Isomorphism on a torsion group - automorphism or endomorphism?
up vote
-1
down vote
favorite
Let $f:Gto G$ be a surjection from a torsion group $G$ onto itself.
Let the kernel have infinite cardinality: $lvertker(f)rvert=aleph_0$
What category of function on groups is this?
To my mind this cannot be a group automorphism because at the very least an automorphism must map the identity onto itself in order to satisfy $f(acdot b)=f(a)circ f(b)$.
Clearly $f$ does not yield distinct inverses.
Is $f$ therefore an endomorphism instead?
In part, I'm asking whether an endomorphism on a torsion group can have a kernel with infinite cardinality - what with the elements themselves having finite order and all.
I'm asking with half an eye on groups of intermediate growth such as the Grigorchuk group, and any torsion group structure on which variants of the Collatz function might be an endomorphism.
group-theory group-homomorphism torsion-groups
asked Nov 20 at 10:50
Robert Frost
4,1841039
|
show 7 more comments
up vote
-1
down vote
favorite
Let $f:Gto G$ be a surjection from a torsion group $G$ onto itself.
Let the kernel have infinite cardinality: $lvertker(f)rvert=aleph_0$
What category of function on groups is this?
To my mind this cannot be a group automorphism because at the very least an automorphism must map the identity onto itself in order to satisfy $f(acdot b)=f(a)circ f(b)$.
Clearly $f$ does not yield distinct inverses.
Is $f$ therefore an endomorphism instead?
In part, I'm asking whether an endomorphism on a torsion group can have a kernel with infinite cardinality - what with the elements themselves having finite order and all.
I'm asking with half an eye on groups of intermediate growth such as the Grigorchuk group, and any torsion group structure on which variants of the Collatz function might be an endomorphism.
group-theory group-homomorphism torsion-groups
asked Nov 20 at 10:50
Robert Frost
4,1841039
1
Well, the trivial homomorphism is an endomorphism with infinite cardinality if the group is infinite...
– DonAntonio
Nov 20 at 11:05
2
Anyway if you actualy meant "epimorphism" then there still can be an epimorphism $Gto G$ with infinite kernel. Take $G=bigoplus_{i=1}^{infty} mathbb{Z}_2$ and map $(x_1,x_2,x_3,x_4,ldots)$ to $(x_1, x_3, x_5, x_7,ldots)$.
– freakish
Nov 20 at 11:11
1
So you are looking for non-Hopfian torsion groups. It's easy to construct examples such as the direct product of infinitely many copies of the same finite group. The question would be more interesting if you assumed that $G$ is finitely generated. I don't know immediately whether there are any such examples.
– Derek Holt
Nov 20 at 11:13
1
Grigorchuck group would not work since every quotient is finite. I don't know of any finitely generated non hopfian torsion groups(they could exist, but it might be an open problem if they do too)
– Paul Plummer
Nov 20 at 11:35
1
@RobertFrost Yes, formally $mathbb{Z}_2=mathbb{Z}/2mathbb{Z}$ is the group of integers modulo $2$. The finite group of $2$ elements.
– freakish
Nov 20 at 12:23
|
show 7 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $f:Gto G$ be a surjection from a torsion group $G$ onto itself.
Let the kernel have infinite cardinality: $lvertker(f)rvert=aleph_0$
What category of function on groups is this?
To my mind this cannot be a group automorphism because at the very least an automorphism must map the identity onto itself in order to satisfy $f(acdot b)=f(a)circ f(b)$.
Clearly $f$ does not yield distinct inverses.
Is $f$ therefore an endomorphism instead?
In part, I'm asking whether an endomorphism on a torsion group can have a kernel with infinite cardinality - what with the elements themselves having finite order and all.
I'm asking with half an eye on groups of intermediate growth such as the Grigorchuk group, and any torsion group structure on which variants of the Collatz function might be an endomorphism.
group-theory group-homomorphism torsion-groups
asked Nov 20 at 10:50
Robert Frost
4,1841039
Let $f:Gto G$ be a surjection from a torsion group $G$ onto itself.
Let the kernel have infinite cardinality: $lvertker(f)rvert=aleph_0$
What category of function on groups is this?
To my mind this cannot be a group automorphism because at the very least an automorphism must map the identity onto itself in order to satisfy $f(acdot b)=f(a)circ f(b)$.
Clearly $f$ does not yield distinct inverses.
Is $f$ therefore an endomorphism instead?
In part, I'm asking whether an endomorphism on a torsion group can have a kernel with infinite cardinality - what with the elements themselves having finite order and all.
I'm asking with half an eye on groups of intermediate growth such as the Grigorchuk group, and any torsion group structure on which variants of the Collatz function might be an endomorphism.
group-theory group-homomorphism torsion-groups
group-theory group-homomorphism torsion-groups
asked Nov 20 at 10:50
Robert Frost
4,1841039
asked Nov 20 at 10:50
Robert Frost
4,1841039
edited Nov 20 at 12:13
asked Nov 20 at 10:50
Robert Frost
4,1841039
asked Nov 20 at 10:50
Robert Frost
4,1841039
asked Nov 20 at 10:50
Robert Frost
4,1841039
4,1841039
1
Well, the trivial homomorphism is an endomorphism with infinite cardinality if the group is infinite...
– DonAntonio
Nov 20 at 11:05
2
Anyway if you actualy meant "epimorphism" then there still can be an epimorphism $Gto G$ with infinite kernel. Take $G=bigoplus_{i=1}^{infty} mathbb{Z}_2$ and map $(x_1,x_2,x_3,x_4,ldots)$ to $(x_1, x_3, x_5, x_7,ldots)$.
– freakish
Nov 20 at 11:11
1
So you are looking for non-Hopfian torsion groups. It's easy to construct examples such as the direct product of infinitely many copies of the same finite group. The question would be more interesting if you assumed that $G$ is finitely generated. I don't know immediately whether there are any such examples.
– Derek Holt
Nov 20 at 11:13
1
Grigorchuck group would not work since every quotient is finite. I don't know of any finitely generated non hopfian torsion groups(they could exist, but it might be an open problem if they do too)
– Paul Plummer
Nov 20 at 11:35
1
@RobertFrost Yes, formally $mathbb{Z}_2=mathbb{Z}/2mathbb{Z}$ is the group of integers modulo $2$. The finite group of $2$ elements.
– freakish
Nov 20 at 12:23
|
show 7 more comments
1
Well, the trivial homomorphism is an endomorphism with infinite cardinality if the group is infinite...
– DonAntonio
Nov 20 at 11:05
2
Anyway if you actualy meant "epimorphism" then there still can be an epimorphism $Gto G$ with infinite kernel. Take $G=bigoplus_{i=1}^{infty} mathbb{Z}_2$ and map $(x_1,x_2,x_3,x_4,ldots)$ to $(x_1, x_3, x_5, x_7,ldots)$.
– freakish
Nov 20 at 11:11
1
So you are looking for non-Hopfian torsion groups. It's easy to construct examples such as the direct product of infinitely many copies of the same finite group. The question would be more interesting if you assumed that $G$ is finitely generated. I don't know immediately whether there are any such examples.
– Derek Holt
Nov 20 at 11:13
1
Grigorchuck group would not work since every quotient is finite. I don't know of any finitely generated non hopfian torsion groups(they could exist, but it might be an open problem if they do too)
– Paul Plummer
Nov 20 at 11:35
1
@RobertFrost Yes, formally $mathbb{Z}_2=mathbb{Z}/2mathbb{Z}$ is the group of integers modulo $2$. The finite group of $2$ elements.
– freakish
Nov 20 at 12:23
1
1
Well, the trivial homomorphism is an endomorphism with infinite cardinality if the group is infinite...
– DonAntonio
Nov 20 at 11:05
Well, the trivial homomorphism is an endomorphism with infinite cardinality if the group is infinite...
– DonAntonio
Nov 20 at 11:05
2
2
Anyway if you actualy meant "epimorphism" then there still can be an epimorphism $Gto G$ with infinite kernel. Take $G=bigoplus_{i=1}^{infty} mathbb{Z}_2$ and map $(x_1,x_2,x_3,x_4,ldots)$ to $(x_1, x_3, x_5, x_7,ldots)$.
– freakish
Nov 20 at 11:11
Anyway if you actualy meant "epimorphism" then there still can be an epimorphism $Gto G$ with infinite kernel. Take $G=bigoplus_{i=1}^{infty} mathbb{Z}_2$ and map $(x_1,x_2,x_3,x_4,ldots)$ to $(x_1, x_3, x_5, x_7,ldots)$.
– freakish
Nov 20 at 11:11
1
1
So you are looking for non-Hopfian torsion groups. It's easy to construct examples such as the direct product of infinitely many copies of the same finite group. The question would be more interesting if you assumed that $G$ is finitely generated. I don't know immediately whether there are any such examples.
– Derek Holt
Nov 20 at 11:13
So you are looking for non-Hopfian torsion groups. It's easy to construct examples such as the direct product of infinitely many copies of the same finite group. The question would be more interesting if you assumed that $G$ is finitely generated. I don't know immediately whether there are any such examples.
– Derek Holt
Nov 20 at 11:13
1
1
Grigorchuck group would not work since every quotient is finite. I don't know of any finitely generated non hopfian torsion groups(they could exist, but it might be an open problem if they do too)
– Paul Plummer
Nov 20 at 11:35
Grigorchuck group would not work since every quotient is finite. I don't know of any finitely generated non hopfian torsion groups(they could exist, but it might be an open problem if they do too)
– Paul Plummer
Nov 20 at 11:35
1
1
@RobertFrost Yes, formally $mathbb{Z}_2=mathbb{Z}/2mathbb{Z}$ is the group of integers modulo $2$. The finite group of $2$ elements.
– freakish
Nov 20 at 12:23
@RobertFrost Yes, formally $mathbb{Z}_2=mathbb{Z}/2mathbb{Z}$ is the group of integers modulo $2$. The finite group of $2$ elements.
– freakish
Nov 20 at 12:23
|
show 7 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
I don't think there is an accepted name to those types of functions except for saying what they are: surjective homomorphism with some specified size of kernel. You can say surjective endomorphism, or epic endomorphsims(which sounds sort of cool). The endomorphism means it is a morphism between the same object. In the category of groups surjective morphisms and epimorphism/epic morphism are the same(in general they are nor the same thing).
You mention your motiviation is studying torsion groups of intermediate growth(so finitely generated) with a function like the above. As far as I can tell, it is not known if such torsion f.g. groups exist even without the condition of intermediate growth. The Grigorchuck group does not have any homomorphism like that since it is just infinite which is an infinite group where all proper quotients are finite.
Without finitely generated condition there are groups like that, for example $ bigoplus_{i in mathbb N} mathbb Z / 2 mathbb Z$.
answered 2 days ago
Paul Plummer
4,81421950
Thanks. The critical thing I drew from the comments which isn't in this answer was that this is a non-isomorphic epimorphism and therefore the group would be non Hopfian. I suspect the object I'm trying to construct is more likely akin to your infinitely generated example though.
– Robert Frost
2 days ago
1
Sure, that is the definition of non-hopfian group. I didn't bring it up since you didn't ask about it in the question(just what to call such f), although it is good to mention. There are plenty of finitely generated non-hopfian groups, just don't know of any torsion examples(and I think it is probably an open question if such exist).
– Paul Plummer
2 days ago
A direct product was mentioned but I guess what you show here is an infinite sum. Then I guess either would suffice?
– Robert Frost
2 days ago
1
@RobertFrost You could do something similar with direct products, but the easy maps give uncountable kernel(I actually wouldn't be surprised if any such mapping must have uncountable kernel or finite kernel when dealing with infinite products).
– Paul Plummer
2 days ago
1
@RobertFrost Another word about infinite product case is that there is a natural topology(the product topology) to put on such groups and frequently homomorphisms between such groups are forced to be continuous in this topology. This is called automatic continuity. As an example, it can be used to prove infinite product of integers is not free abelian (even though in the finite case they are). This continuity frequently means kernels will be very large or small.
– Paul Plummer
2 days ago
|
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I don't think there is an accepted name to those types of functions except for saying what they are: surjective homomorphism with some specified size of kernel. You can say surjective endomorphism, or epic endomorphsims(which sounds sort of cool). The endomorphism means it is a morphism between the same object. In the category of groups surjective morphisms and epimorphism/epic morphism are the same(in general they are nor the same thing).
You mention your motiviation is studying torsion groups of intermediate growth(so finitely generated) with a function like the above. As far as I can tell, it is not known if such torsion f.g. groups exist even without the condition of intermediate growth. The Grigorchuck group does not have any homomorphism like that since it is just infinite which is an infinite group where all proper quotients are finite.
Without finitely generated condition there are groups like that, for example $ bigoplus_{i in mathbb N} mathbb Z / 2 mathbb Z$.
answered 2 days ago
Paul Plummer
4,81421950
Thanks. The critical thing I drew from the comments which isn't in this answer was that this is a non-isomorphic epimorphism and therefore the group would be non Hopfian. I suspect the object I'm trying to construct is more likely akin to your infinitely generated example though.
– Robert Frost
2 days ago
1
Sure, that is the definition of non-hopfian group. I didn't bring it up since you didn't ask about it in the question(just what to call such f), although it is good to mention. There are plenty of finitely generated non-hopfian groups, just don't know of any torsion examples(and I think it is probably an open question if such exist).
– Paul Plummer
2 days ago
A direct product was mentioned but I guess what you show here is an infinite sum. Then I guess either would suffice?
– Robert Frost
2 days ago
1
@RobertFrost You could do something similar with direct products, but the easy maps give uncountable kernel(I actually wouldn't be surprised if any such mapping must have uncountable kernel or finite kernel when dealing with infinite products).
– Paul Plummer
2 days ago
1
@RobertFrost Another word about infinite product case is that there is a natural topology(the product topology) to put on such groups and frequently homomorphisms between such groups are forced to be continuous in this topology. This is called automatic continuity. As an example, it can be used to prove infinite product of integers is not free abelian (even though in the finite case they are). This continuity frequently means kernels will be very large or small.
– Paul Plummer
2 days ago
|
show 2 more comments
up vote
1
down vote
accepted
I don't think there is an accepted name to those types of functions except for saying what they are: surjective homomorphism with some specified size of kernel. You can say surjective endomorphism, or epic endomorphsims(which sounds sort of cool). The endomorphism means it is a morphism between the same object. In the category of groups surjective morphisms and epimorphism/epic morphism are the same(in general they are nor the same thing).
You mention your motiviation is studying torsion groups of intermediate growth(so finitely generated) with a function like the above. As far as I can tell, it is not known if such torsion f.g. groups exist even without the condition of intermediate growth. The Grigorchuck group does not have any homomorphism like that since it is just infinite which is an infinite group where all proper quotients are finite.
Without finitely generated condition there are groups like that, for example $ bigoplus_{i in mathbb N} mathbb Z / 2 mathbb Z$.
answered 2 days ago
Paul Plummer
4,81421950
Thanks. The critical thing I drew from the comments which isn't in this answer was that this is a non-isomorphic epimorphism and therefore the group would be non Hopfian. I suspect the object I'm trying to construct is more likely akin to your infinitely generated example though.
– Robert Frost
2 days ago
1
Sure, that is the definition of non-hopfian group. I didn't bring it up since you didn't ask about it in the question(just what to call such f), although it is good to mention. There are plenty of finitely generated non-hopfian groups, just don't know of any torsion examples(and I think it is probably an open question if such exist).
– Paul Plummer
2 days ago
A direct product was mentioned but I guess what you show here is an infinite sum. Then I guess either would suffice?
– Robert Frost
2 days ago
1
@RobertFrost You could do something similar with direct products, but the easy maps give uncountable kernel(I actually wouldn't be surprised if any such mapping must have uncountable kernel or finite kernel when dealing with infinite products).
– Paul Plummer
2 days ago
1
@RobertFrost Another word about infinite product case is that there is a natural topology(the product topology) to put on such groups and frequently homomorphisms between such groups are forced to be continuous in this topology. This is called automatic continuity. As an example, it can be used to prove infinite product of integers is not free abelian (even though in the finite case they are). This continuity frequently means kernels will be very large or small.
– Paul Plummer
2 days ago
|
show 2 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I don't think there is an accepted name to those types of functions except for saying what they are: surjective homomorphism with some specified size of kernel. You can say surjective endomorphism, or epic endomorphsims(which sounds sort of cool). The endomorphism means it is a morphism between the same object. In the category of groups surjective morphisms and epimorphism/epic morphism are the same(in general they are nor the same thing).
You mention your motiviation is studying torsion groups of intermediate growth(so finitely generated) with a function like the above. As far as I can tell, it is not known if such torsion f.g. groups exist even without the condition of intermediate growth. The Grigorchuck group does not have any homomorphism like that since it is just infinite which is an infinite group where all proper quotients are finite.
Without finitely generated condition there are groups like that, for example $ bigoplus_{i in mathbb N} mathbb Z / 2 mathbb Z$.
answered 2 days ago
Paul Plummer
4,81421950
I don't think there is an accepted name to those types of functions except for saying what they are: surjective homomorphism with some specified size of kernel. You can say surjective endomorphism, or epic endomorphsims(which sounds sort of cool). The endomorphism means it is a morphism between the same object. In the category of groups surjective morphisms and epimorphism/epic morphism are the same(in general they are nor the same thing).
You mention your motiviation is studying torsion groups of intermediate growth(so finitely generated) with a function like the above. As far as I can tell, it is not known if such torsion f.g. groups exist even without the condition of intermediate growth. The Grigorchuck group does not have any homomorphism like that since it is just infinite which is an infinite group where all proper quotients are finite.
Without finitely generated condition there are groups like that, for example $ bigoplus_{i in mathbb N} mathbb Z / 2 mathbb Z$.
answered 2 days ago
Paul Plummer
4,81421950
answered 2 days ago
Paul Plummer
4,81421950
answered 2 days ago
Paul Plummer
4,81421950
answered 2 days ago
Paul Plummer
4,81421950
4,81421950
Thanks. The critical thing I drew from the comments which isn't in this answer was that this is a non-isomorphic epimorphism and therefore the group would be non Hopfian. I suspect the object I'm trying to construct is more likely akin to your infinitely generated example though.
– Robert Frost
2 days ago
1
Sure, that is the definition of non-hopfian group. I didn't bring it up since you didn't ask about it in the question(just what to call such f), although it is good to mention. There are plenty of finitely generated non-hopfian groups, just don't know of any torsion examples(and I think it is probably an open question if such exist).
– Paul Plummer
2 days ago
A direct product was mentioned but I guess what you show here is an infinite sum. Then I guess either would suffice?
– Robert Frost
2 days ago
1
@RobertFrost You could do something similar with direct products, but the easy maps give uncountable kernel(I actually wouldn't be surprised if any such mapping must have uncountable kernel or finite kernel when dealing with infinite products).
– Paul Plummer
2 days ago
1
@RobertFrost Another word about infinite product case is that there is a natural topology(the product topology) to put on such groups and frequently homomorphisms between such groups are forced to be continuous in this topology. This is called automatic continuity. As an example, it can be used to prove infinite product of integers is not free abelian (even though in the finite case they are). This continuity frequently means kernels will be very large or small.
– Paul Plummer
2 days ago
|
show 2 more comments
Thanks. The critical thing I drew from the comments which isn't in this answer was that this is a non-isomorphic epimorphism and therefore the group would be non Hopfian. I suspect the object I'm trying to construct is more likely akin to your infinitely generated example though.
– Robert Frost
2 days ago
1
Sure, that is the definition of non-hopfian group. I didn't bring it up since you didn't ask about it in the question(just what to call such f), although it is good to mention. There are plenty of finitely generated non-hopfian groups, just don't know of any torsion examples(and I think it is probably an open question if such exist).
– Paul Plummer
2 days ago
A direct product was mentioned but I guess what you show here is an infinite sum. Then I guess either would suffice?
– Robert Frost
2 days ago
1
@RobertFrost You could do something similar with direct products, but the easy maps give uncountable kernel(I actually wouldn't be surprised if any such mapping must have uncountable kernel or finite kernel when dealing with infinite products).
– Paul Plummer
2 days ago
1
@RobertFrost Another word about infinite product case is that there is a natural topology(the product topology) to put on such groups and frequently homomorphisms between such groups are forced to be continuous in this topology. This is called automatic continuity. As an example, it can be used to prove infinite product of integers is not free abelian (even though in the finite case they are). This continuity frequently means kernels will be very large or small.
– Paul Plummer
2 days ago
Thanks. The critical thing I drew from the comments which isn't in this answer was that this is a non-isomorphic epimorphism and therefore the group would be non Hopfian. I suspect the object I'm trying to construct is more likely akin to your infinitely generated example though.
– Robert Frost
2 days ago
Thanks. The critical thing I drew from the comments which isn't in this answer was that this is a non-isomorphic epimorphism and therefore the group would be non Hopfian. I suspect the object I'm trying to construct is more likely akin to your infinitely generated example though.
– Robert Frost
2 days ago
1
1
Sure, that is the definition of non-hopfian group. I didn't bring it up since you didn't ask about it in the question(just what to call such f), although it is good to mention. There are plenty of finitely generated non-hopfian groups, just don't know of any torsion examples(and I think it is probably an open question if such exist).
– Paul Plummer
2 days ago
Sure, that is the definition of non-hopfian group. I didn't bring it up since you didn't ask about it in the question(just what to call such f), although it is good to mention. There are plenty of finitely generated non-hopfian groups, just don't know of any torsion examples(and I think it is probably an open question if such exist).
– Paul Plummer
2 days ago
A direct product was mentioned but I guess what you show here is an infinite sum. Then I guess either would suffice?
– Robert Frost
2 days ago
A direct product was mentioned but I guess what you show here is an infinite sum. Then I guess either would suffice?
– Robert Frost
2 days ago
1
1
@RobertFrost You could do something similar with direct products, but the easy maps give uncountable kernel(I actually wouldn't be surprised if any such mapping must have uncountable kernel or finite kernel when dealing with infinite products).
– Paul Plummer
2 days ago
@RobertFrost You could do something similar with direct products, but the easy maps give uncountable kernel(I actually wouldn't be surprised if any such mapping must have uncountable kernel or finite kernel when dealing with infinite products).
– Paul Plummer
2 days ago
1
1
@RobertFrost Another word about infinite product case is that there is a natural topology(the product topology) to put on such groups and frequently homomorphisms between such groups are forced to be continuous in this topology. This is called automatic continuity. As an example, it can be used to prove infinite product of integers is not free abelian (even though in the finite case they are). This continuity frequently means kernels will be very large or small.
– Paul Plummer
2 days ago
@RobertFrost Another word about infinite product case is that there is a natural topology(the product topology) to put on such groups and frequently homomorphisms between such groups are forced to be continuous in this topology. This is called automatic continuity. As an example, it can be used to prove infinite product of integers is not free abelian (even though in the finite case they are). This continuity frequently means kernels will be very large or small.
– Paul Plummer
2 days ago
|
show 2 more comments
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006180%2fisomorphism-on-a-torsion-group-automorphism-or-endomorphism%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Well, the trivial homomorphism is an endomorphism with infinite cardinality if the group is infinite...
– DonAntonio
Nov 20 at 11:05
2
Anyway if you actualy meant "epimorphism" then there still can be an epimorphism $Gto G$ with infinite kernel. Take $G=bigoplus_{i=1}^{infty} mathbb{Z}_2$ and map $(x_1,x_2,x_3,x_4,ldots)$ to $(x_1, x_3, x_5, x_7,ldots)$.
– freakish
Nov 20 at 11:11
1
So you are looking for non-Hopfian torsion groups. It's easy to construct examples such as the direct product of infinitely many copies of the same finite group. The question would be more interesting if you assumed that $G$ is finitely generated. I don't know immediately whether there are any such examples.
– Derek Holt
Nov 20 at 11:13
1
Grigorchuck group would not work since every quotient is finite. I don't know of any finitely generated non hopfian torsion groups(they could exist, but it might be an open problem if they do too)
– Paul Plummer
Nov 20 at 11:35
1
@RobertFrost Yes, formally $mathbb{Z}_2=mathbb{Z}/2mathbb{Z}$ is the group of integers modulo $2$. The finite group of $2$ elements.
– freakish
Nov 20 at 12:23