(x)+h(x)=f(x)+g(x)+h(x)\ (f+(g+h))(x)=f(x)+(g+h)(x)=f(x)+g(x)+h(x)\ \end{rcases}\implies \checkmark

$$Commutativity:$$

\forall f,g\in R^M:f+g=g+f

$$\begin{rcases} (f+g)(x)=f(x)+g(x)\\ (g+f)(x)=g(x)+f(x)\\ \end{rcases}\implies \checkmark$$ Identity element: $$\begin{gathered} \exists e\in R^M \text{ s.t. } \forall f\in R^M:e+f=f+e=f\\ \theta\\ (e+f)(x)=f(x)\Leftrightarrow e(x)+f(x)=f(x)\Leftrightarrow e(x)=0=\theta(x) \end{gathered}

Symmetrical element:

Subgroup Associativity $\forall f,g,h\in R^M:(f\cdot g)\cdot h=f\cdot (g\cdot h)$

$$\begin{array}{l}((f\cdot g)\cdot h)(x)=(f\cdot g)(x)\cdot h(x)=f(x)\cdot g(x)\cdot g(x)\\ (f\cdot (g\cdot h))(x)=f(x)\cdot (g\cdot h)(x)=f(x)\cdot g(x)\cdot h(x)\end{array}\}⟹(R^{M\ast },\cdot )\text{ semigroup}$$

Distributivity: $\forall f,g,h\in R^M:f\cdot (g+h)=(f\cdot g)+(f\cdot h)$

Exercise 3

$H=\{z\in \mathbb{C}||z|=1\}$ Prove that $h\le ({\mathbb{C}}^{\ast },\cdot )$, $H\nleqq (\mathbb{C},+)$

$$H\le ({\mathbb{C}}^{\ast },\cdot )\iff \{\begin{array}{l}H\ne \varnothing \\ \forall x,y\in H:x\cdot y\in H\end{array}$$

Let $z=1\in H⟹|z|=1\in M$

$$\begin{array}{c}\forall x,y\in M:x+y\in M\\ |x\cdot y|=|x|\cdot |y|=1\in M\end{array}$$

Symmetrical elements: $\forall$

Exercise 5

$n\in \mathbb{N},n\ge 2$

1. $G{L}_n(\mathbb{C})=\{A\in M_n(\mathbb{C}|detA\ne 0\}$ stable subset of $(M_n(\mathbb{C}),\cdot )$

2. $G{L}_n(\mathbb{C},\cdot )$ group

3. $S{L}_n(\mathbb{C})=\{A\in M_n(\mathbb{C})|detA=1\}$ subgroup of $(G{L}_n(\mathbb{C}),\cdot )$

4. $\forall A,V\in G{L}_n(\mathbb{C}):A\cdot B\in G{L}_n(\mathbb{C})$

det A\not= 0, det B\not = 0\\ det(A\cdot B)=det A\cdot det B\not = 0\\ \end{rcases}\implies GL_n(\mathbb{C}) \text{ stable subset}

2. "$\cdot$" on $M_n(\mathbb{C})\text{ is associative}$ Let $I_n$ identity element for $(M_n(\mathbb{C}),\cdot )$ $det{I}_n=1\ne 0⟹I_n\in G{L}_n(\mathbb{C})$ $\forall A\in M_n(\mathbb{C})\exists A^{-1}\in M_n(\mathbb{C})$ $det{A}^{-1}\ne 0$ $det{I}_N=1=det$
3. SL

Important

The first exercise in algebra exams is giving definitions and examples

Exercise 7

Exercise 9

$n\in \mathbb{N},n\ge 2$ $({\mathbb{Z}}_n,+,\cdot )$ ring $\hat{a}\in {\mathbb{Z}}_n^{\ast }$ i. $\hat{a}$ invertible $\iff (a,n)=1$ ii. $({\mathbb{Z}}_n,+,\cdot )$ field $\leftrightarrow$ n prime\

i. $\hat{a}$ inv. $\iff \exists {\hat{a}}^{-1}\text{ s.t. }\hat{a}\cdot {\hat{a}}^{-1}\iff \exists {\hat{a}}^{-1}\in {\mathbb{Z}}_n^{\ast }\text{ s.t. }\hat{a}\cdot {\hat{a}}^{-1}-\hat{1}=\hat{0}$ $\exists {\hat{a}}^{-1}\in {\mathbb{Z}}_n^{\ast }n|a\cdot a^{-1}-1$