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Is it possible to demonstrate two groups are isomorphic without specifying an isomorphism between them?
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Usually, when there are two finite groups of small order, we can check if they are isomorphic by trying to impose an isomorphism.
But suppose we have two very large finite groups, what are some conditions on the two groups I can scrutinise so as to conclude an isomorphism exists between them or not? One condition I can think of is they must have the same order. My another crude guess is
If an isomorphism $phi$ exists between two finite groups $G_1$ and
$G_2$ , then $phi$ must map element of order $a$ in $G_1$ to element
of order $a$ in $G_2$ too.
So there are two main questions in this post:
1.What are some conditions on the two groups one can scrutinise so as to conclude an isomorphism exists between them or not
2.Whether my guess is true, and whether there is a counterexample.
group-theory finite-groups group-isomorphism
asked 3 hours ago
hephaes
927
add a comment |Â
up vote
4
down vote
favorite
Usually, when there are two finite groups of small order, we can check if they are isomorphic by trying to impose an isomorphism.
But suppose we have two very large finite groups, what are some conditions on the two groups I can scrutinise so as to conclude an isomorphism exists between them or not? One condition I can think of is they must have the same order. My another crude guess is
If an isomorphism $phi$ exists between two finite groups $G_1$ and
$G_2$ , then $phi$ must map element of order $a$ in $G_1$ to element
of order $a$ in $G_2$ too.
So there are two main questions in this post:
1.What are some conditions on the two groups one can scrutinise so as to conclude an isomorphism exists between them or not
2.Whether my guess is true, and whether there is a counterexample.
group-theory finite-groups group-isomorphism
asked 3 hours ago
hephaes
927
1
Incidentally, it's worth mentioning that (assuming the axiom of choice) the groups $(mathbbR,+)$ and $(mathbbC,+)$ are isomorphic(!) but provably no easily-describably isomorphism between the two exists (e.g. they aren't Borel isomorphic).
– Noah Schweber
2 hours ago
@Moritz I agree that you can use presentations to prove groups isomorphic, but I don't see how you conclude that the answer to the question is yes. By finding a presentation you are effectively finding a specific isomorphism.
– Derek Holt
1 hour ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Usually, when there are two finite groups of small order, we can check if they are isomorphic by trying to impose an isomorphism.
But suppose we have two very large finite groups, what are some conditions on the two groups I can scrutinise so as to conclude an isomorphism exists between them or not? One condition I can think of is they must have the same order. My another crude guess is
If an isomorphism $phi$ exists between two finite groups $G_1$ and
$G_2$ , then $phi$ must map element of order $a$ in $G_1$ to element
of order $a$ in $G_2$ too.
So there are two main questions in this post:
1.What are some conditions on the two groups one can scrutinise so as to conclude an isomorphism exists between them or not
2.Whether my guess is true, and whether there is a counterexample.
group-theory finite-groups group-isomorphism
asked 3 hours ago
hephaes
927
Usually, when there are two finite groups of small order, we can check if they are isomorphic by trying to impose an isomorphism.
But suppose we have two very large finite groups, what are some conditions on the two groups I can scrutinise so as to conclude an isomorphism exists between them or not? One condition I can think of is they must have the same order. My another crude guess is
If an isomorphism $phi$ exists between two finite groups $G_1$ and
$G_2$ , then $phi$ must map element of order $a$ in $G_1$ to element
of order $a$ in $G_2$ too.
So there are two main questions in this post:
1.What are some conditions on the two groups one can scrutinise so as to conclude an isomorphism exists between them or not
2.Whether my guess is true, and whether there is a counterexample.
group-theory finite-groups group-isomorphism
group-theory finite-groups group-isomorphism
asked 3 hours ago
hephaes
927
asked 3 hours ago
hephaes
927
asked 3 hours ago
hephaes
927
asked 3 hours ago
hephaes
927
asked 3 hours ago
hephaes
927
927
1
Incidentally, it's worth mentioning that (assuming the axiom of choice) the groups $(mathbbR,+)$ and $(mathbbC,+)$ are isomorphic(!) but provably no easily-describably isomorphism between the two exists (e.g. they aren't Borel isomorphic).
– Noah Schweber
2 hours ago
@Moritz I agree that you can use presentations to prove groups isomorphic, but I don't see how you conclude that the answer to the question is yes. By finding a presentation you are effectively finding a specific isomorphism.
– Derek Holt
1 hour ago
add a comment |Â
1
Incidentally, it's worth mentioning that (assuming the axiom of choice) the groups $(mathbbR,+)$ and $(mathbbC,+)$ are isomorphic(!) but provably no easily-describably isomorphism between the two exists (e.g. they aren't Borel isomorphic).
– Noah Schweber
2 hours ago
@Moritz I agree that you can use presentations to prove groups isomorphic, but I don't see how you conclude that the answer to the question is yes. By finding a presentation you are effectively finding a specific isomorphism.
– Derek Holt
1 hour ago
1
1
Incidentally, it's worth mentioning that (assuming the axiom of choice) the groups $(mathbbR,+)$ and $(mathbbC,+)$ are isomorphic(!) but provably no easily-describably isomorphism between the two exists (e.g. they aren't Borel isomorphic).
– Noah Schweber
2 hours ago
Incidentally, it's worth mentioning that (assuming the axiom of choice) the groups $(mathbbR,+)$ and $(mathbbC,+)$ are isomorphic(!) but provably no easily-describably isomorphism between the two exists (e.g. they aren't Borel isomorphic).
– Noah Schweber
2 hours ago
@Moritz I agree that you can use presentations to prove groups isomorphic, but I don't see how you conclude that the answer to the question is yes. By finding a presentation you are effectively finding a specific isomorphism.
– Derek Holt
1 hour ago
@Moritz I agree that you can use presentations to prove groups isomorphic, but I don't see how you conclude that the answer to the question is yes. By finding a presentation you are effectively finding a specific isomorphism.
– Derek Holt
1 hour ago
add a comment |Â
2 Answers
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up vote
3
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Perhaps what you are seeking are "classification theorems", namely theorems which have the following format: If $G,H$ are two finite groups having property $P$ then $G$ is isomorphic to $H$ if and only if $G$ and $H$ have the same properties $Q_1,...,Q_n$.
For example, the classification of finite cyclic groups says that if $G$ and $H$ are two finite cyclic groups then $G$ is isomorphic to $H$ if and only if $G$ and $H$ have the same order. So if you know the two groups are cyclic, all you have to do is count how many elements they have to determine whether they are isomorphic.
The next step up is the classification of finite abelian groups. First one proves a theorem: for every finite abelian group $G$ there exists a direct sum decomposition of the form $G = G_1 oplus G_2 oplus cdots G_K$ such that each $G_i$ is a finite cyclic group of some order $d_i$, and such that the sequence of group orders is linearly ordered by divisibility: $d_1$ divides $d_2$ which divides $d_3$ which divides... which divides $d_K$. There's a word for this sequence $d_i$ which I don't know, so I'll just temporarily call it the "divisor sequence" of the group $G$. The classification theorem says that if $G$ and $H$ are finite abelian groups then $G$ is isomorphic to $H$ if and only if they have the exact same divisor sequence.
There are a lot of examples of theorems like this, of more and more power applying to more and more general classes of finite groups. They get harder and harder to prove, as the class becomes more and more general.
Keep in mind, though, that there is an important reality hidden in the proofs: when you trace through the proof of existence of an isomorphism between two finite groups, you will almost always find the construction of an isomorphism. (In finite group theory, I doubt there's a single exception to this rule. In infinite group theory, there are probably many exceptions in which the proof of existence of an isomorphism leads instead to some existence axiom such as the axiom of choice).
answered 2 hours ago
Lee Mosher
46.5k33681
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up vote
0
down vote
- You already mentioned the order of the group and the order of elements in the group. There are several other easy invariants we can use. For example, is there an isomorphism between $S_4$ and $D_6times D_6$? Both groups have the same order, namely $24$. But $S_4$ has trivial center, whereas $D_6times D_6$ has not. So there cannot be any isomorphism between them!
Your question in 1. is: "so as to conclude an isomorphism exists between them or not".
It is often very easy to show that two groups are not isomorphic. But then we have decided the question.
- Your guess is true.
answered 45 mins ago
Dietrich Burde
75.7k64185
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Perhaps what you are seeking are "classification theorems", namely theorems which have the following format: If $G,H$ are two finite groups having property $P$ then $G$ is isomorphic to $H$ if and only if $G$ and $H$ have the same properties $Q_1,...,Q_n$.
For example, the classification of finite cyclic groups says that if $G$ and $H$ are two finite cyclic groups then $G$ is isomorphic to $H$ if and only if $G$ and $H$ have the same order. So if you know the two groups are cyclic, all you have to do is count how many elements they have to determine whether they are isomorphic.
The next step up is the classification of finite abelian groups. First one proves a theorem: for every finite abelian group $G$ there exists a direct sum decomposition of the form $G = G_1 oplus G_2 oplus cdots G_K$ such that each $G_i$ is a finite cyclic group of some order $d_i$, and such that the sequence of group orders is linearly ordered by divisibility: $d_1$ divides $d_2$ which divides $d_3$ which divides... which divides $d_K$. There's a word for this sequence $d_i$ which I don't know, so I'll just temporarily call it the "divisor sequence" of the group $G$. The classification theorem says that if $G$ and $H$ are finite abelian groups then $G$ is isomorphic to $H$ if and only if they have the exact same divisor sequence.
There are a lot of examples of theorems like this, of more and more power applying to more and more general classes of finite groups. They get harder and harder to prove, as the class becomes more and more general.
Keep in mind, though, that there is an important reality hidden in the proofs: when you trace through the proof of existence of an isomorphism between two finite groups, you will almost always find the construction of an isomorphism. (In finite group theory, I doubt there's a single exception to this rule. In infinite group theory, there are probably many exceptions in which the proof of existence of an isomorphism leads instead to some existence axiom such as the axiom of choice).
answered 2 hours ago
Lee Mosher
46.5k33681
add a comment |Â
up vote
3
down vote
Perhaps what you are seeking are "classification theorems", namely theorems which have the following format: If $G,H$ are two finite groups having property $P$ then $G$ is isomorphic to $H$ if and only if $G$ and $H$ have the same properties $Q_1,...,Q_n$.
For example, the classification of finite cyclic groups says that if $G$ and $H$ are two finite cyclic groups then $G$ is isomorphic to $H$ if and only if $G$ and $H$ have the same order. So if you know the two groups are cyclic, all you have to do is count how many elements they have to determine whether they are isomorphic.
The next step up is the classification of finite abelian groups. First one proves a theorem: for every finite abelian group $G$ there exists a direct sum decomposition of the form $G = G_1 oplus G_2 oplus cdots G_K$ such that each $G_i$ is a finite cyclic group of some order $d_i$, and such that the sequence of group orders is linearly ordered by divisibility: $d_1$ divides $d_2$ which divides $d_3$ which divides... which divides $d_K$. There's a word for this sequence $d_i$ which I don't know, so I'll just temporarily call it the "divisor sequence" of the group $G$. The classification theorem says that if $G$ and $H$ are finite abelian groups then $G$ is isomorphic to $H$ if and only if they have the exact same divisor sequence.
There are a lot of examples of theorems like this, of more and more power applying to more and more general classes of finite groups. They get harder and harder to prove, as the class becomes more and more general.
Keep in mind, though, that there is an important reality hidden in the proofs: when you trace through the proof of existence of an isomorphism between two finite groups, you will almost always find the construction of an isomorphism. (In finite group theory, I doubt there's a single exception to this rule. In infinite group theory, there are probably many exceptions in which the proof of existence of an isomorphism leads instead to some existence axiom such as the axiom of choice).
answered 2 hours ago
Lee Mosher
46.5k33681
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Perhaps what you are seeking are "classification theorems", namely theorems which have the following format: If $G,H$ are two finite groups having property $P$ then $G$ is isomorphic to $H$ if and only if $G$ and $H$ have the same properties $Q_1,...,Q_n$.
For example, the classification of finite cyclic groups says that if $G$ and $H$ are two finite cyclic groups then $G$ is isomorphic to $H$ if and only if $G$ and $H$ have the same order. So if you know the two groups are cyclic, all you have to do is count how many elements they have to determine whether they are isomorphic.
The next step up is the classification of finite abelian groups. First one proves a theorem: for every finite abelian group $G$ there exists a direct sum decomposition of the form $G = G_1 oplus G_2 oplus cdots G_K$ such that each $G_i$ is a finite cyclic group of some order $d_i$, and such that the sequence of group orders is linearly ordered by divisibility: $d_1$ divides $d_2$ which divides $d_3$ which divides... which divides $d_K$. There's a word for this sequence $d_i$ which I don't know, so I'll just temporarily call it the "divisor sequence" of the group $G$. The classification theorem says that if $G$ and $H$ are finite abelian groups then $G$ is isomorphic to $H$ if and only if they have the exact same divisor sequence.
There are a lot of examples of theorems like this, of more and more power applying to more and more general classes of finite groups. They get harder and harder to prove, as the class becomes more and more general.
Keep in mind, though, that there is an important reality hidden in the proofs: when you trace through the proof of existence of an isomorphism between two finite groups, you will almost always find the construction of an isomorphism. (In finite group theory, I doubt there's a single exception to this rule. In infinite group theory, there are probably many exceptions in which the proof of existence of an isomorphism leads instead to some existence axiom such as the axiom of choice).
answered 2 hours ago
Lee Mosher
46.5k33681
Perhaps what you are seeking are "classification theorems", namely theorems which have the following format: If $G,H$ are two finite groups having property $P$ then $G$ is isomorphic to $H$ if and only if $G$ and $H$ have the same properties $Q_1,...,Q_n$.
For example, the classification of finite cyclic groups says that if $G$ and $H$ are two finite cyclic groups then $G$ is isomorphic to $H$ if and only if $G$ and $H$ have the same order. So if you know the two groups are cyclic, all you have to do is count how many elements they have to determine whether they are isomorphic.
The next step up is the classification of finite abelian groups. First one proves a theorem: for every finite abelian group $G$ there exists a direct sum decomposition of the form $G = G_1 oplus G_2 oplus cdots G_K$ such that each $G_i$ is a finite cyclic group of some order $d_i$, and such that the sequence of group orders is linearly ordered by divisibility: $d_1$ divides $d_2$ which divides $d_3$ which divides... which divides $d_K$. There's a word for this sequence $d_i$ which I don't know, so I'll just temporarily call it the "divisor sequence" of the group $G$. The classification theorem says that if $G$ and $H$ are finite abelian groups then $G$ is isomorphic to $H$ if and only if they have the exact same divisor sequence.
There are a lot of examples of theorems like this, of more and more power applying to more and more general classes of finite groups. They get harder and harder to prove, as the class becomes more and more general.
Keep in mind, though, that there is an important reality hidden in the proofs: when you trace through the proof of existence of an isomorphism between two finite groups, you will almost always find the construction of an isomorphism. (In finite group theory, I doubt there's a single exception to this rule. In infinite group theory, there are probably many exceptions in which the proof of existence of an isomorphism leads instead to some existence axiom such as the axiom of choice).
answered 2 hours ago
Lee Mosher
46.5k33681
answered 2 hours ago
Lee Mosher
46.5k33681
answered 2 hours ago
Lee Mosher
46.5k33681
answered 2 hours ago
Lee Mosher
46.5k33681
46.5k33681
add a comment |Â
add a comment |Â
up vote
0
down vote
- You already mentioned the order of the group and the order of elements in the group. There are several other easy invariants we can use. For example, is there an isomorphism between $S_4$ and $D_6times D_6$? Both groups have the same order, namely $24$. But $S_4$ has trivial center, whereas $D_6times D_6$ has not. So there cannot be any isomorphism between them!
Your question in 1. is: "so as to conclude an isomorphism exists between them or not".
It is often very easy to show that two groups are not isomorphic. But then we have decided the question.
- Your guess is true.
answered 45 mins ago
Dietrich Burde
75.7k64185
add a comment |Â
up vote
0
down vote
- You already mentioned the order of the group and the order of elements in the group. There are several other easy invariants we can use. For example, is there an isomorphism between $S_4$ and $D_6times D_6$? Both groups have the same order, namely $24$. But $S_4$ has trivial center, whereas $D_6times D_6$ has not. So there cannot be any isomorphism between them!
Your question in 1. is: "so as to conclude an isomorphism exists between them or not".
It is often very easy to show that two groups are not isomorphic. But then we have decided the question.
- Your guess is true.
answered 45 mins ago
Dietrich Burde
75.7k64185
add a comment |Â
up vote
0
down vote
up vote
0
down vote
- You already mentioned the order of the group and the order of elements in the group. There are several other easy invariants we can use. For example, is there an isomorphism between $S_4$ and $D_6times D_6$? Both groups have the same order, namely $24$. But $S_4$ has trivial center, whereas $D_6times D_6$ has not. So there cannot be any isomorphism between them!
Your question in 1. is: "so as to conclude an isomorphism exists between them or not".
It is often very easy to show that two groups are not isomorphic. But then we have decided the question.
- Your guess is true.
answered 45 mins ago
Dietrich Burde
75.7k64185
- You already mentioned the order of the group and the order of elements in the group. There are several other easy invariants we can use. For example, is there an isomorphism between $S_4$ and $D_6times D_6$? Both groups have the same order, namely $24$. But $S_4$ has trivial center, whereas $D_6times D_6$ has not. So there cannot be any isomorphism between them!
Your question in 1. is: "so as to conclude an isomorphism exists between them or not".
It is often very easy to show that two groups are not isomorphic. But then we have decided the question.
- Your guess is true.
answered 45 mins ago
Dietrich Burde
75.7k64185
edited 21 mins ago
answered 45 mins ago
Dietrich Burde
75.7k64185
answered 45 mins ago
Dietrich Burde
75.7k64185
answered 45 mins ago
Dietrich Burde
75.7k64185
75.7k64185
add a comment |Â
add a comment |Â
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1
Incidentally, it's worth mentioning that (assuming the axiom of choice) the groups $(mathbbR,+)$ and $(mathbbC,+)$ are isomorphic(!) but provably no easily-describably isomorphism between the two exists (e.g. they aren't Borel isomorphic).
– Noah Schweber
2 hours ago
@Moritz I agree that you can use presentations to prove groups isomorphic, but I don't see how you conclude that the answer to the question is yes. By finding a presentation you are effectively finding a specific isomorphism.
– Derek Holt
1 hour ago