Aufgaben:Exercise 2.1: Group, Ring, Field - Revision history

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	[[Category:Channel Coding: Exercises\|^2.1 Some Basics of Algebra^]]		[[Category:Channel Coding: Exercises\|^2.1 Some Basics of Algebra^]]
	[[de:Aufgaben:2.~~1_Gruppe~~,~~_Ring~~,~~_Körper~~]]		[[de:Aufgaben:Aufgabe 2.1: Gruppe, Ring, Körper]]

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	An essential property of the Galois field  ${\rm GF}(q) = \{\hspace{0.1cm}z_0,\hspace{0.1cm} z_1,\hspace{0.05cm}\text{...} \hspace{0.1cm} , \hspace{0.1cm}z_{q-1}\}$  is that it has at least one primitive element:  An element  $z_i ≠ 0$  is said to be  "primitive"  if the following condition is satisfied  $(k$  is an integer$)$:		An essential property of the Galois field  ${\rm GF}(q) = \{\hspace{0.1cm}z_0,\hspace{0.1cm} z_1,\hspace{0.05cm}\text{...} \hspace{0.1cm} , \hspace{0.1cm}z_{q-1}\}$  is that it has at least one primitive element:  An element  $z_i ≠ 0$  is said to be  "primitive"  if the following condition is satisfied  $(k$  is an integer$)$:
	:$$z_i^k \hspace{0.15cm}{\rm mod}\hspace{0.15cm}q = \left\{ \begin{array}{c} \ne 1\\		:$$z_i^k \hspace{0.15cm}{\rm mod}\hspace{0.15cm}q = \left\{ \begin{array}{c} \ne 1\\1 \end{array} \right.\quad\begin{array}{*{1}c} {\rm for} \hspace{0.15cm}1 \le k < q-1\\ {\rm for} \hspace{0.15cm}k = q-1 \\ \end{array}\hspace{0.35cm} \Rightarrow \hspace{0.35cm} z_i \hspace{0.15cm}{\rm is \hspace{0.15cm}a\hspace{0.15cm} primitive \hspace{0.15cm}element}\hspace{0.05cm}. $$
	1 \end{array} \right.\quad
	\begin{array}{*{1}c} {\rm for} \hspace{0.15cm}1 \le k < q-1
	\\ {\rm for} \hspace{0.15cm}k = q-1 \\ \end{array}
	\hspace{0.35cm} \Rightarrow \hspace{0.35cm} z_i \hspace{0.15cm}{\rm is \hspace{0.15cm}a\hspace{0.15cm} primitive \hspace{0.15cm}element}
	\hspace{0.05cm}. $$

	Only for a primitive element  $z_i$		Only for a primitive element  $z_i$
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	'''(4)'''  The zero element is never a primitive element.  Also  $z_1 = 1$  is not a primitive element,  because then with  $q = 3$  would have to hold:		'''(4)'''  The zero element is never a primitive element.  Also  $z_1 = 1$  is not a primitive element,  because then with  $q = 3$  would have to hold:
	:$$z_1^1 \hspace{0.15cm}{\rm mod\hspace{0.15cm}} 3\hspace{0.1cm} \ne 1 \hspace{0.05cm},\hspace{0.3cm}z_1^2 \hspace{0.15cm}{\rm mod\hspace{0.15cm}} 3\hspace{0.1cm} = 1		:$$z_1^1 \hspace{0.15cm}{\rm mod\hspace{0.15cm}} 3\hspace{0.1cm} \ne 1 \hspace{0.05cm},\hspace{0.3cm}z_1^2 \hspace{0.15cm}{\rm mod\hspace{0.15cm}} 3\hspace{0.1cm} = 1\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	On the other hand,  $z_2 = 2$  is a primitive element because of		On the other hand,  $z_2 = 2$  is a primitive element because of

	:$$2^1 \hspace{0.15cm}{\rm mod\hspace{0.15cm}} 3\hspace{0.1cm} = 2 \hspace{0.05cm},\hspace{0.3cm}2^2 \hspace{0.15cm}{\rm mod\hspace{0.15cm}} 3\hspace{0.1cm} = 1		:$$2^1 \hspace{0.15cm}{\rm mod\hspace{0.15cm}} 3\hspace{0.1cm} = 2 \hspace{0.05cm},\hspace{0.3cm}2^2 \hspace{0.15cm}{\rm mod\hspace{0.15cm}} 3\hspace{0.1cm} = 1\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	The correct solution is  <u>solution suggestion 3</u>.		The correct solution is  <u>solution suggestion 3</u>.

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	[[Category:Channel Coding: Exercises\|^2.1 Some Basics of Algebra^]]		[[Category:Channel Coding: Exercises\|^2.1 Some Basics of Algebra^]]
			[[de:Aufgaben:2.1_Gruppe,_Ring,_Körper]]

Guenter at 13:17, 27 August 2022

2022-08-27T13:17:54Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  With the number set $Z_3 = \{0, \, 1, \, 2\}$ describes		'''(1)'''  With the number set  $Z_3 = \{0, \, 1, \, 2\}$  describes
	* the table A3 the additive group $(Z_3, \, +)$,		* the table A3 the additive group  $(Z_3, \, +)$,
	* the table M3 the multiplicative group $(Z_3, \, \cdot)$.		* the table M3 the multiplicative group  $(Z_3, \, \cdot)$.


	⇒   <u>Solution suggestion 1 and 2</u>. ~~The solution suggestions 3 and 4 do not apply here, because only one arithmetic operation (addition or multiplication) is defined for each group.~~		⇒   <u>Solution suggestion 1 and 2</u>.

			The solution suggestions 3 and 4 do not apply here,  because only one arithmetic operation  (addition or multiplication)  is defined for each group.


	~~'''(2)'''  On an algebraic ring, in contrast, two arithmetic operations are defined. Thus, the <u>proposed solutions 3 and 4</u> are correct:~~
	* Tables A3 and M3 describe the ring $(Z_3, \, +, \, \cdot)$.
	* Tables A4 and M4 describe the ring $(Z_4, \, +, \, \cdot)$.

			'''(2)'''  On an algebraic ring,  in contrast,  two arithmetic operations are defined.  Thus,  the  <u>proposed solutions 3 and 4</u>  are correct:
			* Tables A3 and M3 describe together the ring  $(Z_3, \, +, \, \cdot)$.
			* Tables A4 and M4 describe together the ring  $(Z_4, \, +, \, \cdot)$.

	~~In contrast, A3 and M4 do not describe a ring because they refer to different quantities.~~

			In contrast,  A3 and M4 do not describe a ring because they refer to different quantities.


	~~'''(3)'''  Every field is also a ring, but not every ring is also a field:~~
	*For the latter, division is also defined and for each element there is also the '''multiplicative inverse'''.
	*A '''finite number ring''' of order $q$ (that is, with $q$ elements) is a field only if $q$ is a prime number.
	*This is then also called a '''Galois field''' ${\rm GF}(q)$.

			'''(3)'''  Every field is also a ring,  but not every ring is also a field:
			*For the latter,  division is also defined and for each element there is also the  '''multiplicative inverse'''.
			*A  '''finite number ring'''  of order  $q$  (that is,  with  $q$  elements)  is a field only if  $q$  is a prime number.
			*This is then also called a '''Galois field'''  ${\rm GF}(q)$.

	~~So the correct answer is <u>answer 3</u>:~~
	*The arithmetic operations according to tables A3 and M3 together result in the Galois field ${\rm GF}(3)$.
	*In contrast, tables A4 (addition) and M4 multiplication) together with the set $\{0, \, 1, \, 2, \, 3\}$ do not result in the Galois field ${\rm GF}(4)$.
	*A condition for a Galois field is that for each element $z_i$ there exists a multiplicative inverse ${\rm Inv}_{\rm M}{(z_i)}$ such that the equation $z_i \cdot {\rm Inv}_{\rm M}(z_i) = 1$ is satisfied.
	*According to table M4, however, $\rm Inv_M(2)$ does not exist. There is no "$1$" in the third row.

			So the correct answer is  <u>answer 3</u>:
			*The arithmetic operations according to tables A3 and M3 together result in the Galois field  ${\rm GF}(3)$.
			*In contrast,  tables A4 (addition) and M4 multiplication) together with the set  $\{0, \, 1, \, 2, \, 3\}$  do not result in the Galois field  ${\rm GF}(4)$.
			*A condition for a Galois field is that for each element  $z_i$  there exists a multiplicative inverse  ${\rm Inv}_{\rm M}{(z_i)}$  such that the equation  $z_i \cdot {\rm Inv}_{\rm M}(z_i) = 1$  is satisfied.
			*According to table M4,  however, $\rm Inv_M(2)$  does not exist.  There is no  "$1$"  in the third row.

	For example, a Galois field $\rm GF(4)$ is obtained by extending the binary set $\{0, \, 1\}$ to the set $\{0, \, 1, \, \alpha, \, 1+\alpha\}$. For more details, see the [[Channel_Coding/Extension_Field#GF.2822.29_.E2.80.93_Example_of_an_extension.C3.B6rpers\|"example of an extension field"]] page.

			For example,  a Galois field  $\rm GF(4)$  is obtained by extending the binary set  $\{0, \, 1\}$  to the set  $\{0, \, 1, \, \alpha, \, 1+\alpha\}$.  For more details, see the page  [[Channel_Coding/Extension_Field#GF.2822.29_.E2.80.93_Example_of_an_extension.C3.B6rpers\|"Example of an extension field"]] page.

	'''(4)'''  The zero element is never a primitive element. Also $z_1 = 1$ is not a primitive element, because then $q = 3$ would have to hold:
			'''(4)'''  The zero element is never a primitive element.  Also  $z_1 = 1$  is not a primitive element,  because then with  $q = 3$  would have to hold:
	:$$z_1^1 \hspace{0.15cm}{\rm mod\hspace{0.15cm}} 3\hspace{0.1cm} \ne 1 \hspace{0.05cm},\hspace{0.3cm}z_1^2 \hspace{0.15cm}{\rm mod\hspace{0.15cm}} 3\hspace{0.1cm} = 1		:$$z_1^1 \hspace{0.15cm}{\rm mod\hspace{0.15cm}} 3\hspace{0.1cm} \ne 1 \hspace{0.05cm},\hspace{0.3cm}z_1^2 \hspace{0.15cm}{\rm mod\hspace{0.15cm}} 3\hspace{0.1cm} = 1
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	On the other hand, $z_2 = 2$ is a primitive element because of		On the other hand,  $z_2 = 2$  is a primitive element because of

	:$$2^1 \hspace{0.15cm}{\rm mod\hspace{0.15cm}} 3\hspace{0.1cm} = 2 \hspace{0.05cm},\hspace{0.3cm}2^2 \hspace{0.15cm}{\rm mod\hspace{0.15cm}} 3\hspace{0.1cm} = 1		:$$2^1 \hspace{0.15cm}{\rm mod\hspace{0.15cm}} 3\hspace{0.1cm} = 2 \hspace{0.05cm},\hspace{0.3cm}2^2 \hspace{0.15cm}{\rm mod\hspace{0.15cm}} 3\hspace{0.1cm} = 1
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	The correct solution is <u>solution suggestion 3</u>.		The correct solution is  <u>solution suggestion 3</u>.



	'''(5)'''  The set $\{0, \, 1, \, 2, \, 3\}$ has <u>no primitive element</u> and accordingly does not satisfy the requirements of a Galois field:		'''(5)'''  The set  $\{0, \, 1, \, 2, \, 3\}$  has <u>no primitive element</u> and accordingly does not satisfy the requirements of a Galois field:
	:$$z_1 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1\text{:}\hspace{0.35cm} 1^1 = 1\hspace{0.05cm},\hspace{0.1cm}1^2 = 1\hspace{0.05cm},\hspace{0.1cm}1^3 = 1\hspace{0.05cm},$$		:$$z_1 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1\text{:}\hspace{0.35cm} 1^1 = 1\hspace{0.05cm},\hspace{0.1cm}1^2 = 1\hspace{0.05cm},\hspace{0.1cm}1^3 = 1\hspace{0.05cm},$$
	:$$z_2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 2\text{:}\hspace{0.35cm} 2^1 = 2\hspace{0.05cm},\hspace{0.1cm}2^2 \hspace{0.15cm}{\rm mod}\hspace{0.15cm}4 = 0 \hspace{0.05cm},\hspace{0.1cm}2^3 \hspace{0.15cm}{\rm mod}\hspace{0.15cm}4 = 0\hspace{0.05cm},$$		:$$z_2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 2\text{:}\hspace{0.35cm} 2^1 = 2\hspace{0.05cm},\hspace{0.1cm}2^2 \hspace{0.15cm}{\rm mod}\hspace{0.15cm}4 = 0 \hspace{0.05cm},\hspace{0.1cm}2^3 \hspace{0.15cm}{\rm mod}\hspace{0.15cm}4 = 0\hspace{0.05cm},$$

Guenter at 12:37, 27 August 2022

2022-08-27T12:37:11Z

← Older revision		Revision as of 14:37, 27 August 2022
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	{{quiz-Header\|Buchseite=Channel_Coding/Some_Basics_of_Algebra}}		{{quiz-Header\|Buchseite=Channel_Coding/Some_Basics_of_Algebra}}

	[[File:EN_KC_A_2_1.png\|right\|frame\|"Addition" and "Multiplication" for  $q = 3$  and  $q = 4$.]]		[[File:EN_KC_A_2_1.png\|right\|frame\|"Addition" and "Multiplication" <br>     for  $q = 3$  and  $q = 4$.]]
	In the  [[Channel_Coding/Some_Basics_of_Algebra\|"theory section"]]  to this chapter, various algebraic terms were defined. For what follows, we assume that all sets consist of  $q$  elements each, where here either  $q = 3$  or  $q = 4$  ~~shall hold.~~ Then holds:		In the  [[Channel_Coding/Some_Basics_of_Algebra\|"theory section"]]  to this chapter,  various algebraic terms were defined.  For what follows,  we assume that all sets consist of  $q$  elements each,  where here either  $q = 3$  or  $q = 4$.  Then holds:
	* A  [[Channel_Coding/Some_Basics_of_Algebra#Definition_and_examples_of_an_algebraic_group\|algebraic group]]  is a finite set  $G = \{0, \, 1, \hspace{0.05cm}\text{...} \hspace{0.1cm} , \ q-1\}$  together with a linking rule defined between all elements. An additive group is denoted by  $(G, \ +)$  a multiplicative one by  $(G, \ \cdot)$.		* An  [[Channel_Coding/Some_Basics_of_Algebra#Definition_and_examples_of_an_algebraic_group\|"algebraic group"]]  is a finite set  $G = \{0, \, 1, \hspace{0.05cm}\text{...} \hspace{0.1cm} , \ q-1\}$  together with a linking rule defined between all elements.  An  "additive group"  is denoted by  $(G, \ +)$,  a multiplicative one by  $(G, \ \cdot)$.

	* A  [[Channel_Coding/Some_Basics_of_Algebra#Definition_and_examples_of_an_algebraic_ring\|"algebraic ring"]]  denotes a set  $R = \{0, \, 1, \hspace{0.05cm}\text{...} \hspace{0.1cm} , \ q-1\}$  together with two arithmetic operations defined therein, namely addition  ("$+$")  and multiplication  ("$\hspace{0.05cm}\cdot\hspace{0.05cm}$").		* An  [[Channel_Coding/Some_Basics_of_Algebra#Definition_and_examples_of_an_algebraic_ring\|"algebraic ring"]]  denotes a set  $R = \{0, \, 1, \hspace{0.05cm}\text{...} \hspace{0.1cm} , \ q-1\}$  together with two arithmetic operations defined therein,  namely addition  ("$+$")  and multiplication  ("$\hspace{0.05cm}\cdot\hspace{0.05cm}$").

	* A  [[Channel_Coding/Some_Basics_of_Algebra#Group.2C_ring.2C_field_-_basic_algebraic_concepts\|"algebraic field"]]  is a ring where division is additionally allowed and the commutative law is satisfied.		* An  [[Channel_Coding/Some_Basics_of_Algebra#Group.2C_ring.2C_field_-_basic_algebraic_concepts\|"algebraic field"]]  is a ring where division is additionally allowed and the commutative law is satisfied.


	Since we consider here finite sets only, a field is at the same time a Galois field  ${\rm GF}(q)$  of order  $q$.		Since we consider here finite sets only,  a  "field"  is at the same time a  [[Channel_Coding/Some_Basics_of_Algebra#Definition_of_a_Galois_field\|"Galois field"]]  ${\rm GF}(q)$  of order  $q$.

	An essential property of the Galois field  ${\rm GF}(q) = \{\hspace{0.1cm}z_0,\hspace{0.1cm} z_1,\hspace{0.05cm}\text{...} \hspace{0.1cm} , \hspace{0.1cm}z_{q-1}\}$  is that it has at least one primitive element. An element  $z_i ≠ 0$  is said to be primitive if the following condition is satisfied  $(k$ is integer$)$:		An essential property of the Galois field  ${\rm GF}(q) = \{\hspace{0.1cm}z_0,\hspace{0.1cm} z_1,\hspace{0.05cm}\text{...} \hspace{0.1cm} , \hspace{0.1cm}z_{q-1}\}$  is that it has at least one primitive element:  An element  $z_i ≠ 0$  is said to be  "primitive"  if the following condition is satisfied  $(k$  is an integer$)$:
	:$$z_i^k \hspace{0.15cm}{\rm mod}\hspace{0.15cm}q = \left\{ \begin{array}{c} \ne 1\\		:$$z_i^k \hspace{0.15cm}{\rm mod}\hspace{0.15cm}q = \left\{ \begin{array}{c} \ne 1\\
	1 \end{array} \right.\quad		1 \end{array} \right.\quad
	\begin{array}{*{1}c} {\rm ~~f\ddot{u}r~~} \hspace{0.15cm}1 \le k < q-1		\begin{array}{*{1}c} {\rm for} \hspace{0.15cm}1 \le k < q-1
	\\ {\rm ~~f\ddot{u}r~~} \hspace{0.15cm}k = q-1 \\ \end{array}		\\ {\rm for} \hspace{0.15cm}k = q-1 \\ \end{array}
	\hspace{0.35cm} \Rightarrow \hspace{0.35cm} z_i \hspace{0.15cm}{\rm ~~ist~~ \hspace{0.15cm}~~ein~~\hspace{0.15cm} ~~primitives~~ \hspace{0.15cm}~~Element~~}		\hspace{0.35cm} \Rightarrow \hspace{0.35cm} z_i \hspace{0.15cm}{\rm is \hspace{0.15cm}a\hspace{0.15cm} primitive \hspace{0.15cm}element}
	\hspace{0.05cm}. $$		\hspace{0.05cm}. $$

	Only for a primitive element  $z_i$  the arithmetic oparation  $z_i^k$  $($with  $k = 1, \, 2, \, 3, \, \hspace{0.05cm}\text{...} )$ yields all elements of the Galois field except the zero element $z_0 = 0$.		Only for a primitive element  $z_i$
			*the arithmetic oparation  $z_i^k$  $($with  $k = 1, \, 2, \, 3, \, \hspace{0.05cm}\text{...} )$  yields all elements of the Galois field
			*except the zero element $z_0 = 0$.


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			Hints:
			* The exercise covers the topic of the chapter  [[Channel_Coding/Some_Basics_of_Algebra\| "Some basics of algebra"]].

	~~Hints:~~		* Note that for group, ring and field with each  $q$  elements,  the arithmetic operations  "$+$"  and  "$\hspace{0.05cm}\cdot\hspace{0.05cm}$"  are to be understood modulo  $q$.
	* The exercise covers the topic of the chapter  [[Channel_Coding/Some_Basics_of_Algebra\| "some basics of algebra"]].
	* Note that for group, ring and field with each  $q$  elements, the arithmetic operations "$+$" and "$\hspace{0.05cm}\cdot\hspace{0.05cm}$" are to be understood modulo  $q$~~  respectively~~.


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	===Questions===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{Which of the given tables describe a group?		{Which of the given tables describe a  '''group'''?
	\|type="[]"}		\|type="[]"}
	+ Table  $\rm A3$,		+ Table  $\rm A3$,
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	- Table  $\rm A4$  and table  $\rm M4$  together.		- Table  $\rm A4$  and table  $\rm M4$  together.

	{Which of the given tables describe a ring?		{Which of the given tables describe a  '''ring'''?
	\|type="[]"}		\|type="[]"}
	- Table  $\rm A3$,		- Table  $\rm A3$,
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	- Table  $\rm A3$  and table  $\rm M4$  together.		- Table  $\rm A3$  and table  $\rm M4$  together.

	{Which of the tables describe a ~~field or a~~ Galois field?		{Which of the tables describe a  '''(Galois) field'''?
	\|type="[]"}		\|type="[]"}
	- Table  $\rm A3$,		- Table  $\rm A3$,
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	- Table  $\rm A4$  and table  $\rm M4$  together.		- Table  $\rm A4$  and table  $\rm M4$  together.

	{Which elements of the set  $\{0, \, 1, \, 2\} \ \Rightarrow \ q = 3$  are primitive?		{Which elements of the set  $\{0, \, 1, \, 2\} \ \Rightarrow \ q = 3$  are  "primitive"?
	\|type="[]"}		\|type="[]"}
	- $z_0 = 0$,		- $z_0 = 0$,
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	+ $z_2 = 2.$		+ $z_2 = 2.$

	{Which elements of the set  $\{0, \, 1, \, 2, \, 3\} \ \Rightarrow \ q = 4$  are primitive?		{Which elements of the set  $\{0, \, 1, \, 2, \, 3\} \ \Rightarrow \ q = 4$  are  "primitive"?
	\|type="[]"}		\|type="[]"}
	- $z_0 = 0$,		- $z_0 = 0$,

Guenter at 14:41, 24 August 2022

2022-08-24T14:41:43Z

← Older revision		Revision as of 16:41, 24 August 2022
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	[[Category:Channel Coding: Exercises\|^2.1 ~~Einige Grundlagen der~~ Algebra^]]		[[Category:Channel Coding: Exercises\|^2.1 Some Basics of Algebra^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 2.1: Gruppe, Ring, Körper to Aufgaben:Exercise 2.1: Group, Ring, Field

2022-08-24T14:36:39Z

Guenter moved page Aufgaben:Aufgabe 2.1: Gruppe, Ring, Körper to Aufgaben:Exercise 2.1: Group, Ring, Field

← Older revision	Revision as of 16:36, 24 August 2022
(No difference)

Noah at 13:24, 24 August 2022

2022-08-24T13:24:49Z

← Older revision		Revision as of 15:24, 24 August 2022
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	Hints:		Hints:
	* The exercise covers the topic of the chapter  [[Channel_Coding/Some_Basics_of_Algebra\| some basics of algebra]].		* The exercise covers the topic of the chapter  [[Channel_Coding/Some_Basics_of_Algebra\| "some basics of algebra"]].
	* Note that for group, ring and field with each  $q$  elements, the arithmetic operations "$+$" and "$\hspace{0.05cm}\cdot\hspace{0.05cm}$" are to be understood modulo  $q$  respectively.		* Note that for group, ring and field with each  $q$  elements, the arithmetic operations "$+$" and "$\hspace{0.05cm}\cdot\hspace{0.05cm}$" are to be understood modulo  $q$  respectively.

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	For example, a Galois field $\rm GF(4)$ is obtained by extending the binary set $\{0, \, 1\}$ to the set $\{0, \, 1, \, \alpha, \, 1+\alpha\}$. For more details, see the [[Channel_Coding/Extension_Field#GF.2822.29_.E2.80.93_Example_of_an_extension.C3.B6rpers\|~~Example~~ of an extension field]] page.		For example, a Galois field $\rm GF(4)$ is obtained by extending the binary set $\{0, \, 1\}$ to the set $\{0, \, 1, \, \alpha, \, 1+\alpha\}$. For more details, see the [[Channel_Coding/Extension_Field#GF.2822.29_.E2.80.93_Example_of_an_extension.C3.B6rpers\|"example of an extension field"]] page.

Noah at 13:24, 24 August 2022

2022-08-24T13:24:15Z

Show changes

Guenter at 12:45, 22 August 2022

2022-08-22T12:45:52Z

← Older revision		Revision as of 14:45, 22 August 2022
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	{{quiz-Header\|Buchseite=Kanalcodierung/Einige Grundlagen der Algebra}}		{{quiz-Header\|Buchseite=Kanalcodierung/Einige Grundlagen der Algebra}}

	[[File:~~P_ID2490__KC_A_2_1~~.png\|right\|frame\|"Addition" und "Multiplikation" für  $q = 3$  und  $q = 4$]]		[[File:EN_KC_A_2_1.png\|right\|frame\|"Addition" und "Multiplikation" für  $q = 3$  und  $q = 4$]]
	Im  [[Channel_Coding/Einige_Grundlagen_der_Algebra\|Theorieteil]]  zu diesem Kapitel wurden verschiedene algebraische Begriffe definiert. Für das Folgende setzen wir voraus, dass alle Mengen aus jeweils  $q$  Elementen bestehen, wobei hier entweder  $q = 3$  oder  $q = 4$  gelten soll. Dann gilt:		Im  [[Channel_Coding/Einige_Grundlagen_der_Algebra\|Theorieteil]]  zu diesem Kapitel wurden verschiedene algebraische Begriffe definiert. Für das Folgende setzen wir voraus, dass alle Mengen aus jeweils  $q$  Elementen bestehen, wobei hier entweder  $q = 3$  oder  $q = 4$  gelten soll. Dann gilt:
	* Eine  [[Channel_Coding/Einige_Grundlagen_der_Algebra#Definition_und_Beispiele_einer_algebraischen_Gruppe\|algebraische Gruppe]]  ist eine endliche Menge  $G = \{0, \, 1, \hspace{0.05cm}\text{...} \hspace{0.1cm} , \ q-1\}$  zusammen mit einer zwischen allen Elementen definierten Verknüpfungsvorschrift. Eine additive Gruppe wird mit  $(G, \ +)$  bezeichnet, eine multiplikative mit  $(G, \ \cdot)$.		* Eine  [[Channel_Coding/Einige_Grundlagen_der_Algebra#Definition_und_Beispiele_einer_algebraischen_Gruppe\|algebraische Gruppe]]  ist eine endliche Menge  $G = \{0, \, 1, \hspace{0.05cm}\text{...} \hspace{0.1cm} , \ q-1\}$  zusammen mit einer zwischen allen Elementen definierten Verknüpfungsvorschrift. Eine additive Gruppe wird mit  $(G, \ +)$  bezeichnet, eine multiplikative mit  $(G, \ \cdot)$.

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	{{quiz-Header\|Buchseite=Kanalcodierung/Einige Grundlagen der Algebra}}		{{quiz-Header\|Buchseite=Kanalcodierung/Einige Grundlagen der Algebra}}

	[[File:~~P_ID2490__KC_A_2_1~~.png\|right\|frame\|"Addition" und "Multiplikation" für  $q = 3$  und  $q = 4$]]		[[File:EN_KC_A_2_1.png\|right\|frame\|"Addition" und "Multiplikation" für  $q = 3$  und  $q = 4$]]
	Im  [[Channel_Coding/Einige_Grundlagen_der_Algebra\|Theorieteil]]  zu diesem Kapitel wurden verschiedene algebraische Begriffe definiert. Für das Folgende setzen wir voraus, dass alle Mengen aus jeweils  $q$  Elementen bestehen, wobei hier entweder  $q = 3$  oder  $q = 4$  gelten soll. Dann gilt:		Im  [[Channel_Coding/Einige_Grundlagen_der_Algebra\|Theorieteil]]  zu diesem Kapitel wurden verschiedene algebraische Begriffe definiert. Für das Folgende setzen wir voraus, dass alle Mengen aus jeweils  $q$  Elementen bestehen, wobei hier entweder  $q = 3$  oder  $q = 4$  gelten soll. Dann gilt:
	* Eine  [[Channel_Coding/Einige_Grundlagen_der_Algebra#Definition_und_Beispiele_einer_algebraischen_Gruppe\|algebraische Gruppe]]  ist eine endliche Menge  $G = \{0, \, 1, \hspace{0.05cm}\text{...} \hspace{0.1cm} , \ q-1\}$  zusammen mit einer zwischen allen Elementen definierten Verknüpfungsvorschrift. Eine additive Gruppe wird mit  $(G, \ +)$  bezeichnet, eine multiplikative mit  $(G, \ \cdot)$.		* Eine  [[Channel_Coding/Einige_Grundlagen_der_Algebra#Definition_und_Beispiele_einer_algebraischen_Gruppe\|algebraische Gruppe]]  ist eine endliche Menge  $G = \{0, \, 1, \hspace{0.05cm}\text{...} \hspace{0.1cm} , \ q-1\}$  zusammen mit einer zwischen allen Elementen definierten Verknüpfungsvorschrift. Eine additive Gruppe wird mit  $(G, \ +)$  bezeichnet, eine multiplikative mit  $(G, \ \cdot)$.