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-3

$begingroup$

Let $P$ be a subgroup of the group $G$ and $Q$ be a subgroup of the group $H$. How would you Prove that:
$P times Q$ is a subgroup of the direct product $G times H$.

group-theory normal-subgroups

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edited yesterday

Vinyl_cape_jawa

3,32211333

asked yesterday

George HallGeorge Hall

11

New contributor

George Hall is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by verret, Mark, Vinyl_cape_jawa, Shaun, José Carlos Santos yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – verret, Mark, Vinyl_cape_jawa, Shaun, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What did you try? Let's start from the beginning: do you know what is the definition of a subgroup?
  $endgroup$
  – Mark
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Using the Subgroup test, I can prove that the group is closed under taking inverses but I do not understand how to show that it is closed under the binary operation
  $endgroup$
  – George Hall
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Let's suppose $(p_1,q_1)$ and $(p_2,q_2)$ are elements in $Ptimes Q$. What is their product by definition?
  $endgroup$
  – Mark
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (p1p2) and (q1,q2) ?
  $endgroup$
  – George Hall
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The operation is multiplying in every coordinate. So it is $(p_1p_2,q_1q_2)$. Well, can you see why is it in $Ptimes Q$?
  $endgroup$
  – Mark
  yesterday
  

  
  
  
  

 | 
show 2 more comments

-3

$begingroup$

Let $P$ be a subgroup of the group $G$ and $Q$ be a subgroup of the group $H$. How would you Prove that:
$P times Q$ is a subgroup of the direct product $G times H$.

group-theory normal-subgroups

share|cite|improve this question

edited yesterday

Vinyl_cape_jawa

3,32211333

asked yesterday

George HallGeorge Hall

11

New contributor

George Hall is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by verret, Mark, Vinyl_cape_jawa, Shaun, José Carlos Santos yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – verret, Mark, Vinyl_cape_jawa, Shaun, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What did you try? Let's start from the beginning: do you know what is the definition of a subgroup?
  $endgroup$
  – Mark
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Using the Subgroup test, I can prove that the group is closed under taking inverses but I do not understand how to show that it is closed under the binary operation
  $endgroup$
  – George Hall
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Let's suppose $(p_1,q_1)$ and $(p_2,q_2)$ are elements in $Ptimes Q$. What is their product by definition?
  $endgroup$
  – Mark
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (p1p2) and (q1,q2) ?
  $endgroup$
  – George Hall
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The operation is multiplying in every coordinate. So it is $(p_1p_2,q_1q_2)$. Well, can you see why is it in $Ptimes Q$?
  $endgroup$
  – Mark
  yesterday
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Let $P$ be a subgroup of the group $G$ and $Q$ be a subgroup of the group $H$. How would you Prove that:
$P times Q$ is a subgroup of the direct product $G times H$.

group-theory normal-subgroups

share|cite|improve this question

edited yesterday

Vinyl_cape_jawa

3,32211333

asked yesterday

George HallGeorge Hall

11

New contributor

George Hall is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited yesterday

Vinyl_cape_jawa

3,32211333

asked yesterday

George HallGeorge Hall

11

New contributor

George Hall is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited yesterday

Vinyl_cape_jawa

3,32211333

asked yesterday

George HallGeorge Hall

11

New contributor

George Hall is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited yesterday

Vinyl_cape_jawa

3,32211333

edited yesterday

Vinyl_cape_jawa

3,32211333

asked yesterday

George HallGeorge Hall

11

New contributor

George Hall is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

George HallGeorge Hall

11

2 Answers
2

active

oldest

votes

0

$begingroup$

If $P leq G$ and $Q le H$, we want to show that $P times Q le G times H$.

First, $P times Q = {(p,q) : p in P wedge q in Q }$, while $G times H = {(g,h) : g in G wedge h in H}$.

The product is defined by $(p_1,q_1)cdot(p_2,q_2) = (p_1p_2,q_1q_2)$.

Let's check the group axoims:

G1: For all $(p_1,q_1),(p_2,q_2) in P times Q$, $(p_1,q_1)cdot(p_2,q_2) in P times Q$.

Since $P$ and $Q$ are groups, $p_1p_2 in P$ and $q_1q_2 in Q$, so $(p_1p_2,q_1q_2) in P times Q$.

G2: Associativity is inherited from $P$ and $Q$.

$[(p_1,q_1)cdot(p_2,q_2)]cdot(p_3,q_3) = (p_1,q_1) cdot [(p_2,q_2)cdot(p_3,q_3)]$

because $(p_1p_2)p_3=p_1(p_2p_3)$ and $(q_1q_2)q_3 = q_1(q_2q_3)$

G3: The identity element is $(e_P,e_Q) in P times Q$,

where $e_P in P$ and $e_Q in Q$ are the identity elements of $P$ and $Q$.

G4: The inverse of $(p,q) in P times Q$ is $(p^{-1},q^{-1}) in P times Q$

where $p^{-1} in P$ is the inverse of $p in P$ and $q^{-1} in Q$ is the inverse of $q in Q$.

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answered yesterday

Fly by NightFly by Night

26k32978

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add a comment | 

0

$begingroup$

The fastest way would be using the following fact:

Let $G$ be a group and let $H$ be a subset of $G$. Then $H$ is a subgroup of $G$ $iff$ $H$ is nonempty and $ab^{-1}in H$, for all $a,bin H$

In your case we have to show the following:

• $Ptimes Q$ is nonempty




• $(p_1,q_1)cdot (p_2,q_2)^{-1}in Ptimes Q$ for all $(p_1,q_1),(p_2,q_2)in Ptimes Q$

Firstly $P$ and $Q$ are subgroups themselves so they at least should have an identity element hence they are nonempty, whence their direct product is nonempty.

The second requirement follows from the fact that $P$ and $Q$ were subgroups themselves.

share|cite|improve this answer

answered yesterday

Vinyl_cape_jawaVinyl_cape_jawa

3,32211333

$endgroup$

add a comment | 

0

$begingroup$

If $P leq G$ and $Q le H$, we want to show that $P times Q le G times H$.

First, $P times Q = {(p,q) : p in P wedge q in Q }$, while $G times H = {(g,h) : g in G wedge h in H}$.

The product is defined by $(p_1,q_1)cdot(p_2,q_2) = (p_1p_2,q_1q_2)$.

Let's check the group axoims:

G1: For all $(p_1,q_1),(p_2,q_2) in P times Q$, $(p_1,q_1)cdot(p_2,q_2) in P times Q$.

Since $P$ and $Q$ are groups, $p_1p_2 in P$ and $q_1q_2 in Q$, so $(p_1p_2,q_1q_2) in P times Q$.

G2: Associativity is inherited from $P$ and $Q$.

$[(p_1,q_1)cdot(p_2,q_2)]cdot(p_3,q_3) = (p_1,q_1) cdot [(p_2,q_2)cdot(p_3,q_3)]$

because $(p_1p_2)p_3=p_1(p_2p_3)$ and $(q_1q_2)q_3 = q_1(q_2q_3)$

G3: The identity element is $(e_P,e_Q) in P times Q$,

where $e_P in P$ and $e_Q in Q$ are the identity elements of $P$ and $Q$.

G4: The inverse of $(p,q) in P times Q$ is $(p^{-1},q^{-1}) in P times Q$

where $p^{-1} in P$ is the inverse of $p in P$ and $q^{-1} in Q$ is the inverse of $q in Q$.

share|cite|improve this answer

answered yesterday

Fly by NightFly by Night

26k32978

$endgroup$

add a comment | 

0

$begingroup$

If $P leq G$ and $Q le H$, we want to show that $P times Q le G times H$.

First, $P times Q = {(p,q) : p in P wedge q in Q }$, while $G times H = {(g,h) : g in G wedge h in H}$.

The product is defined by $(p_1,q_1)cdot(p_2,q_2) = (p_1p_2,q_1q_2)$.

Let's check the group axoims:

G1: For all $(p_1,q_1),(p_2,q_2) in P times Q$, $(p_1,q_1)cdot(p_2,q_2) in P times Q$.

Since $P$ and $Q$ are groups, $p_1p_2 in P$ and $q_1q_2 in Q$, so $(p_1p_2,q_1q_2) in P times Q$.

G2: Associativity is inherited from $P$ and $Q$.

$[(p_1,q_1)cdot(p_2,q_2)]cdot(p_3,q_3) = (p_1,q_1) cdot [(p_2,q_2)cdot(p_3,q_3)]$

because $(p_1p_2)p_3=p_1(p_2p_3)$ and $(q_1q_2)q_3 = q_1(q_2q_3)$

G3: The identity element is $(e_P,e_Q) in P times Q$,

where $e_P in P$ and $e_Q in Q$ are the identity elements of $P$ and $Q$.

G4: The inverse of $(p,q) in P times Q$ is $(p^{-1},q^{-1}) in P times Q$

where $p^{-1} in P$ is the inverse of $p in P$ and $q^{-1} in Q$ is the inverse of $q in Q$.

share|cite|improve this answer

answered yesterday

Fly by NightFly by Night

26k32978

$endgroup$

add a comment | 

$begingroup$

If $P leq G$ and $Q le H$, we want to show that $P times Q le G times H$.

First, $P times Q = {(p,q) : p in P wedge q in Q }$, while $G times H = {(g,h) : g in G wedge h in H}$.

The product is defined by $(p_1,q_1)cdot(p_2,q_2) = (p_1p_2,q_1q_2)$.

Let's check the group axoims:

G1: For all $(p_1,q_1),(p_2,q_2) in P times Q$, $(p_1,q_1)cdot(p_2,q_2) in P times Q$.

Since $P$ and $Q$ are groups, $p_1p_2 in P$ and $q_1q_2 in Q$, so $(p_1p_2,q_1q_2) in P times Q$.

G2: Associativity is inherited from $P$ and $Q$.

$[(p_1,q_1)cdot(p_2,q_2)]cdot(p_3,q_3) = (p_1,q_1) cdot [(p_2,q_2)cdot(p_3,q_3)]$

because $(p_1p_2)p_3=p_1(p_2p_3)$ and $(q_1q_2)q_3 = q_1(q_2q_3)$

G3: The identity element is $(e_P,e_Q) in P times Q$,

where $e_P in P$ and $e_Q in Q$ are the identity elements of $P$ and $Q$.

G4: The inverse of $(p,q) in P times Q$ is $(p^{-1},q^{-1}) in P times Q$

where $p^{-1} in P$ is the inverse of $p in P$ and $q^{-1} in Q$ is the inverse of $q in Q$.

share|cite|improve this answer

answered yesterday

Fly by NightFly by Night

26k32978

$endgroup$

share|cite|improve this answer

answered yesterday

Fly by NightFly by Night

26k32978

answered yesterday

Fly by NightFly by Night

26k32978

answered yesterday

Fly by NightFly by Night

26k32978

0

$begingroup$

The fastest way would be using the following fact:

Let $G$ be a group and let $H$ be a subset of $G$. Then $H$ is a subgroup of $G$ $iff$ $H$ is nonempty and $ab^{-1}in H$, for all $a,bin H$

In your case we have to show the following:

• $Ptimes Q$ is nonempty




• $(p_1,q_1)cdot (p_2,q_2)^{-1}in Ptimes Q$ for all $(p_1,q_1),(p_2,q_2)in Ptimes Q$

Firstly $P$ and $Q$ are subgroups themselves so they at least should have an identity element hence they are nonempty, whence their direct product is nonempty.

The second requirement follows from the fact that $P$ and $Q$ were subgroups themselves.

share|cite|improve this answer

answered yesterday

Vinyl_cape_jawaVinyl_cape_jawa

3,32211333

$endgroup$

add a comment | 

0

$begingroup$

The fastest way would be using the following fact:

Let $G$ be a group and let $H$ be a subset of $G$. Then $H$ is a subgroup of $G$ $iff$ $H$ is nonempty and $ab^{-1}in H$, for all $a,bin H$

In your case we have to show the following:

• $Ptimes Q$ is nonempty




• $(p_1,q_1)cdot (p_2,q_2)^{-1}in Ptimes Q$ for all $(p_1,q_1),(p_2,q_2)in Ptimes Q$

Firstly $P$ and $Q$ are subgroups themselves so they at least should have an identity element hence they are nonempty, whence their direct product is nonempty.

The second requirement follows from the fact that $P$ and $Q$ were subgroups themselves.

share|cite|improve this answer

answered yesterday

Vinyl_cape_jawaVinyl_cape_jawa

3,32211333

$endgroup$

add a comment | 

$begingroup$

The fastest way would be using the following fact:

Let $G$ be a group and let $H$ be a subset of $G$. Then $H$ is a subgroup of $G$ $iff$ $H$ is nonempty and $ab^{-1}in H$, for all $a,bin H$

In your case we have to show the following:

• $Ptimes Q$ is nonempty




• $(p_1,q_1)cdot (p_2,q_2)^{-1}in Ptimes Q$ for all $(p_1,q_1),(p_2,q_2)in Ptimes Q$

Firstly $P$ and $Q$ are subgroups themselves so they at least should have an identity element hence they are nonempty, whence their direct product is nonempty.

The second requirement follows from the fact that $P$ and $Q$ were subgroups themselves.

share|cite|improve this answer

answered yesterday

Vinyl_cape_jawaVinyl_cape_jawa

3,32211333

$endgroup$

share|cite|improve this answer

answered yesterday

Vinyl_cape_jawaVinyl_cape_jawa

3,32211333

answered yesterday

Vinyl_cape_jawaVinyl_cape_jawa

3,32211333

answered yesterday

Vinyl_cape_jawaVinyl_cape_jawa

3,32211333