Bsrgvty

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extract specific cheracters from each line

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Showing $mathbb Z_4 times mathbb Z_2$ to be a group

Solid whose full symmetry group corresponds to $A_4timesmathbb Z_2$Group extension of $mathbb Z_4$ by $mathbb Z_2$Question about method of finding homomorphisms from $mathbb Z_4$ to $mathbb Z_2 times mathbb Z_2$Regular representations of $Z_2 times Z_2$ and $Z_4$ and their decompositiononto homomorphic mapping from $Z_4 $ to $Z_2 times Z_2$Showing an isomorphism exists between either $mathbb Z_4$ or $mathbb Z_2 times mathbb Z_2$Find a subgroup of $Bbb Z_4oplusBbb Z_2$ not of the form $Hoplus K$ for some $Hle Bbb Z_4, Kle Bbb Z_2$.

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

1

$begingroup$

I've been posed the question:

Let $P$ be the pairs $(a,b)$ where $a in Bbb Z_4$, and $b in Bbb Z_2$

An operation, $*$, is defined by: $$(a,b)*(c,d)=(a+c pmod 4, b+d pmod 2)$$ for all $(a,c),(b,d)in P$

How do I show that this is a group?

I know how to do this with multiplication tables by working through the axioms but I don't know how to apply these to this question, nor if that's the best approach

abstract-algebra group-theory

share|cite|improve this question

edited Apr 15 at 8:51

YuiTo Cheng

3,29788 gold badges1717 silver badges4949 bronze badges

asked Apr 15 at 7:29

woddalwoddal

2144 bronze badges

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Check the three Groups axioms : closure, associativity, identity.
  $endgroup$
  – Mauro ALLEGRANZA
  Apr 15 at 7:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  1st : is it true that $(a,b) * (c,d) in mathbb Z$ ?
  $endgroup$
  – Mauro ALLEGRANZA
  Apr 15 at 7:35
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  2nd : is it true that : $[(a,b) * (c,d)] * (e,f) = (a,b) * [(c,d) * (e,f)]$ ?
  $endgroup$
  – Mauro ALLEGRANZA
  Apr 15 at 7:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  3rd : is it true that there is $(e_1,e_2)$ such that $(a,b) * (e_1,e_2) = ldots$ ?
  $endgroup$
  – Mauro ALLEGRANZA
  Apr 15 at 7:37
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @MauroALLEGRANZA You forgot inverses. Closure is usually not listed as one of the three group axioms, but rather a required property of $*$ before we even speak about the group axioms.
  $endgroup$
  – Arthur
  Apr 15 at 7:43
  
  

  
  
  
  

add a comment | 

1

$begingroup$

I've been posed the question:

Let $P$ be the pairs $(a,b)$ where $a in Bbb Z_4$, and $b in Bbb Z_2$

An operation, $*$, is defined by: $$(a,b)*(c,d)=(a+c pmod 4, b+d pmod 2)$$ for all $(a,c),(b,d)in P$

How do I show that this is a group?

I know how to do this with multiplication tables by working through the axioms but I don't know how to apply these to this question, nor if that's the best approach

abstract-algebra group-theory

share|cite|improve this question

edited Apr 15 at 8:51

YuiTo Cheng

3,29788 gold badges1717 silver badges4949 bronze badges

asked Apr 15 at 7:29

woddalwoddal

2144 bronze badges

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Check the three Groups axioms : closure, associativity, identity.
  $endgroup$
  – Mauro ALLEGRANZA
  Apr 15 at 7:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  1st : is it true that $(a,b) * (c,d) in mathbb Z$ ?
  $endgroup$
  – Mauro ALLEGRANZA
  Apr 15 at 7:35
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  2nd : is it true that : $[(a,b) * (c,d)] * (e,f) = (a,b) * [(c,d) * (e,f)]$ ?
  $endgroup$
  – Mauro ALLEGRANZA
  Apr 15 at 7:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  3rd : is it true that there is $(e_1,e_2)$ such that $(a,b) * (e_1,e_2) = ldots$ ?
  $endgroup$
  – Mauro ALLEGRANZA
  Apr 15 at 7:37
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @MauroALLEGRANZA You forgot inverses. Closure is usually not listed as one of the three group axioms, but rather a required property of $*$ before we even speak about the group axioms.
  $endgroup$
  – Arthur
  Apr 15 at 7:43
  
  

  
  
  
  

add a comment | 

$begingroup$

I've been posed the question:

Let $P$ be the pairs $(a,b)$ where $a in Bbb Z_4$, and $b in Bbb Z_2$

An operation, $*$, is defined by: $$(a,b)*(c,d)=(a+c pmod 4, b+d pmod 2)$$ for all $(a,c),(b,d)in P$

How do I show that this is a group?

I know how to do this with multiplication tables by working through the axioms but I don't know how to apply these to this question, nor if that's the best approach

abstract-algebra group-theory

share|cite|improve this question

edited Apr 15 at 8:51

YuiTo Cheng

3,29788 gold badges1717 silver badges4949 bronze badges

asked Apr 15 at 7:29

woddalwoddal

2144 bronze badges

$endgroup$

share|cite|improve this question

edited Apr 15 at 8:51

YuiTo Cheng

3,29788 gold badges1717 silver badges4949 bronze badges

asked Apr 15 at 7:29

woddalwoddal

2144 bronze badges

share|cite|improve this question

edited Apr 15 at 8:51

YuiTo Cheng

3,29788 gold badges1717 silver badges4949 bronze badges

asked Apr 15 at 7:29

woddalwoddal

2144 bronze badges

edited Apr 15 at 8:51

YuiTo Cheng

3,29788 gold badges1717 silver badges4949 bronze badges

edited Apr 15 at 8:51

YuiTo Cheng

3,29788 gold badges1717 silver badges4949 bronze badges

asked Apr 15 at 7:29

woddalwoddal

2144 bronze badges

asked Apr 15 at 7:29

woddalwoddal

2144 bronze badges

2 Answers
2

active

oldest

votes

3

$begingroup$

Associativity follows from that of $(Bbb Z_4, +_4)$ and of $(Bbb Z_2, +_2)$.

The identity is $(0pmod 4, 0pmod 2)$. (Why?)

The inverse of $(a,b)$ under $*$ is given by $(-a, -b)$ since $$beginalign(a,b)*(-a, -b)&=(a+(-a)pmod 4, b+(-b)pmod 2)\
&=(0pmod 4, 0pmod 2).
endalign$$

Closure follows from the closure of $Bbb Z_4$ under $+_4$ and of $Bbb Z_2$ under $+_2$.

share|cite|improve this answer

answered Apr 15 at 8:01

ShaunShaun

12k1212 gold badges3737 silver badges9292 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

In general, the group $Gtimes H$ (for groups $G,H$) with elements in the cartesian product (the set) of $G$ and $H$, and the group operation defined component-wise: The multiplication of $(g_1,h_1),(g_2,h_2)in Gtimes H$ being defined by $(g_1,h_1)(g_2,h_2)=(g_1g_2,h_1h_2)$ always gives a group. This is because the associativity follows from the group structure of $G$ and $H$, and the identity is $(1_G,1_H)$. You can check the other group axioms, which all follow from the fact that $G,H$ are groups. The group $Gtimes H$ is called the direct product of $G$ and $H$.

In particular, you just need that $mathbb Z_4$ and $mathbb Z_2$ to be groups to prove your conclusion in this case.

share|cite|improve this answer

answered Apr 15 at 8:06

YiFanYiFan

6,70822 gold badges1212 silver badges3434 bronze badges

$endgroup$

add a comment | 

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3

$begingroup$

Associativity follows from that of $(Bbb Z_4, +_4)$ and of $(Bbb Z_2, +_2)$.

The identity is $(0pmod 4, 0pmod 2)$. (Why?)

The inverse of $(a,b)$ under $*$ is given by $(-a, -b)$ since $$beginalign(a,b)*(-a, -b)&=(a+(-a)pmod 4, b+(-b)pmod 2)\
&=(0pmod 4, 0pmod 2).
endalign$$

Closure follows from the closure of $Bbb Z_4$ under $+_4$ and of $Bbb Z_2$ under $+_2$.

share|cite|improve this answer

answered Apr 15 at 8:01

ShaunShaun

12k1212 gold badges3737 silver badges9292 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

Associativity follows from that of $(Bbb Z_4, +_4)$ and of $(Bbb Z_2, +_2)$.

The identity is $(0pmod 4, 0pmod 2)$. (Why?)

The inverse of $(a,b)$ under $*$ is given by $(-a, -b)$ since $$beginalign(a,b)*(-a, -b)&=(a+(-a)pmod 4, b+(-b)pmod 2)\
&=(0pmod 4, 0pmod 2).
endalign$$

Closure follows from the closure of $Bbb Z_4$ under $+_4$ and of $Bbb Z_2$ under $+_2$.

share|cite|improve this answer

answered Apr 15 at 8:01

ShaunShaun

12k1212 gold badges3737 silver badges9292 bronze badges

$endgroup$

add a comment | 

$begingroup$

Associativity follows from that of $(Bbb Z_4, +_4)$ and of $(Bbb Z_2, +_2)$.

The identity is $(0pmod 4, 0pmod 2)$. (Why?)

The inverse of $(a,b)$ under $*$ is given by $(-a, -b)$ since $$beginalign(a,b)*(-a, -b)&=(a+(-a)pmod 4, b+(-b)pmod 2)\
&=(0pmod 4, 0pmod 2).
endalign$$

Closure follows from the closure of $Bbb Z_4$ under $+_4$ and of $Bbb Z_2$ under $+_2$.

share|cite|improve this answer

answered Apr 15 at 8:01

ShaunShaun

12k1212 gold badges3737 silver badges9292 bronze badges

$endgroup$

share|cite|improve this answer

answered Apr 15 at 8:01

ShaunShaun

12k1212 gold badges3737 silver badges9292 bronze badges

answered Apr 15 at 8:01

ShaunShaun

12k1212 gold badges3737 silver badges9292 bronze badges

answered Apr 15 at 8:01

ShaunShaun

12k1212 gold badges3737 silver badges9292 bronze badges

3

$begingroup$

In general, the group $Gtimes H$ (for groups $G,H$) with elements in the cartesian product (the set) of $G$ and $H$, and the group operation defined component-wise: The multiplication of $(g_1,h_1),(g_2,h_2)in Gtimes H$ being defined by $(g_1,h_1)(g_2,h_2)=(g_1g_2,h_1h_2)$ always gives a group. This is because the associativity follows from the group structure of $G$ and $H$, and the identity is $(1_G,1_H)$. You can check the other group axioms, which all follow from the fact that $G,H$ are groups. The group $Gtimes H$ is called the direct product of $G$ and $H$.

In particular, you just need that $mathbb Z_4$ and $mathbb Z_2$ to be groups to prove your conclusion in this case.

share|cite|improve this answer

answered Apr 15 at 8:06

YiFanYiFan

6,70822 gold badges1212 silver badges3434 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

In general, the group $Gtimes H$ (for groups $G,H$) with elements in the cartesian product (the set) of $G$ and $H$, and the group operation defined component-wise: The multiplication of $(g_1,h_1),(g_2,h_2)in Gtimes H$ being defined by $(g_1,h_1)(g_2,h_2)=(g_1g_2,h_1h_2)$ always gives a group. This is because the associativity follows from the group structure of $G$ and $H$, and the identity is $(1_G,1_H)$. You can check the other group axioms, which all follow from the fact that $G,H$ are groups. The group $Gtimes H$ is called the direct product of $G$ and $H$.

In particular, you just need that $mathbb Z_4$ and $mathbb Z_2$ to be groups to prove your conclusion in this case.

share|cite|improve this answer

answered Apr 15 at 8:06

YiFanYiFan

6,70822 gold badges1212 silver badges3434 bronze badges

$endgroup$

add a comment | 

$begingroup$

In general, the group $Gtimes H$ (for groups $G,H$) with elements in the cartesian product (the set) of $G$ and $H$, and the group operation defined component-wise: The multiplication of $(g_1,h_1),(g_2,h_2)in Gtimes H$ being defined by $(g_1,h_1)(g_2,h_2)=(g_1g_2,h_1h_2)$ always gives a group. This is because the associativity follows from the group structure of $G$ and $H$, and the identity is $(1_G,1_H)$. You can check the other group axioms, which all follow from the fact that $G,H$ are groups. The group $Gtimes H$ is called the direct product of $G$ and $H$.

In particular, you just need that $mathbb Z_4$ and $mathbb Z_2$ to be groups to prove your conclusion in this case.

share|cite|improve this answer

answered Apr 15 at 8:06

YiFanYiFan

6,70822 gold badges1212 silver badges3434 bronze badges

$endgroup$

share|cite|improve this answer

answered Apr 15 at 8:06

YiFanYiFan

6,70822 gold badges1212 silver badges3434 bronze badges

answered Apr 15 at 8:06

YiFanYiFan

6,70822 gold badges1212 silver badges3434 bronze badges

answered Apr 15 at 8:06

YiFanYiFan

6,70822 gold badges1212 silver badges3434 bronze badges