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I have a doubt about something: if $G$ is a group, what can we say about the free group $rm Free(U(G))$ of $U(G)$, where $U$ is the forgetful functor from $Grp$ to $Set$?
I'm confused because if I'm not wrong: every group is a quotient of a free group by some relations, so when one "forgets" about the group structure of $G$:
do we forget about these relations (in which case one would not recover $G$ by taking the free group over $U(G)$ in general)
or are these relations actual equalities in the carrier set (e.g. if we have the group relation $ab=c$, then $ab$ and $c$ are really the same element in the carrier set), in which case I have the impression that $rm Free(U(G))$ is isomorphic to $G$?
I really feel like there's something wrong:
on the one hand, I suppose that $rm Free(U(G))$ shouldn't be isomorphic to $G$ in general (and that to recover $G$ from $rm Free(U(G))$ one would need to "evaluate" the lists of elements of $U(G)$ that are in $rm Free(U(G))$ with the group operation of $G$)
but on the other hand, I feel like the group relations are actual equalities in the carrier set, so that taking the free group over this carrier set should yield a group isomorphic to the original one
Any help would be very much appreciated!
group-theory category-theory free-groups
asked Jul 23 at 23:06
Cooke4
255
add a comment |Â
up vote
1
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favorite
I have a doubt about something: if $G$ is a group, what can we say about the free group $rm Free(U(G))$ of $U(G)$, where $U$ is the forgetful functor from $Grp$ to $Set$?
I'm confused because if I'm not wrong: every group is a quotient of a free group by some relations, so when one "forgets" about the group structure of $G$:
do we forget about these relations (in which case one would not recover $G$ by taking the free group over $U(G)$ in general)
or are these relations actual equalities in the carrier set (e.g. if we have the group relation $ab=c$, then $ab$ and $c$ are really the same element in the carrier set), in which case I have the impression that $rm Free(U(G))$ is isomorphic to $G$?
I really feel like there's something wrong:
on the one hand, I suppose that $rm Free(U(G))$ shouldn't be isomorphic to $G$ in general (and that to recover $G$ from $rm Free(U(G))$ one would need to "evaluate" the lists of elements of $U(G)$ that are in $rm Free(U(G))$ with the group operation of $G$)
but on the other hand, I feel like the group relations are actual equalities in the carrier set, so that taking the free group over this carrier set should yield a group isomorphic to the original one
Any help would be very much appreciated!
group-theory category-theory free-groups
asked Jul 23 at 23:06
Cooke4
255
No, one does not forget the relations. If you present a group by generators and relations you are still describing some set (namely a certain quotient of a free group) and that set is the underlying set $U(G)$. For example, $mathbbZ$ can be presented by a single generator with no relations. $U(mathbbZ)$ is a countable set, and so the free group on $U(mathbbZ)$ is the free group on countably many generators.
– Qiaochu Yuan
Jul 23 at 23:24
Thanks! Simple yet effective example, how had I not seen that? :-)
– Cooke4
Jul 23 at 23:35
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a doubt about something: if $G$ is a group, what can we say about the free group $rm Free(U(G))$ of $U(G)$, where $U$ is the forgetful functor from $Grp$ to $Set$?
I'm confused because if I'm not wrong: every group is a quotient of a free group by some relations, so when one "forgets" about the group structure of $G$:
do we forget about these relations (in which case one would not recover $G$ by taking the free group over $U(G)$ in general)
or are these relations actual equalities in the carrier set (e.g. if we have the group relation $ab=c$, then $ab$ and $c$ are really the same element in the carrier set), in which case I have the impression that $rm Free(U(G))$ is isomorphic to $G$?
I really feel like there's something wrong:
on the one hand, I suppose that $rm Free(U(G))$ shouldn't be isomorphic to $G$ in general (and that to recover $G$ from $rm Free(U(G))$ one would need to "evaluate" the lists of elements of $U(G)$ that are in $rm Free(U(G))$ with the group operation of $G$)
but on the other hand, I feel like the group relations are actual equalities in the carrier set, so that taking the free group over this carrier set should yield a group isomorphic to the original one
Any help would be very much appreciated!
group-theory category-theory free-groups
asked Jul 23 at 23:06
Cooke4
255
I have a doubt about something: if $G$ is a group, what can we say about the free group $rm Free(U(G))$ of $U(G)$, where $U$ is the forgetful functor from $Grp$ to $Set$?
I'm confused because if I'm not wrong: every group is a quotient of a free group by some relations, so when one "forgets" about the group structure of $G$:
do we forget about these relations (in which case one would not recover $G$ by taking the free group over $U(G)$ in general)
or are these relations actual equalities in the carrier set (e.g. if we have the group relation $ab=c$, then $ab$ and $c$ are really the same element in the carrier set), in which case I have the impression that $rm Free(U(G))$ is isomorphic to $G$?
I really feel like there's something wrong:
on the one hand, I suppose that $rm Free(U(G))$ shouldn't be isomorphic to $G$ in general (and that to recover $G$ from $rm Free(U(G))$ one would need to "evaluate" the lists of elements of $U(G)$ that are in $rm Free(U(G))$ with the group operation of $G$)
but on the other hand, I feel like the group relations are actual equalities in the carrier set, so that taking the free group over this carrier set should yield a group isomorphic to the original one
Any help would be very much appreciated!
group-theory category-theory free-groups
asked Jul 23 at 23:06
Cooke4
255
edited Jul 24 at 17:32
asked Jul 23 at 23:06
Cooke4
255
asked Jul 23 at 23:06
Cooke4
255
asked Jul 23 at 23:06
Cooke4
255
255
No, one does not forget the relations. If you present a group by generators and relations you are still describing some set (namely a certain quotient of a free group) and that set is the underlying set $U(G)$. For example, $mathbbZ$ can be presented by a single generator with no relations. $U(mathbbZ)$ is a countable set, and so the free group on $U(mathbbZ)$ is the free group on countably many generators.
– Qiaochu Yuan
Jul 23 at 23:24
Thanks! Simple yet effective example, how had I not seen that? :-)
– Cooke4
Jul 23 at 23:35
add a comment |Â
No, one does not forget the relations. If you present a group by generators and relations you are still describing some set (namely a certain quotient of a free group) and that set is the underlying set $U(G)$. For example, $mathbbZ$ can be presented by a single generator with no relations. $U(mathbbZ)$ is a countable set, and so the free group on $U(mathbbZ)$ is the free group on countably many generators.
– Qiaochu Yuan
Jul 23 at 23:24
Thanks! Simple yet effective example, how had I not seen that? :-)
– Cooke4
Jul 23 at 23:35
No, one does not forget the relations. If you present a group by generators and relations you are still describing some set (namely a certain quotient of a free group) and that set is the underlying set $U(G)$. For example, $mathbbZ$ can be presented by a single generator with no relations. $U(mathbbZ)$ is a countable set, and so the free group on $U(mathbbZ)$ is the free group on countably many generators.
– Qiaochu Yuan
Jul 23 at 23:24
No, one does not forget the relations. If you present a group by generators and relations you are still describing some set (namely a certain quotient of a free group) and that set is the underlying set $U(G)$. For example, $mathbbZ$ can be presented by a single generator with no relations. $U(mathbbZ)$ is a countable set, and so the free group on $U(mathbbZ)$ is the free group on countably many generators.
– Qiaochu Yuan
Jul 23 at 23:24
Thanks! Simple yet effective example, how had I not seen that? :-)
– Cooke4
Jul 23 at 23:35
Thanks! Simple yet effective example, how had I not seen that? :-)
– Cooke4
Jul 23 at 23:35
add a comment |Â
1 Answer
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No, $mathrmFree(U(G)) neq G$. For example, if $G$ is the trivial group, then $U(G)$ is a one element set, so $mathrmFree(U(G)) cong mathbb Z.$ In general, $mathrmFree(U(G))$ is much larger than $G$.
Another example. Let $G=mathbb Z/2mathbb Z=a,b$, equipped with a binary operation $ab=ba=b$, $aa=bb=a$. Then $U(G)=0,1$, just as a set. In $mathrmFree(U(G))$, the same multiplication structure does not hold. Every element of $mathrmFree(U(G))$ is a string of symbols taken from the list
$$
a,a^-1,b,b^-1.
$$
where the only relations that hold are $aa^-1$ and $bb^-1$ are equal to the identity.
answered Jul 23 at 23:24
Mike Earnest
15.1k11644
Thank you! So we can get back $G$ from $rm Free(U(G))$ by quotienting by your equations ($ab=ba=b$ and $aa=bb=a$) again I suppose?
– Cooke4
Jul 23 at 23:37
1
@Cooke4 No you can't get back $G$ from $operatornameFree(U(G))$ in general. For example both $mathbb Z_2times mathbb Z_2$ and $mathbb Z_4$ have an underlying set of $4$ elements, so the free groups $operatornameFree(U(mathbb Z_2times mathbb Z_2))$ and $operatornameFree(U(mathbb Z_4))$ are isomorphic. Broadly speaking, $operatornameFree(U(G))$ is only telling "I'm the free group on $operatornameCard(U(G))$ elements", but the multiplicative law of $G$ is a priori beyond recovery.
– Pece
Jul 24 at 7:31
@Pece Thanks for you answer! I see, but then if you quotient $rm Free(U(G))$ by the appropriate relations, can't you recover $G$?
– Cooke4
Jul 24 at 15:39
@Cooke4 Yes, but you have to know the relation somehow, which is pretty much the same as knowing the group. Categorically speaking, $operatornameFree$ is left adjoint to $U$, so you have a counit morphism $epsilon_G:operatornameFree(U(G)) to G$ and the normal subgroup generated by the "appriopriate relations" is $ker (epsilon_G)$.
– Pece
Jul 24 at 16:29
@Pece Oh I see, and then by quotienting by the kernel of the counit you get back $G$ I guess! Thanks for your explanation!
– Cooke4
Jul 24 at 17:29
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
No, $mathrmFree(U(G)) neq G$. For example, if $G$ is the trivial group, then $U(G)$ is a one element set, so $mathrmFree(U(G)) cong mathbb Z.$ In general, $mathrmFree(U(G))$ is much larger than $G$.
Another example. Let $G=mathbb Z/2mathbb Z=a,b$, equipped with a binary operation $ab=ba=b$, $aa=bb=a$. Then $U(G)=0,1$, just as a set. In $mathrmFree(U(G))$, the same multiplication structure does not hold. Every element of $mathrmFree(U(G))$ is a string of symbols taken from the list
$$
a,a^-1,b,b^-1.
$$
where the only relations that hold are $aa^-1$ and $bb^-1$ are equal to the identity.
answered Jul 23 at 23:24
Mike Earnest
15.1k11644
Thank you! So we can get back $G$ from $rm Free(U(G))$ by quotienting by your equations ($ab=ba=b$ and $aa=bb=a$) again I suppose?
– Cooke4
Jul 23 at 23:37
1
@Cooke4 No you can't get back $G$ from $operatornameFree(U(G))$ in general. For example both $mathbb Z_2times mathbb Z_2$ and $mathbb Z_4$ have an underlying set of $4$ elements, so the free groups $operatornameFree(U(mathbb Z_2times mathbb Z_2))$ and $operatornameFree(U(mathbb Z_4))$ are isomorphic. Broadly speaking, $operatornameFree(U(G))$ is only telling "I'm the free group on $operatornameCard(U(G))$ elements", but the multiplicative law of $G$ is a priori beyond recovery.
– Pece
Jul 24 at 7:31
@Pece Thanks for you answer! I see, but then if you quotient $rm Free(U(G))$ by the appropriate relations, can't you recover $G$?
– Cooke4
Jul 24 at 15:39
@Cooke4 Yes, but you have to know the relation somehow, which is pretty much the same as knowing the group. Categorically speaking, $operatornameFree$ is left adjoint to $U$, so you have a counit morphism $epsilon_G:operatornameFree(U(G)) to G$ and the normal subgroup generated by the "appriopriate relations" is $ker (epsilon_G)$.
– Pece
Jul 24 at 16:29
@Pece Oh I see, and then by quotienting by the kernel of the counit you get back $G$ I guess! Thanks for your explanation!
– Cooke4
Jul 24 at 17:29
 |Â
show 2 more comments
up vote
4
down vote
accepted
No, $mathrmFree(U(G)) neq G$. For example, if $G$ is the trivial group, then $U(G)$ is a one element set, so $mathrmFree(U(G)) cong mathbb Z.$ In general, $mathrmFree(U(G))$ is much larger than $G$.
Another example. Let $G=mathbb Z/2mathbb Z=a,b$, equipped with a binary operation $ab=ba=b$, $aa=bb=a$. Then $U(G)=0,1$, just as a set. In $mathrmFree(U(G))$, the same multiplication structure does not hold. Every element of $mathrmFree(U(G))$ is a string of symbols taken from the list
$$
a,a^-1,b,b^-1.
$$
where the only relations that hold are $aa^-1$ and $bb^-1$ are equal to the identity.
answered Jul 23 at 23:24
Mike Earnest
15.1k11644
Thank you! So we can get back $G$ from $rm Free(U(G))$ by quotienting by your equations ($ab=ba=b$ and $aa=bb=a$) again I suppose?
– Cooke4
Jul 23 at 23:37
1
@Cooke4 No you can't get back $G$ from $operatornameFree(U(G))$ in general. For example both $mathbb Z_2times mathbb Z_2$ and $mathbb Z_4$ have an underlying set of $4$ elements, so the free groups $operatornameFree(U(mathbb Z_2times mathbb Z_2))$ and $operatornameFree(U(mathbb Z_4))$ are isomorphic. Broadly speaking, $operatornameFree(U(G))$ is only telling "I'm the free group on $operatornameCard(U(G))$ elements", but the multiplicative law of $G$ is a priori beyond recovery.
– Pece
Jul 24 at 7:31
@Pece Thanks for you answer! I see, but then if you quotient $rm Free(U(G))$ by the appropriate relations, can't you recover $G$?
– Cooke4
Jul 24 at 15:39
@Cooke4 Yes, but you have to know the relation somehow, which is pretty much the same as knowing the group. Categorically speaking, $operatornameFree$ is left adjoint to $U$, so you have a counit morphism $epsilon_G:operatornameFree(U(G)) to G$ and the normal subgroup generated by the "appriopriate relations" is $ker (epsilon_G)$.
– Pece
Jul 24 at 16:29
@Pece Oh I see, and then by quotienting by the kernel of the counit you get back $G$ I guess! Thanks for your explanation!
– Cooke4
Jul 24 at 17:29
 |Â
show 2 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
No, $mathrmFree(U(G)) neq G$. For example, if $G$ is the trivial group, then $U(G)$ is a one element set, so $mathrmFree(U(G)) cong mathbb Z.$ In general, $mathrmFree(U(G))$ is much larger than $G$.
Another example. Let $G=mathbb Z/2mathbb Z=a,b$, equipped with a binary operation $ab=ba=b$, $aa=bb=a$. Then $U(G)=0,1$, just as a set. In $mathrmFree(U(G))$, the same multiplication structure does not hold. Every element of $mathrmFree(U(G))$ is a string of symbols taken from the list
$$
a,a^-1,b,b^-1.
$$
where the only relations that hold are $aa^-1$ and $bb^-1$ are equal to the identity.
answered Jul 23 at 23:24
Mike Earnest
15.1k11644
No, $mathrmFree(U(G)) neq G$. For example, if $G$ is the trivial group, then $U(G)$ is a one element set, so $mathrmFree(U(G)) cong mathbb Z.$ In general, $mathrmFree(U(G))$ is much larger than $G$.
Another example. Let $G=mathbb Z/2mathbb Z=a,b$, equipped with a binary operation $ab=ba=b$, $aa=bb=a$. Then $U(G)=0,1$, just as a set. In $mathrmFree(U(G))$, the same multiplication structure does not hold. Every element of $mathrmFree(U(G))$ is a string of symbols taken from the list
$$
a,a^-1,b,b^-1.
$$
where the only relations that hold are $aa^-1$ and $bb^-1$ are equal to the identity.
answered Jul 23 at 23:24
Mike Earnest
15.1k11644
answered Jul 23 at 23:24
Mike Earnest
15.1k11644
answered Jul 23 at 23:24
Mike Earnest
15.1k11644
answered Jul 23 at 23:24
Mike Earnest
15.1k11644
15.1k11644
Thank you! So we can get back $G$ from $rm Free(U(G))$ by quotienting by your equations ($ab=ba=b$ and $aa=bb=a$) again I suppose?
– Cooke4
Jul 23 at 23:37
1
@Cooke4 No you can't get back $G$ from $operatornameFree(U(G))$ in general. For example both $mathbb Z_2times mathbb Z_2$ and $mathbb Z_4$ have an underlying set of $4$ elements, so the free groups $operatornameFree(U(mathbb Z_2times mathbb Z_2))$ and $operatornameFree(U(mathbb Z_4))$ are isomorphic. Broadly speaking, $operatornameFree(U(G))$ is only telling "I'm the free group on $operatornameCard(U(G))$ elements", but the multiplicative law of $G$ is a priori beyond recovery.
– Pece
Jul 24 at 7:31
@Pece Thanks for you answer! I see, but then if you quotient $rm Free(U(G))$ by the appropriate relations, can't you recover $G$?
– Cooke4
Jul 24 at 15:39
@Cooke4 Yes, but you have to know the relation somehow, which is pretty much the same as knowing the group. Categorically speaking, $operatornameFree$ is left adjoint to $U$, so you have a counit morphism $epsilon_G:operatornameFree(U(G)) to G$ and the normal subgroup generated by the "appriopriate relations" is $ker (epsilon_G)$.
– Pece
Jul 24 at 16:29
@Pece Oh I see, and then by quotienting by the kernel of the counit you get back $G$ I guess! Thanks for your explanation!
– Cooke4
Jul 24 at 17:29
 |Â
show 2 more comments
Thank you! So we can get back $G$ from $rm Free(U(G))$ by quotienting by your equations ($ab=ba=b$ and $aa=bb=a$) again I suppose?
– Cooke4
Jul 23 at 23:37
1
@Cooke4 No you can't get back $G$ from $operatornameFree(U(G))$ in general. For example both $mathbb Z_2times mathbb Z_2$ and $mathbb Z_4$ have an underlying set of $4$ elements, so the free groups $operatornameFree(U(mathbb Z_2times mathbb Z_2))$ and $operatornameFree(U(mathbb Z_4))$ are isomorphic. Broadly speaking, $operatornameFree(U(G))$ is only telling "I'm the free group on $operatornameCard(U(G))$ elements", but the multiplicative law of $G$ is a priori beyond recovery.
– Pece
Jul 24 at 7:31
@Pece Thanks for you answer! I see, but then if you quotient $rm Free(U(G))$ by the appropriate relations, can't you recover $G$?
– Cooke4
Jul 24 at 15:39
@Cooke4 Yes, but you have to know the relation somehow, which is pretty much the same as knowing the group. Categorically speaking, $operatornameFree$ is left adjoint to $U$, so you have a counit morphism $epsilon_G:operatornameFree(U(G)) to G$ and the normal subgroup generated by the "appriopriate relations" is $ker (epsilon_G)$.
– Pece
Jul 24 at 16:29
@Pece Oh I see, and then by quotienting by the kernel of the counit you get back $G$ I guess! Thanks for your explanation!
– Cooke4
Jul 24 at 17:29
Thank you! So we can get back $G$ from $rm Free(U(G))$ by quotienting by your equations ($ab=ba=b$ and $aa=bb=a$) again I suppose?
– Cooke4
Jul 23 at 23:37
Thank you! So we can get back $G$ from $rm Free(U(G))$ by quotienting by your equations ($ab=ba=b$ and $aa=bb=a$) again I suppose?
– Cooke4
Jul 23 at 23:37
1
1
@Cooke4 No you can't get back $G$ from $operatornameFree(U(G))$ in general. For example both $mathbb Z_2times mathbb Z_2$ and $mathbb Z_4$ have an underlying set of $4$ elements, so the free groups $operatornameFree(U(mathbb Z_2times mathbb Z_2))$ and $operatornameFree(U(mathbb Z_4))$ are isomorphic. Broadly speaking, $operatornameFree(U(G))$ is only telling "I'm the free group on $operatornameCard(U(G))$ elements", but the multiplicative law of $G$ is a priori beyond recovery.
– Pece
Jul 24 at 7:31
@Cooke4 No you can't get back $G$ from $operatornameFree(U(G))$ in general. For example both $mathbb Z_2times mathbb Z_2$ and $mathbb Z_4$ have an underlying set of $4$ elements, so the free groups $operatornameFree(U(mathbb Z_2times mathbb Z_2))$ and $operatornameFree(U(mathbb Z_4))$ are isomorphic. Broadly speaking, $operatornameFree(U(G))$ is only telling "I'm the free group on $operatornameCard(U(G))$ elements", but the multiplicative law of $G$ is a priori beyond recovery.
– Pece
Jul 24 at 7:31
@Pece Thanks for you answer! I see, but then if you quotient $rm Free(U(G))$ by the appropriate relations, can't you recover $G$?
– Cooke4
Jul 24 at 15:39
@Pece Thanks for you answer! I see, but then if you quotient $rm Free(U(G))$ by the appropriate relations, can't you recover $G$?
– Cooke4
Jul 24 at 15:39
@Cooke4 Yes, but you have to know the relation somehow, which is pretty much the same as knowing the group. Categorically speaking, $operatornameFree$ is left adjoint to $U$, so you have a counit morphism $epsilon_G:operatornameFree(U(G)) to G$ and the normal subgroup generated by the "appriopriate relations" is $ker (epsilon_G)$.
– Pece
Jul 24 at 16:29
@Cooke4 Yes, but you have to know the relation somehow, which is pretty much the same as knowing the group. Categorically speaking, $operatornameFree$ is left adjoint to $U$, so you have a counit morphism $epsilon_G:operatornameFree(U(G)) to G$ and the normal subgroup generated by the "appriopriate relations" is $ker (epsilon_G)$.
– Pece
Jul 24 at 16:29
@Pece Oh I see, and then by quotienting by the kernel of the counit you get back $G$ I guess! Thanks for your explanation!
– Cooke4
Jul 24 at 17:29
@Pece Oh I see, and then by quotienting by the kernel of the counit you get back $G$ I guess! Thanks for your explanation!
– Cooke4
Jul 24 at 17:29
 |Â
show 2 more comments
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No, one does not forget the relations. If you present a group by generators and relations you are still describing some set (namely a certain quotient of a free group) and that set is the underlying set $U(G)$. For example, $mathbbZ$ can be presented by a single generator with no relations. $U(mathbbZ)$ is a countable set, and so the free group on $U(mathbbZ)$ is the free group on countably many generators.
– Qiaochu Yuan
Jul 23 at 23:24
Thanks! Simple yet effective example, how had I not seen that? :-)
– Cooke4
Jul 23 at 23:35