Mixing
Proving $(Bbb R,+)$ has no proper subgroup isomorphic to itself
$begingroup$
A captioned image with the text "My love for you is like a group which has a proper subgroup isomorphic to itself" was recently posted in a group chat I'm in. Ignoring the argument about the semantics of the language used (I am of the firm opinion that "itself" is in reference to the parent group, not the subgroup) this got me wondering about examples of groups which have this property. That is to say, a group $G$ that has a proper subgroup $H$ such that $G$ is isomorphic to $H$.
A few examples came to mind such as $(Bbb Z,+) > (2Bbb Z,+)$ and $(Bbb Q^*,times) > (K,times)$ where $K={frac{a}{b}~:~a,binBbb Zsetminus{0},~gcd(a,2)=gcd(b,2)=1}$ and similar. (Note: subgroup, not subring)
I got to thinking about $(Bbb R,+)$ however and wondering whether or not the real numbers have a subgroup isomorphic to the reals with respect to addition. My gut feeling is no, but I am at a loss as to how to prove this or come up with an example. Proper subgroups certainly exist, such as how $(Bbb Q,+)<(overline{Bbb Q},+)<(Bbb R,+)$, and so by specifying that our proper subgroup $H$ doesn't contain some element $x$, be it rational or irrational, that doesn't preclude $H$ from existing.
So then, I ask you, does $(Bbb R,+)$ have a proper subgroup isomorphic to $(Bbb R,+)$? How does one prove it doesn't (if it doesn't)? Does it even have a proper subgroup which is uncountable? (All examples I can think of are countable)
abstract-algebra group-theory group-isomorphism
asked Jan 17 at 6:52
JMoravitzJMoravitz
47.9k33886
$endgroup$
add a comment |
$begingroup$
A captioned image with the text "My love for you is like a group which has a proper subgroup isomorphic to itself" was recently posted in a group chat I'm in. Ignoring the argument about the semantics of the language used (I am of the firm opinion that "itself" is in reference to the parent group, not the subgroup) this got me wondering about examples of groups which have this property. That is to say, a group $G$ that has a proper subgroup $H$ such that $G$ is isomorphic to $H$.
A few examples came to mind such as $(Bbb Z,+) > (2Bbb Z,+)$ and $(Bbb Q^*,times) > (K,times)$ where $K={frac{a}{b}~:~a,binBbb Zsetminus{0},~gcd(a,2)=gcd(b,2)=1}$ and similar. (Note: subgroup, not subring)
I got to thinking about $(Bbb R,+)$ however and wondering whether or not the real numbers have a subgroup isomorphic to the reals with respect to addition. My gut feeling is no, but I am at a loss as to how to prove this or come up with an example. Proper subgroups certainly exist, such as how $(Bbb Q,+)<(overline{Bbb Q},+)<(Bbb R,+)$, and so by specifying that our proper subgroup $H$ doesn't contain some element $x$, be it rational or irrational, that doesn't preclude $H$ from existing.
So then, I ask you, does $(Bbb R,+)$ have a proper subgroup isomorphic to $(Bbb R,+)$? How does one prove it doesn't (if it doesn't)? Does it even have a proper subgroup which is uncountable? (All examples I can think of are countable)
abstract-algebra group-theory group-isomorphism
asked Jan 17 at 6:52
JMoravitzJMoravitz
47.9k33886
$endgroup$
1
$begingroup$
For the record, $displaystyle {x in mathbb{R}: lim_{n to infty} sin(n! pi x) = 0}$ is a dense subgroup of $mathbb{R}$ which is uncountable. math.stackexchange.com/questions/494366/…
$endgroup$
– stressed out
Jan 17 at 7:05
$begingroup$
A group $G$ which contains no proper subgroup $H< G$ with $Hcong G$ is called coHopfian (so you are looking for non-coHopfian groups). For example, a group which splits as a free-product is non-coHopfian. A massive result of Sela from '97 (Structure and Rigidity in (Gromov) Hyperbolic Groups and Discrete Groups in Rank 1 Lie Groups II, GAFA, MR1466338) was that a torsion-free hyperbolic group is coHopfian if and only if it does not split as a free product. Reinfeldt-Weidmann (preprint) have proven this for all hyperbolic groups.
$endgroup$
– user1729
Jan 18 at 11:38
$begingroup$
Then "almost every" finitely presented group is coHopfian, so your examples are few and far between: There is a formal definition, due essentially to Gromov, of a "random" group, and these groups are all torsion-free hyperbolic, and hence coHopfian by Sela's result. These papers are hard. A more readable, and mostly self-contained proof that "almost all" groups are coHopfian in Kapovich, Schupp, Genericity, the Arzhantseva-Olʹshanskii method and the isomorphism problem for one-relator groups, Math. Ann. (2005) MR2107437.
$endgroup$
– user1729
Jan 18 at 11:46
$begingroup$
(The other take-home point is that people care about this property - Sela's paper was in GAFA which is a highly prestigious journal, and Sela is a big shot. The Kapovich-Schupp paper was in Mathematische Annalen, one of the big, classic journals. Einstein used to be an editor of Mathematische Annalen.)
$endgroup$
– user1729
Jan 18 at 11:48
add a comment |
$begingroup$
A captioned image with the text "My love for you is like a group which has a proper subgroup isomorphic to itself" was recently posted in a group chat I'm in. Ignoring the argument about the semantics of the language used (I am of the firm opinion that "itself" is in reference to the parent group, not the subgroup) this got me wondering about examples of groups which have this property. That is to say, a group $G$ that has a proper subgroup $H$ such that $G$ is isomorphic to $H$.
A few examples came to mind such as $(Bbb Z,+) > (2Bbb Z,+)$ and $(Bbb Q^*,times) > (K,times)$ where $K={frac{a}{b}~:~a,binBbb Zsetminus{0},~gcd(a,2)=gcd(b,2)=1}$ and similar. (Note: subgroup, not subring)
I got to thinking about $(Bbb R,+)$ however and wondering whether or not the real numbers have a subgroup isomorphic to the reals with respect to addition. My gut feeling is no, but I am at a loss as to how to prove this or come up with an example. Proper subgroups certainly exist, such as how $(Bbb Q,+)<(overline{Bbb Q},+)<(Bbb R,+)$, and so by specifying that our proper subgroup $H$ doesn't contain some element $x$, be it rational or irrational, that doesn't preclude $H$ from existing.
So then, I ask you, does $(Bbb R,+)$ have a proper subgroup isomorphic to $(Bbb R,+)$? How does one prove it doesn't (if it doesn't)? Does it even have a proper subgroup which is uncountable? (All examples I can think of are countable)
abstract-algebra group-theory group-isomorphism
asked Jan 17 at 6:52
JMoravitzJMoravitz
47.9k33886
$endgroup$
A captioned image with the text "My love for you is like a group which has a proper subgroup isomorphic to itself" was recently posted in a group chat I'm in. Ignoring the argument about the semantics of the language used (I am of the firm opinion that "itself" is in reference to the parent group, not the subgroup) this got me wondering about examples of groups which have this property. That is to say, a group $G$ that has a proper subgroup $H$ such that $G$ is isomorphic to $H$.
A few examples came to mind such as $(Bbb Z,+) > (2Bbb Z,+)$ and $(Bbb Q^*,times) > (K,times)$ where $K={frac{a}{b}~:~a,binBbb Zsetminus{0},~gcd(a,2)=gcd(b,2)=1}$ and similar. (Note: subgroup, not subring)
I got to thinking about $(Bbb R,+)$ however and wondering whether or not the real numbers have a subgroup isomorphic to the reals with respect to addition. My gut feeling is no, but I am at a loss as to how to prove this or come up with an example. Proper subgroups certainly exist, such as how $(Bbb Q,+)<(overline{Bbb Q},+)<(Bbb R,+)$, and so by specifying that our proper subgroup $H$ doesn't contain some element $x$, be it rational or irrational, that doesn't preclude $H$ from existing.
So then, I ask you, does $(Bbb R,+)$ have a proper subgroup isomorphic to $(Bbb R,+)$? How does one prove it doesn't (if it doesn't)? Does it even have a proper subgroup which is uncountable? (All examples I can think of are countable)
abstract-algebra group-theory group-isomorphism
abstract-algebra group-theory group-isomorphism
asked Jan 17 at 6:52
JMoravitzJMoravitz
47.9k33886
asked Jan 17 at 6:52
JMoravitzJMoravitz
47.9k33886
asked Jan 17 at 6:52
JMoravitzJMoravitz
47.9k33886
asked Jan 17 at 6:52
JMoravitzJMoravitz
47.9k33886
asked Jan 17 at 6:52
JMoravitzJMoravitz
47.9k33886
47.9k33886
1
$begingroup$
For the record, $displaystyle {x in mathbb{R}: lim_{n to infty} sin(n! pi x) = 0}$ is a dense subgroup of $mathbb{R}$ which is uncountable. math.stackexchange.com/questions/494366/…
$endgroup$
– stressed out
Jan 17 at 7:05
$begingroup$
A group $G$ which contains no proper subgroup $H< G$ with $Hcong G$ is called coHopfian (so you are looking for non-coHopfian groups). For example, a group which splits as a free-product is non-coHopfian. A massive result of Sela from '97 (Structure and Rigidity in (Gromov) Hyperbolic Groups and Discrete Groups in Rank 1 Lie Groups II, GAFA, MR1466338) was that a torsion-free hyperbolic group is coHopfian if and only if it does not split as a free product. Reinfeldt-Weidmann (preprint) have proven this for all hyperbolic groups.
$endgroup$
– user1729
Jan 18 at 11:38
$begingroup$
Then "almost every" finitely presented group is coHopfian, so your examples are few and far between: There is a formal definition, due essentially to Gromov, of a "random" group, and these groups are all torsion-free hyperbolic, and hence coHopfian by Sela's result. These papers are hard. A more readable, and mostly self-contained proof that "almost all" groups are coHopfian in Kapovich, Schupp, Genericity, the Arzhantseva-Olʹshanskii method and the isomorphism problem for one-relator groups, Math. Ann. (2005) MR2107437.
$endgroup$
– user1729
Jan 18 at 11:46
$begingroup$
(The other take-home point is that people care about this property - Sela's paper was in GAFA which is a highly prestigious journal, and Sela is a big shot. The Kapovich-Schupp paper was in Mathematische Annalen, one of the big, classic journals. Einstein used to be an editor of Mathematische Annalen.)
$endgroup$
– user1729
Jan 18 at 11:48
add a comment |
1
$begingroup$
For the record, $displaystyle {x in mathbb{R}: lim_{n to infty} sin(n! pi x) = 0}$ is a dense subgroup of $mathbb{R}$ which is uncountable. math.stackexchange.com/questions/494366/…
$endgroup$
– stressed out
Jan 17 at 7:05
$begingroup$
A group $G$ which contains no proper subgroup $H< G$ with $Hcong G$ is called coHopfian (so you are looking for non-coHopfian groups). For example, a group which splits as a free-product is non-coHopfian. A massive result of Sela from '97 (Structure and Rigidity in (Gromov) Hyperbolic Groups and Discrete Groups in Rank 1 Lie Groups II, GAFA, MR1466338) was that a torsion-free hyperbolic group is coHopfian if and only if it does not split as a free product. Reinfeldt-Weidmann (preprint) have proven this for all hyperbolic groups.
$endgroup$
– user1729
Jan 18 at 11:38
$begingroup$
Then "almost every" finitely presented group is coHopfian, so your examples are few and far between: There is a formal definition, due essentially to Gromov, of a "random" group, and these groups are all torsion-free hyperbolic, and hence coHopfian by Sela's result. These papers are hard. A more readable, and mostly self-contained proof that "almost all" groups are coHopfian in Kapovich, Schupp, Genericity, the Arzhantseva-Olʹshanskii method and the isomorphism problem for one-relator groups, Math. Ann. (2005) MR2107437.
$endgroup$
– user1729
Jan 18 at 11:46
$begingroup$
(The other take-home point is that people care about this property - Sela's paper was in GAFA which is a highly prestigious journal, and Sela is a big shot. The Kapovich-Schupp paper was in Mathematische Annalen, one of the big, classic journals. Einstein used to be an editor of Mathematische Annalen.)
$endgroup$
– user1729
Jan 18 at 11:48
1
1
$begingroup$
For the record, $displaystyle {x in mathbb{R}: lim_{n to infty} sin(n! pi x) = 0}$ is a dense subgroup of $mathbb{R}$ which is uncountable. math.stackexchange.com/questions/494366/…
$endgroup$
– stressed out
Jan 17 at 7:05
$begingroup$
For the record, $displaystyle {x in mathbb{R}: lim_{n to infty} sin(n! pi x) = 0}$ is a dense subgroup of $mathbb{R}$ which is uncountable. math.stackexchange.com/questions/494366/…
$endgroup$
– stressed out
Jan 17 at 7:05
$begingroup$
A group $G$ which contains no proper subgroup $H< G$ with $Hcong G$ is called coHopfian (so you are looking for non-coHopfian groups). For example, a group which splits as a free-product is non-coHopfian. A massive result of Sela from '97 (Structure and Rigidity in (Gromov) Hyperbolic Groups and Discrete Groups in Rank 1 Lie Groups II, GAFA, MR1466338) was that a torsion-free hyperbolic group is coHopfian if and only if it does not split as a free product. Reinfeldt-Weidmann (preprint) have proven this for all hyperbolic groups.
$endgroup$
– user1729
Jan 18 at 11:38
$begingroup$
A group $G$ which contains no proper subgroup $H< G$ with $Hcong G$ is called coHopfian (so you are looking for non-coHopfian groups). For example, a group which splits as a free-product is non-coHopfian. A massive result of Sela from '97 (Structure and Rigidity in (Gromov) Hyperbolic Groups and Discrete Groups in Rank 1 Lie Groups II, GAFA, MR1466338) was that a torsion-free hyperbolic group is coHopfian if and only if it does not split as a free product. Reinfeldt-Weidmann (preprint) have proven this for all hyperbolic groups.
$endgroup$
– user1729
Jan 18 at 11:38
$begingroup$
Then "almost every" finitely presented group is coHopfian, so your examples are few and far between: There is a formal definition, due essentially to Gromov, of a "random" group, and these groups are all torsion-free hyperbolic, and hence coHopfian by Sela's result. These papers are hard. A more readable, and mostly self-contained proof that "almost all" groups are coHopfian in Kapovich, Schupp, Genericity, the Arzhantseva-Olʹshanskii method and the isomorphism problem for one-relator groups, Math. Ann. (2005) MR2107437.
$endgroup$
– user1729
Jan 18 at 11:46
$begingroup$
Then "almost every" finitely presented group is coHopfian, so your examples are few and far between: There is a formal definition, due essentially to Gromov, of a "random" group, and these groups are all torsion-free hyperbolic, and hence coHopfian by Sela's result. These papers are hard. A more readable, and mostly self-contained proof that "almost all" groups are coHopfian in Kapovich, Schupp, Genericity, the Arzhantseva-Olʹshanskii method and the isomorphism problem for one-relator groups, Math. Ann. (2005) MR2107437.
$endgroup$
– user1729
Jan 18 at 11:46
$begingroup$
(The other take-home point is that people care about this property - Sela's paper was in GAFA which is a highly prestigious journal, and Sela is a big shot. The Kapovich-Schupp paper was in Mathematische Annalen, one of the big, classic journals. Einstein used to be an editor of Mathematische Annalen.)
$endgroup$
– user1729
Jan 18 at 11:48
$begingroup$
(The other take-home point is that people care about this property - Sela's paper was in GAFA which is a highly prestigious journal, and Sela is a big shot. The Kapovich-Schupp paper was in Mathematische Annalen, one of the big, classic journals. Einstein used to be an editor of Mathematische Annalen.)
$endgroup$
– user1729
Jan 18 at 11:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There is if you accept AC.
$Bbb R$ is a vector space over $Bbb Q$, so has a vector space
basis (Hamel basis) as a vector space over $Bbb Q$. Removing a generator,
the remaining generators generate a proper $Bbb Q$-subspace isomorphic to $Bbb R$
as a vector space, and so as an Abelian group.
answered Jan 17 at 7:16
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
1
$begingroup$
Might as well mention here that (assuming AC) $(mathbb R,+)$ is isomorphic to $(mathbb C,+)$ as well as (obviously) to a proper subgroup of $(mathbb C,+)$.
$endgroup$
– bof
Jan 17 at 7:31
add a comment |
$begingroup$
Consider $mathbb{R}$ as a vector space over the rationals and some
basis $B$. Then $B$ is Infinite. Let $bin B$. Then there is a bijection between $B$ and $B':=Bsetminus{b}$ implying that the vector spaces generated by $B$ and $B'$ are isomorphic. Thus they are also isomorphic as abelian groups. Finally note that the space generated by $B'$ is strictly contained in the space generated by $B$ which is $mathbb{R}$.
answered Jan 17 at 7:27
Jens SchwaigerJens Schwaiger
1,566138
$endgroup$
$begingroup$
When writing the answer I was uns wäre of the answer by Lord Shark the Unknown.
$endgroup$
– Jens Schwaiger
Jan 17 at 7:30
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is if you accept AC.
$Bbb R$ is a vector space over $Bbb Q$, so has a vector space
basis (Hamel basis) as a vector space over $Bbb Q$. Removing a generator,
the remaining generators generate a proper $Bbb Q$-subspace isomorphic to $Bbb R$
as a vector space, and so as an Abelian group.
answered Jan 17 at 7:16
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
1
$begingroup$
Might as well mention here that (assuming AC) $(mathbb R,+)$ is isomorphic to $(mathbb C,+)$ as well as (obviously) to a proper subgroup of $(mathbb C,+)$.
$endgroup$
– bof
Jan 17 at 7:31
add a comment |
$begingroup$
There is if you accept AC.
$Bbb R$ is a vector space over $Bbb Q$, so has a vector space
basis (Hamel basis) as a vector space over $Bbb Q$. Removing a generator,
the remaining generators generate a proper $Bbb Q$-subspace isomorphic to $Bbb R$
as a vector space, and so as an Abelian group.
answered Jan 17 at 7:16
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
1
$begingroup$
Might as well mention here that (assuming AC) $(mathbb R,+)$ is isomorphic to $(mathbb C,+)$ as well as (obviously) to a proper subgroup of $(mathbb C,+)$.
$endgroup$
– bof
Jan 17 at 7:31
add a comment |
$begingroup$
There is if you accept AC.
$Bbb R$ is a vector space over $Bbb Q$, so has a vector space
basis (Hamel basis) as a vector space over $Bbb Q$. Removing a generator,
the remaining generators generate a proper $Bbb Q$-subspace isomorphic to $Bbb R$
as a vector space, and so as an Abelian group.
answered Jan 17 at 7:16
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
There is if you accept AC.
$Bbb R$ is a vector space over $Bbb Q$, so has a vector space
basis (Hamel basis) as a vector space over $Bbb Q$. Removing a generator,
the remaining generators generate a proper $Bbb Q$-subspace isomorphic to $Bbb R$
as a vector space, and so as an Abelian group.
answered Jan 17 at 7:16
Lord Shark the UnknownLord Shark the Unknown
105k1160132
answered Jan 17 at 7:16
Lord Shark the UnknownLord Shark the Unknown
105k1160132
answered Jan 17 at 7:16
Lord Shark the UnknownLord Shark the Unknown
105k1160132
answered Jan 17 at 7:16
Lord Shark the UnknownLord Shark the Unknown
105k1160132
105k1160132
1
$begingroup$
Might as well mention here that (assuming AC) $(mathbb R,+)$ is isomorphic to $(mathbb C,+)$ as well as (obviously) to a proper subgroup of $(mathbb C,+)$.
$endgroup$
– bof
Jan 17 at 7:31
add a comment |
1
$begingroup$
Might as well mention here that (assuming AC) $(mathbb R,+)$ is isomorphic to $(mathbb C,+)$ as well as (obviously) to a proper subgroup of $(mathbb C,+)$.
$endgroup$
– bof
Jan 17 at 7:31
1
1
$begingroup$
Might as well mention here that (assuming AC) $(mathbb R,+)$ is isomorphic to $(mathbb C,+)$ as well as (obviously) to a proper subgroup of $(mathbb C,+)$.
$endgroup$
– bof
Jan 17 at 7:31
$begingroup$
Might as well mention here that (assuming AC) $(mathbb R,+)$ is isomorphic to $(mathbb C,+)$ as well as (obviously) to a proper subgroup of $(mathbb C,+)$.
$endgroup$
– bof
Jan 17 at 7:31
add a comment |
$begingroup$
Consider $mathbb{R}$ as a vector space over the rationals and some
basis $B$. Then $B$ is Infinite. Let $bin B$. Then there is a bijection between $B$ and $B':=Bsetminus{b}$ implying that the vector spaces generated by $B$ and $B'$ are isomorphic. Thus they are also isomorphic as abelian groups. Finally note that the space generated by $B'$ is strictly contained in the space generated by $B$ which is $mathbb{R}$.
answered Jan 17 at 7:27
Jens SchwaigerJens Schwaiger
1,566138
$endgroup$
$begingroup$
When writing the answer I was uns wäre of the answer by Lord Shark the Unknown.
$endgroup$
– Jens Schwaiger
Jan 17 at 7:30
add a comment |
$begingroup$
Consider $mathbb{R}$ as a vector space over the rationals and some
basis $B$. Then $B$ is Infinite. Let $bin B$. Then there is a bijection between $B$ and $B':=Bsetminus{b}$ implying that the vector spaces generated by $B$ and $B'$ are isomorphic. Thus they are also isomorphic as abelian groups. Finally note that the space generated by $B'$ is strictly contained in the space generated by $B$ which is $mathbb{R}$.
answered Jan 17 at 7:27
Jens SchwaigerJens Schwaiger
1,566138
$endgroup$
$begingroup$
When writing the answer I was uns wäre of the answer by Lord Shark the Unknown.
$endgroup$
– Jens Schwaiger
Jan 17 at 7:30
add a comment |
$begingroup$
Consider $mathbb{R}$ as a vector space over the rationals and some
basis $B$. Then $B$ is Infinite. Let $bin B$. Then there is a bijection between $B$ and $B':=Bsetminus{b}$ implying that the vector spaces generated by $B$ and $B'$ are isomorphic. Thus they are also isomorphic as abelian groups. Finally note that the space generated by $B'$ is strictly contained in the space generated by $B$ which is $mathbb{R}$.
answered Jan 17 at 7:27
Jens SchwaigerJens Schwaiger
1,566138
$endgroup$
Consider $mathbb{R}$ as a vector space over the rationals and some
basis $B$. Then $B$ is Infinite. Let $bin B$. Then there is a bijection between $B$ and $B':=Bsetminus{b}$ implying that the vector spaces generated by $B$ and $B'$ are isomorphic. Thus they are also isomorphic as abelian groups. Finally note that the space generated by $B'$ is strictly contained in the space generated by $B$ which is $mathbb{R}$.
answered Jan 17 at 7:27
Jens SchwaigerJens Schwaiger
1,566138
answered Jan 17 at 7:27
Jens SchwaigerJens Schwaiger
1,566138
answered Jan 17 at 7:27
Jens SchwaigerJens Schwaiger
1,566138
answered Jan 17 at 7:27
Jens SchwaigerJens Schwaiger
1,566138
1,566138
$begingroup$
When writing the answer I was uns wäre of the answer by Lord Shark the Unknown.
$endgroup$
– Jens Schwaiger
Jan 17 at 7:30
add a comment |
$begingroup$
When writing the answer I was uns wäre of the answer by Lord Shark the Unknown.
$endgroup$
– Jens Schwaiger
Jan 17 at 7:30
$begingroup$
When writing the answer I was uns wäre of the answer by Lord Shark the Unknown.
$endgroup$
– Jens Schwaiger
Jan 17 at 7:30
$begingroup$
When writing the answer I was uns wäre of the answer by Lord Shark the Unknown.
$endgroup$
– Jens Schwaiger
Jan 17 at 7:30
add a comment |
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1
$begingroup$
For the record, $displaystyle {x in mathbb{R}: lim_{n to infty} sin(n! pi x) = 0}$ is a dense subgroup of $mathbb{R}$ which is uncountable. math.stackexchange.com/questions/494366/…
$endgroup$
– stressed out
Jan 17 at 7:05
$begingroup$
A group $G$ which contains no proper subgroup $H< G$ with $Hcong G$ is called coHopfian (so you are looking for non-coHopfian groups). For example, a group which splits as a free-product is non-coHopfian. A massive result of Sela from '97 (Structure and Rigidity in (Gromov) Hyperbolic Groups and Discrete Groups in Rank 1 Lie Groups II, GAFA, MR1466338) was that a torsion-free hyperbolic group is coHopfian if and only if it does not split as a free product. Reinfeldt-Weidmann (preprint) have proven this for all hyperbolic groups.
$endgroup$
– user1729
Jan 18 at 11:38
$begingroup$
Then "almost every" finitely presented group is coHopfian, so your examples are few and far between: There is a formal definition, due essentially to Gromov, of a "random" group, and these groups are all torsion-free hyperbolic, and hence coHopfian by Sela's result. These papers are hard. A more readable, and mostly self-contained proof that "almost all" groups are coHopfian in Kapovich, Schupp, Genericity, the Arzhantseva-Olʹshanskii method and the isomorphism problem for one-relator groups, Math. Ann. (2005) MR2107437.
$endgroup$
– user1729
Jan 18 at 11:46
$begingroup$
(The other take-home point is that people care about this property - Sela's paper was in GAFA which is a highly prestigious journal, and Sela is a big shot. The Kapovich-Schupp paper was in Mathematische Annalen, one of the big, classic journals. Einstein used to be an editor of Mathematische Annalen.)
$endgroup$
– user1729
Jan 18 at 11:48