Group presentation of a direct product.
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Suppose $G_1=langle X_1|R_1rangle$, $G_2=langle X_2|R_2rangle$, $X_1cap X_2=emptyset$. I want to show that $G_1times G_2=langle X_1cup X_2|R_1cup R_2cup[X_1,X_2] rangle$.
By using universal property of free groups, I have obtained homomorphisms $f_i:F(X_i)rightarrow F(X_1 cup X_2)$. Also I can define homomorphisms $pi_i:F(X_i) rightarrow G_i$ such that $ker(pi_i)= R_i^{F(X_i)}=$ normal closure of $R_i$.
I want to define a group $G = langle X_1cup X_2|Rrangle$ with $R$ satisfying the conditions of direct product, and somehow $R$ contains $[X_1,X_2]$. But I'm kind of stuck here.
group-theory group-presentation direct-product combinatorial-group-theory
edited Dec 3 at 1:11
Shaun
8,115113577
asked Jun 12 '17 at 11:18
Sid Caroline
1,5532514
|
show 2 more comments
up vote
2
down vote
favorite
Suppose $G_1=langle X_1|R_1rangle$, $G_2=langle X_2|R_2rangle$, $X_1cap X_2=emptyset$. I want to show that $G_1times G_2=langle X_1cup X_2|R_1cup R_2cup[X_1,X_2] rangle$.
By using universal property of free groups, I have obtained homomorphisms $f_i:F(X_i)rightarrow F(X_1 cup X_2)$. Also I can define homomorphisms $pi_i:F(X_i) rightarrow G_i$ such that $ker(pi_i)= R_i^{F(X_i)}=$ normal closure of $R_i$.
I want to define a group $G = langle X_1cup X_2|Rrangle$ with $R$ satisfying the conditions of direct product, and somehow $R$ contains $[X_1,X_2]$. But I'm kind of stuck here.
group-theory group-presentation direct-product combinatorial-group-theory
edited Dec 3 at 1:11
Shaun
8,115113577
asked Jun 12 '17 at 11:18
Sid Caroline
1,5532514
4
Try sending $(g,h)$ to $gh$ in the group defined using generators and relations
– leibnewtz
Jun 12 '17 at 11:24
Can you be a little more explicit? Suppose that I define $H=langle X_1 cup X_2 | R_1 cup R_2 cup [X_1,X_2]rangle$, and $G_1 times G_2={(g,h)|gin G_1, h in G_2}$, and I define a function $phi:G_1times G_2 rightarrow H$, sending $(g,h)$ to $gh$, are you saying that I should try to prove $phi$ is an isomorphism?
– Sid Caroline
Jun 12 '17 at 11:54
1
Yup. Try to show first that it's a homomorphism, then that it's injective, and last that it's surjective. At each step you'll have to use the relations in $H$.
– leibnewtz
Jun 12 '17 at 11:56
1
You could define a homomorphism $psi:H to G_1 times G_2$ that restricts to the identity on $X_1$ and $X_2$, and then show that $phi$ and $psi$ are mutually inverse maps. That implies that they are both bijective.
– Derek Holt
Jun 12 '17 at 16:43
1
If $gh=e$, then $g=h^{-1}$. What can you say about $g$ and $h$? Hint: They're words in $X_1$ and $X_2$ respectively. For surjectivity, take an arbitrary word in $X_1 cup X_2$, and see if you can rewrite it in an appropriate sense. Or you could take Derek Holt's approach. Either way should work
– leibnewtz
Jun 12 '17 at 21:57
|
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose $G_1=langle X_1|R_1rangle$, $G_2=langle X_2|R_2rangle$, $X_1cap X_2=emptyset$. I want to show that $G_1times G_2=langle X_1cup X_2|R_1cup R_2cup[X_1,X_2] rangle$.
By using universal property of free groups, I have obtained homomorphisms $f_i:F(X_i)rightarrow F(X_1 cup X_2)$. Also I can define homomorphisms $pi_i:F(X_i) rightarrow G_i$ such that $ker(pi_i)= R_i^{F(X_i)}=$ normal closure of $R_i$.
I want to define a group $G = langle X_1cup X_2|Rrangle$ with $R$ satisfying the conditions of direct product, and somehow $R$ contains $[X_1,X_2]$. But I'm kind of stuck here.
group-theory group-presentation direct-product combinatorial-group-theory
edited Dec 3 at 1:11
Shaun
8,115113577
asked Jun 12 '17 at 11:18
Sid Caroline
1,5532514
Suppose $G_1=langle X_1|R_1rangle$, $G_2=langle X_2|R_2rangle$, $X_1cap X_2=emptyset$. I want to show that $G_1times G_2=langle X_1cup X_2|R_1cup R_2cup[X_1,X_2] rangle$.
By using universal property of free groups, I have obtained homomorphisms $f_i:F(X_i)rightarrow F(X_1 cup X_2)$. Also I can define homomorphisms $pi_i:F(X_i) rightarrow G_i$ such that $ker(pi_i)= R_i^{F(X_i)}=$ normal closure of $R_i$.
I want to define a group $G = langle X_1cup X_2|Rrangle$ with $R$ satisfying the conditions of direct product, and somehow $R$ contains $[X_1,X_2]$. But I'm kind of stuck here.
group-theory group-presentation direct-product combinatorial-group-theory
group-theory group-presentation direct-product combinatorial-group-theory
edited Dec 3 at 1:11
Shaun
8,115113577
asked Jun 12 '17 at 11:18
Sid Caroline
1,5532514
edited Dec 3 at 1:11
Shaun
8,115113577
asked Jun 12 '17 at 11:18
Sid Caroline
1,5532514
edited Dec 3 at 1:11
Shaun
8,115113577
edited Dec 3 at 1:11
Shaun
8,115113577
edited Dec 3 at 1:11
Shaun
8,115113577
8,115113577
asked Jun 12 '17 at 11:18
Sid Caroline
1,5532514
asked Jun 12 '17 at 11:18
Sid Caroline
1,5532514
asked Jun 12 '17 at 11:18
Sid Caroline
1,5532514
1,5532514
4
Try sending $(g,h)$ to $gh$ in the group defined using generators and relations
– leibnewtz
Jun 12 '17 at 11:24
Can you be a little more explicit? Suppose that I define $H=langle X_1 cup X_2 | R_1 cup R_2 cup [X_1,X_2]rangle$, and $G_1 times G_2={(g,h)|gin G_1, h in G_2}$, and I define a function $phi:G_1times G_2 rightarrow H$, sending $(g,h)$ to $gh$, are you saying that I should try to prove $phi$ is an isomorphism?
– Sid Caroline
Jun 12 '17 at 11:54
1
Yup. Try to show first that it's a homomorphism, then that it's injective, and last that it's surjective. At each step you'll have to use the relations in $H$.
– leibnewtz
Jun 12 '17 at 11:56
1
You could define a homomorphism $psi:H to G_1 times G_2$ that restricts to the identity on $X_1$ and $X_2$, and then show that $phi$ and $psi$ are mutually inverse maps. That implies that they are both bijective.
– Derek Holt
Jun 12 '17 at 16:43
1
If $gh=e$, then $g=h^{-1}$. What can you say about $g$ and $h$? Hint: They're words in $X_1$ and $X_2$ respectively. For surjectivity, take an arbitrary word in $X_1 cup X_2$, and see if you can rewrite it in an appropriate sense. Or you could take Derek Holt's approach. Either way should work
– leibnewtz
Jun 12 '17 at 21:57
|
show 2 more comments
4
Try sending $(g,h)$ to $gh$ in the group defined using generators and relations
– leibnewtz
Jun 12 '17 at 11:24
Can you be a little more explicit? Suppose that I define $H=langle X_1 cup X_2 | R_1 cup R_2 cup [X_1,X_2]rangle$, and $G_1 times G_2={(g,h)|gin G_1, h in G_2}$, and I define a function $phi:G_1times G_2 rightarrow H$, sending $(g,h)$ to $gh$, are you saying that I should try to prove $phi$ is an isomorphism?
– Sid Caroline
Jun 12 '17 at 11:54
1
Yup. Try to show first that it's a homomorphism, then that it's injective, and last that it's surjective. At each step you'll have to use the relations in $H$.
– leibnewtz
Jun 12 '17 at 11:56
1
You could define a homomorphism $psi:H to G_1 times G_2$ that restricts to the identity on $X_1$ and $X_2$, and then show that $phi$ and $psi$ are mutually inverse maps. That implies that they are both bijective.
– Derek Holt
Jun 12 '17 at 16:43
1
If $gh=e$, then $g=h^{-1}$. What can you say about $g$ and $h$? Hint: They're words in $X_1$ and $X_2$ respectively. For surjectivity, take an arbitrary word in $X_1 cup X_2$, and see if you can rewrite it in an appropriate sense. Or you could take Derek Holt's approach. Either way should work
– leibnewtz
Jun 12 '17 at 21:57
4
4
Try sending $(g,h)$ to $gh$ in the group defined using generators and relations
– leibnewtz
Jun 12 '17 at 11:24
Try sending $(g,h)$ to $gh$ in the group defined using generators and relations
– leibnewtz
Jun 12 '17 at 11:24
Can you be a little more explicit? Suppose that I define $H=langle X_1 cup X_2 | R_1 cup R_2 cup [X_1,X_2]rangle$, and $G_1 times G_2={(g,h)|gin G_1, h in G_2}$, and I define a function $phi:G_1times G_2 rightarrow H$, sending $(g,h)$ to $gh$, are you saying that I should try to prove $phi$ is an isomorphism?
– Sid Caroline
Jun 12 '17 at 11:54
Can you be a little more explicit? Suppose that I define $H=langle X_1 cup X_2 | R_1 cup R_2 cup [X_1,X_2]rangle$, and $G_1 times G_2={(g,h)|gin G_1, h in G_2}$, and I define a function $phi:G_1times G_2 rightarrow H$, sending $(g,h)$ to $gh$, are you saying that I should try to prove $phi$ is an isomorphism?
– Sid Caroline
Jun 12 '17 at 11:54
1
1
Yup. Try to show first that it's a homomorphism, then that it's injective, and last that it's surjective. At each step you'll have to use the relations in $H$.
– leibnewtz
Jun 12 '17 at 11:56
Yup. Try to show first that it's a homomorphism, then that it's injective, and last that it's surjective. At each step you'll have to use the relations in $H$.
– leibnewtz
Jun 12 '17 at 11:56
1
1
You could define a homomorphism $psi:H to G_1 times G_2$ that restricts to the identity on $X_1$ and $X_2$, and then show that $phi$ and $psi$ are mutually inverse maps. That implies that they are both bijective.
– Derek Holt
Jun 12 '17 at 16:43
You could define a homomorphism $psi:H to G_1 times G_2$ that restricts to the identity on $X_1$ and $X_2$, and then show that $phi$ and $psi$ are mutually inverse maps. That implies that they are both bijective.
– Derek Holt
Jun 12 '17 at 16:43
1
1
If $gh=e$, then $g=h^{-1}$. What can you say about $g$ and $h$? Hint: They're words in $X_1$ and $X_2$ respectively. For surjectivity, take an arbitrary word in $X_1 cup X_2$, and see if you can rewrite it in an appropriate sense. Or you could take Derek Holt's approach. Either way should work
– leibnewtz
Jun 12 '17 at 21:57
If $gh=e$, then $g=h^{-1}$. What can you say about $g$ and $h$? Hint: They're words in $X_1$ and $X_2$ respectively. For surjectivity, take an arbitrary word in $X_1 cup X_2$, and see if you can rewrite it in an appropriate sense. Or you could take Derek Holt's approach. Either way should work
– leibnewtz
Jun 12 '17 at 21:57
|
show 2 more comments
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4
Try sending $(g,h)$ to $gh$ in the group defined using generators and relations
– leibnewtz
Jun 12 '17 at 11:24
Can you be a little more explicit? Suppose that I define $H=langle X_1 cup X_2 | R_1 cup R_2 cup [X_1,X_2]rangle$, and $G_1 times G_2={(g,h)|gin G_1, h in G_2}$, and I define a function $phi:G_1times G_2 rightarrow H$, sending $(g,h)$ to $gh$, are you saying that I should try to prove $phi$ is an isomorphism?
– Sid Caroline
Jun 12 '17 at 11:54
1
Yup. Try to show first that it's a homomorphism, then that it's injective, and last that it's surjective. At each step you'll have to use the relations in $H$.
– leibnewtz
Jun 12 '17 at 11:56
1
You could define a homomorphism $psi:H to G_1 times G_2$ that restricts to the identity on $X_1$ and $X_2$, and then show that $phi$ and $psi$ are mutually inverse maps. That implies that they are both bijective.
– Derek Holt
Jun 12 '17 at 16:43
1
If $gh=e$, then $g=h^{-1}$. What can you say about $g$ and $h$? Hint: They're words in $X_1$ and $X_2$ respectively. For surjectivity, take an arbitrary word in $X_1 cup X_2$, and see if you can rewrite it in an appropriate sense. Or you could take Derek Holt's approach. Either way should work
– leibnewtz
Jun 12 '17 at 21:57