Cayley Graphs of Product groups is product of Cayley Graphs
$begingroup$
Let us have two Groups $G$ and $H$. Then, is the Cayley graph associated with $Gtimes H$, the direct product of the groups a product of the Cayley graphs associated to groups $G$ and $H$? If so, is the graph product a cartesian product?
I think yes, because the product group acts transitively on the product of Cayley graphs, I think? Thanks beforehand.
combinatorics group-theory graph-theory
asked Jan 7 at 10:43
vidyarthividyarthi
2,9741832
$endgroup$
|
show 3 more comments
$begingroup$
Let us have two Groups $G$ and $H$. Then, is the Cayley graph associated with $Gtimes H$, the direct product of the groups a product of the Cayley graphs associated to groups $G$ and $H$? If so, is the graph product a cartesian product?
I think yes, because the product group acts transitively on the product of Cayley graphs, I think? Thanks beforehand.
combinatorics group-theory graph-theory
asked Jan 7 at 10:43
vidyarthividyarthi
2,9741832
$endgroup$
$begingroup$
just write out the definitions...
$endgroup$
– mathworker21
Jan 7 at 10:44
$begingroup$
@mathworker21 you mean, the answer is yes? But, what would be the generating set of the new graph?
$endgroup$
– vidyarthi
Jan 7 at 11:00
$begingroup$
I just assumed your question would be easy. Now I read your question and am confused. First of all, how do you intersect two arbitrary groups? Second, you use the word "is" too many times in the first question. Third, what is the Cayley graph associated with a group? You need a generating set...
$endgroup$
– mathworker21
Jan 7 at 11:11
$begingroup$
@mathworker21 edited the question. Look into it now
$endgroup$
– vidyarthi
Jan 7 at 11:15
1
$begingroup$
I've already clearly said that you also need a generating set.
$endgroup$
– mathworker21
Jan 7 at 11:52
|
show 3 more comments
$begingroup$
Let us have two Groups $G$ and $H$. Then, is the Cayley graph associated with $Gtimes H$, the direct product of the groups a product of the Cayley graphs associated to groups $G$ and $H$? If so, is the graph product a cartesian product?
I think yes, because the product group acts transitively on the product of Cayley graphs, I think? Thanks beforehand.
combinatorics group-theory graph-theory
asked Jan 7 at 10:43
vidyarthividyarthi
2,9741832
$endgroup$
Let us have two Groups $G$ and $H$. Then, is the Cayley graph associated with $Gtimes H$, the direct product of the groups a product of the Cayley graphs associated to groups $G$ and $H$? If so, is the graph product a cartesian product?
I think yes, because the product group acts transitively on the product of Cayley graphs, I think? Thanks beforehand.
combinatorics group-theory graph-theory
combinatorics group-theory graph-theory
asked Jan 7 at 10:43
vidyarthividyarthi
2,9741832
asked Jan 7 at 10:43
vidyarthividyarthi
2,9741832
edited Jan 7 at 11:13
vidyarthi
asked Jan 7 at 10:43
vidyarthividyarthi
2,9741832
asked Jan 7 at 10:43
vidyarthividyarthi
2,9741832
asked Jan 7 at 10:43
vidyarthividyarthi
2,9741832
2,9741832
$begingroup$
just write out the definitions...
$endgroup$
– mathworker21
Jan 7 at 10:44
$begingroup$
@mathworker21 you mean, the answer is yes? But, what would be the generating set of the new graph?
$endgroup$
– vidyarthi
Jan 7 at 11:00
$begingroup$
I just assumed your question would be easy. Now I read your question and am confused. First of all, how do you intersect two arbitrary groups? Second, you use the word "is" too many times in the first question. Third, what is the Cayley graph associated with a group? You need a generating set...
$endgroup$
– mathworker21
Jan 7 at 11:11
$begingroup$
@mathworker21 edited the question. Look into it now
$endgroup$
– vidyarthi
Jan 7 at 11:15
1
$begingroup$
I've already clearly said that you also need a generating set.
$endgroup$
– mathworker21
Jan 7 at 11:52
|
show 3 more comments
$begingroup$
just write out the definitions...
$endgroup$
– mathworker21
Jan 7 at 10:44
$begingroup$
@mathworker21 you mean, the answer is yes? But, what would be the generating set of the new graph?
$endgroup$
– vidyarthi
Jan 7 at 11:00
$begingroup$
I just assumed your question would be easy. Now I read your question and am confused. First of all, how do you intersect two arbitrary groups? Second, you use the word "is" too many times in the first question. Third, what is the Cayley graph associated with a group? You need a generating set...
$endgroup$
– mathworker21
Jan 7 at 11:11
$begingroup$
@mathworker21 edited the question. Look into it now
$endgroup$
– vidyarthi
Jan 7 at 11:15
1
$begingroup$
I've already clearly said that you also need a generating set.
$endgroup$
– mathworker21
Jan 7 at 11:52
$begingroup$
just write out the definitions...
$endgroup$
– mathworker21
Jan 7 at 10:44
$begingroup$
just write out the definitions...
$endgroup$
– mathworker21
Jan 7 at 10:44
$begingroup$
@mathworker21 you mean, the answer is yes? But, what would be the generating set of the new graph?
$endgroup$
– vidyarthi
Jan 7 at 11:00
$begingroup$
@mathworker21 you mean, the answer is yes? But, what would be the generating set of the new graph?
$endgroup$
– vidyarthi
Jan 7 at 11:00
$begingroup$
I just assumed your question would be easy. Now I read your question and am confused. First of all, how do you intersect two arbitrary groups? Second, you use the word "is" too many times in the first question. Third, what is the Cayley graph associated with a group? You need a generating set...
$endgroup$
– mathworker21
Jan 7 at 11:11
$begingroup$
I just assumed your question would be easy. Now I read your question and am confused. First of all, how do you intersect two arbitrary groups? Second, you use the word "is" too many times in the first question. Third, what is the Cayley graph associated with a group? You need a generating set...
$endgroup$
– mathworker21
Jan 7 at 11:11
$begingroup$
@mathworker21 edited the question. Look into it now
$endgroup$
– vidyarthi
Jan 7 at 11:15
$begingroup$
@mathworker21 edited the question. Look into it now
$endgroup$
– vidyarthi
Jan 7 at 11:15
1
1
$begingroup$
I've already clearly said that you also need a generating set.
$endgroup$
– mathworker21
Jan 7 at 11:52
$begingroup$
I've already clearly said that you also need a generating set.
$endgroup$
– mathworker21
Jan 7 at 11:52
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Let $G$ and $H$ be groups, with generating sets $S$ and $T$. Then $(Stimes {e_H})cup ({e_G}times T)$ is a generating set for $Gtimes H$. Moreover, using Wikipedia's notation,
$$
Gamma(Gtimes H, (Stimes {e_H})cup ({e_G}times T))cong Gamma(G,S)times Gamma(H,T)
$$
where the rightmost $times$ is a Cartesian product. This is because the Cayley graph for $Gtimes H$ has an edge of the form $(g,h)to (gs,h)$ and $(g,h)to (g,ht)$, which exactly fits the definition of the Cartesian product of $Gamma(G,S)$ and $Gamma(H,T)$.
It may help to look at some examples, like $(G,S)=(H,T)=(mathbb Z_2,{1})$, where the generating set for $Gtimes H=mathbb Z_2times mathbb Z_2$ is ${(1,0),(0,1)}$.
answered Jan 7 at 17:06
Mike EarnestMike Earnest
23k12051
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Let $G$ and $H$ be groups, with generating sets $S$ and $T$. Then $(Stimes {e_H})cup ({e_G}times T)$ is a generating set for $Gtimes H$. Moreover, using Wikipedia's notation,
$$
Gamma(Gtimes H, (Stimes {e_H})cup ({e_G}times T))cong Gamma(G,S)times Gamma(H,T)
$$
where the rightmost $times$ is a Cartesian product. This is because the Cayley graph for $Gtimes H$ has an edge of the form $(g,h)to (gs,h)$ and $(g,h)to (g,ht)$, which exactly fits the definition of the Cartesian product of $Gamma(G,S)$ and $Gamma(H,T)$.
It may help to look at some examples, like $(G,S)=(H,T)=(mathbb Z_2,{1})$, where the generating set for $Gtimes H=mathbb Z_2times mathbb Z_2$ is ${(1,0),(0,1)}$.
answered Jan 7 at 17:06
Mike EarnestMike Earnest
23k12051
$endgroup$
add a comment |
$begingroup$
Let $G$ and $H$ be groups, with generating sets $S$ and $T$. Then $(Stimes {e_H})cup ({e_G}times T)$ is a generating set for $Gtimes H$. Moreover, using Wikipedia's notation,
$$
Gamma(Gtimes H, (Stimes {e_H})cup ({e_G}times T))cong Gamma(G,S)times Gamma(H,T)
$$
where the rightmost $times$ is a Cartesian product. This is because the Cayley graph for $Gtimes H$ has an edge of the form $(g,h)to (gs,h)$ and $(g,h)to (g,ht)$, which exactly fits the definition of the Cartesian product of $Gamma(G,S)$ and $Gamma(H,T)$.
It may help to look at some examples, like $(G,S)=(H,T)=(mathbb Z_2,{1})$, where the generating set for $Gtimes H=mathbb Z_2times mathbb Z_2$ is ${(1,0),(0,1)}$.
answered Jan 7 at 17:06
Mike EarnestMike Earnest
23k12051
$endgroup$
add a comment |
$begingroup$
Let $G$ and $H$ be groups, with generating sets $S$ and $T$. Then $(Stimes {e_H})cup ({e_G}times T)$ is a generating set for $Gtimes H$. Moreover, using Wikipedia's notation,
$$
Gamma(Gtimes H, (Stimes {e_H})cup ({e_G}times T))cong Gamma(G,S)times Gamma(H,T)
$$
where the rightmost $times$ is a Cartesian product. This is because the Cayley graph for $Gtimes H$ has an edge of the form $(g,h)to (gs,h)$ and $(g,h)to (g,ht)$, which exactly fits the definition of the Cartesian product of $Gamma(G,S)$ and $Gamma(H,T)$.
It may help to look at some examples, like $(G,S)=(H,T)=(mathbb Z_2,{1})$, where the generating set for $Gtimes H=mathbb Z_2times mathbb Z_2$ is ${(1,0),(0,1)}$.
answered Jan 7 at 17:06
Mike EarnestMike Earnest
23k12051
$endgroup$
Let $G$ and $H$ be groups, with generating sets $S$ and $T$. Then $(Stimes {e_H})cup ({e_G}times T)$ is a generating set for $Gtimes H$. Moreover, using Wikipedia's notation,
$$
Gamma(Gtimes H, (Stimes {e_H})cup ({e_G}times T))cong Gamma(G,S)times Gamma(H,T)
$$
where the rightmost $times$ is a Cartesian product. This is because the Cayley graph for $Gtimes H$ has an edge of the form $(g,h)to (gs,h)$ and $(g,h)to (g,ht)$, which exactly fits the definition of the Cartesian product of $Gamma(G,S)$ and $Gamma(H,T)$.
It may help to look at some examples, like $(G,S)=(H,T)=(mathbb Z_2,{1})$, where the generating set for $Gtimes H=mathbb Z_2times mathbb Z_2$ is ${(1,0),(0,1)}$.
answered Jan 7 at 17:06
Mike EarnestMike Earnest
23k12051
answered Jan 7 at 17:06
Mike EarnestMike Earnest
23k12051
answered Jan 7 at 17:06
Mike EarnestMike Earnest
23k12051
answered Jan 7 at 17:06
Mike EarnestMike Earnest
23k12051
23k12051
add a comment |
add a comment |
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$begingroup$
just write out the definitions...
$endgroup$
– mathworker21
Jan 7 at 10:44
$begingroup$
@mathworker21 you mean, the answer is yes? But, what would be the generating set of the new graph?
$endgroup$
– vidyarthi
Jan 7 at 11:00
$begingroup$
I just assumed your question would be easy. Now I read your question and am confused. First of all, how do you intersect two arbitrary groups? Second, you use the word "is" too many times in the first question. Third, what is the Cayley graph associated with a group? You need a generating set...
$endgroup$
– mathworker21
Jan 7 at 11:11
$begingroup$
@mathworker21 edited the question. Look into it now
$endgroup$
– vidyarthi
Jan 7 at 11:15
1
$begingroup$
I've already clearly said that you also need a generating set.
$endgroup$
– mathworker21
Jan 7 at 11:52