Coproduct of $C_2$ and $C_3$ in $mathsf{Grp}$
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I have been self-studying from Aluffi's Algebra: Chapter 0. I am looking at Chapter II section 3, exercise 3.8. Here it is:
As far as I can tell, I need to do the following:
- Describe the projections $pi_2 : C_2 to G$ and $pi_3 : C_3 to G$.
- Show that the universal property for coproducts is satisfied: that is, for any group $A$ and any choice of group homomorphisms $varphi_2 : C_2 to A$ and $varphi_3 : C_3 to A$, there exists a unique group homomorphism $sigma : G to A$.
Giving a more concrete description of the cyclic groups, take $C_2 = {0, 1}$ and $C_3 = {0, 1, 2}$. Then the projections are $pi_2(k) = x^k$ where $k=0,1$, and $pi_3(l) = y^l$ where $l=0,1,2$. I think that this takes care of (1).
Beyond this, it is not clear to me what I should do. Any help would be appreciated!
abstract-algebra group-theory category-theory
asked Nov 21 at 10:08
Josh
657
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up vote
1
down vote
favorite
I have been self-studying from Aluffi's Algebra: Chapter 0. I am looking at Chapter II section 3, exercise 3.8. Here it is:
As far as I can tell, I need to do the following:
- Describe the projections $pi_2 : C_2 to G$ and $pi_3 : C_3 to G$.
- Show that the universal property for coproducts is satisfied: that is, for any group $A$ and any choice of group homomorphisms $varphi_2 : C_2 to A$ and $varphi_3 : C_3 to A$, there exists a unique group homomorphism $sigma : G to A$.
Giving a more concrete description of the cyclic groups, take $C_2 = {0, 1}$ and $C_3 = {0, 1, 2}$. Then the projections are $pi_2(k) = x^k$ where $k=0,1$, and $pi_3(l) = y^l$ where $l=0,1,2$. I think that this takes care of (1).
Beyond this, it is not clear to me what I should do. Any help would be appreciated!
abstract-algebra group-theory category-theory
asked Nov 21 at 10:08
Josh
657
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have been self-studying from Aluffi's Algebra: Chapter 0. I am looking at Chapter II section 3, exercise 3.8. Here it is:
As far as I can tell, I need to do the following:
- Describe the projections $pi_2 : C_2 to G$ and $pi_3 : C_3 to G$.
- Show that the universal property for coproducts is satisfied: that is, for any group $A$ and any choice of group homomorphisms $varphi_2 : C_2 to A$ and $varphi_3 : C_3 to A$, there exists a unique group homomorphism $sigma : G to A$.
Giving a more concrete description of the cyclic groups, take $C_2 = {0, 1}$ and $C_3 = {0, 1, 2}$. Then the projections are $pi_2(k) = x^k$ where $k=0,1$, and $pi_3(l) = y^l$ where $l=0,1,2$. I think that this takes care of (1).
Beyond this, it is not clear to me what I should do. Any help would be appreciated!
abstract-algebra group-theory category-theory
asked Nov 21 at 10:08
Josh
657
I have been self-studying from Aluffi's Algebra: Chapter 0. I am looking at Chapter II section 3, exercise 3.8. Here it is:
As far as I can tell, I need to do the following:
- Describe the projections $pi_2 : C_2 to G$ and $pi_3 : C_3 to G$.
- Show that the universal property for coproducts is satisfied: that is, for any group $A$ and any choice of group homomorphisms $varphi_2 : C_2 to A$ and $varphi_3 : C_3 to A$, there exists a unique group homomorphism $sigma : G to A$.
Giving a more concrete description of the cyclic groups, take $C_2 = {0, 1}$ and $C_3 = {0, 1, 2}$. Then the projections are $pi_2(k) = x^k$ where $k=0,1$, and $pi_3(l) = y^l$ where $l=0,1,2$. I think that this takes care of (1).
Beyond this, it is not clear to me what I should do. Any help would be appreciated!
abstract-algebra group-theory category-theory
abstract-algebra group-theory category-theory
asked Nov 21 at 10:08
Josh
657
asked Nov 21 at 10:08
Josh
657
asked Nov 21 at 10:08
Josh
657
asked Nov 21 at 10:08
Josh
657
asked Nov 21 at 10:08
Josh
657
657
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2 Answers
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You can look at $G$ as a group that has strings like $xyxy^2xy$ or $y^2xyxy$ as elements.
Then $phi:Gto A$ on e.g. $xyxy^2xy$ must be prescribed by:$$xyxy^2xymapstophi_2(x)phi_3(y)phi_2(x)phi_3(y)^2phi_2(x)phi_3(y)$$
This $phi$ must be shown to be a group homomorphism with $phicircpi_2=phi_2$ and $phicircpi_3=phi_3$, and must be shown to be unique in satisfying this.
answered Nov 21 at 10:37
drhab
96.3k543126
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A map out of a group is just given by a map on the generators such that the relations are satisfied. Thus a map from $G$ to $H$ is uniquely determined by two elements of $H$, one which squares and one which cubes to the identity. This is exactly the description of a pair of maps from $C_2$ and $C_3$ to $H$, and the correspondence is induced by maps from $C_2$ and $C_3$ to $H$ taking a generator to a generator, as you say.
answered Nov 21 at 15:52
Kevin Carlson
32.4k23270
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2 Answers
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2 Answers
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up vote
1
down vote
You can look at $G$ as a group that has strings like $xyxy^2xy$ or $y^2xyxy$ as elements.
Then $phi:Gto A$ on e.g. $xyxy^2xy$ must be prescribed by:$$xyxy^2xymapstophi_2(x)phi_3(y)phi_2(x)phi_3(y)^2phi_2(x)phi_3(y)$$
This $phi$ must be shown to be a group homomorphism with $phicircpi_2=phi_2$ and $phicircpi_3=phi_3$, and must be shown to be unique in satisfying this.
answered Nov 21 at 10:37
drhab
96.3k543126
add a comment |
up vote
1
down vote
You can look at $G$ as a group that has strings like $xyxy^2xy$ or $y^2xyxy$ as elements.
Then $phi:Gto A$ on e.g. $xyxy^2xy$ must be prescribed by:$$xyxy^2xymapstophi_2(x)phi_3(y)phi_2(x)phi_3(y)^2phi_2(x)phi_3(y)$$
This $phi$ must be shown to be a group homomorphism with $phicircpi_2=phi_2$ and $phicircpi_3=phi_3$, and must be shown to be unique in satisfying this.
answered Nov 21 at 10:37
drhab
96.3k543126
add a comment |
up vote
1
down vote
up vote
1
down vote
You can look at $G$ as a group that has strings like $xyxy^2xy$ or $y^2xyxy$ as elements.
Then $phi:Gto A$ on e.g. $xyxy^2xy$ must be prescribed by:$$xyxy^2xymapstophi_2(x)phi_3(y)phi_2(x)phi_3(y)^2phi_2(x)phi_3(y)$$
This $phi$ must be shown to be a group homomorphism with $phicircpi_2=phi_2$ and $phicircpi_3=phi_3$, and must be shown to be unique in satisfying this.
answered Nov 21 at 10:37
drhab
96.3k543126
You can look at $G$ as a group that has strings like $xyxy^2xy$ or $y^2xyxy$ as elements.
Then $phi:Gto A$ on e.g. $xyxy^2xy$ must be prescribed by:$$xyxy^2xymapstophi_2(x)phi_3(y)phi_2(x)phi_3(y)^2phi_2(x)phi_3(y)$$
This $phi$ must be shown to be a group homomorphism with $phicircpi_2=phi_2$ and $phicircpi_3=phi_3$, and must be shown to be unique in satisfying this.
answered Nov 21 at 10:37
drhab
96.3k543126
answered Nov 21 at 10:37
drhab
96.3k543126
answered Nov 21 at 10:37
drhab
96.3k543126
answered Nov 21 at 10:37
drhab
96.3k543126
96.3k543126
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add a comment |
up vote
0
down vote
A map out of a group is just given by a map on the generators such that the relations are satisfied. Thus a map from $G$ to $H$ is uniquely determined by two elements of $H$, one which squares and one which cubes to the identity. This is exactly the description of a pair of maps from $C_2$ and $C_3$ to $H$, and the correspondence is induced by maps from $C_2$ and $C_3$ to $H$ taking a generator to a generator, as you say.
answered Nov 21 at 15:52
Kevin Carlson
32.4k23270
add a comment |
up vote
0
down vote
A map out of a group is just given by a map on the generators such that the relations are satisfied. Thus a map from $G$ to $H$ is uniquely determined by two elements of $H$, one which squares and one which cubes to the identity. This is exactly the description of a pair of maps from $C_2$ and $C_3$ to $H$, and the correspondence is induced by maps from $C_2$ and $C_3$ to $H$ taking a generator to a generator, as you say.
answered Nov 21 at 15:52
Kevin Carlson
32.4k23270
add a comment |
up vote
0
down vote
up vote
0
down vote
A map out of a group is just given by a map on the generators such that the relations are satisfied. Thus a map from $G$ to $H$ is uniquely determined by two elements of $H$, one which squares and one which cubes to the identity. This is exactly the description of a pair of maps from $C_2$ and $C_3$ to $H$, and the correspondence is induced by maps from $C_2$ and $C_3$ to $H$ taking a generator to a generator, as you say.
answered Nov 21 at 15:52
Kevin Carlson
32.4k23270
A map out of a group is just given by a map on the generators such that the relations are satisfied. Thus a map from $G$ to $H$ is uniquely determined by two elements of $H$, one which squares and one which cubes to the identity. This is exactly the description of a pair of maps from $C_2$ and $C_3$ to $H$, and the correspondence is induced by maps from $C_2$ and $C_3$ to $H$ taking a generator to a generator, as you say.
answered Nov 21 at 15:52
Kevin Carlson
32.4k23270
answered Nov 21 at 15:52
Kevin Carlson
32.4k23270
answered Nov 21 at 15:52
Kevin Carlson
32.4k23270
answered Nov 21 at 15:52
Kevin Carlson
32.4k23270
32.4k23270
add a comment |
add a comment |
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