Mixing
Existence and construction of isomorphism between finite groups
$begingroup$
Assume I have two finite groups $G$ and $H$ of equal order. Further assume I have found minimal generating sets $A$ and $B$ for a the two groups respectively (of equal size) and additionally (see comments) I know at least one decomposition into generating elements of all $g in G$.
I now want to find out if the two groups are isomorphic give this extra information of the minimal generating sets.
Is there an approach along these lines:
Define a bijective function $f: Ato B$. Extend the function to all of $G$ in the following manner:
For $g in G$ find a decomposition of $g = a_1 + cdots + a_m$ where $a_i in A$ and define $f(g) = sum_{i=1}^m f(a_i)$. If this extension fulfills the homomorphism property $f(g_1 + g_2) = f(g_1) + f(g_2)$ for all $g_1, g_2 in G$ then it is an isomorphism.
Question 1. Is the above statement correct? Do I need to check $f(G) = H$ or is this already implicitly true?
Question 2. Here I need to expand the function for all $g in G$ and check the homomorphism property for all pairs of elements from $G$.
Given the information of two minimal generating sets, can I reduce the amount of checking I have to do?
(Checking only the pairs of generators is obviously not enough since the extension fulfills the homomorphism property for elements from $A$ by construction)
group-theory finite-groups group-isomorphism
edited Jan 29 at 13:50
Shaun
9,869113684
asked Jan 29 at 11:32
elfeckelfeck
1428
$endgroup$
|
show 6 more comments
$begingroup$
Assume I have two finite groups $G$ and $H$ of equal order. Further assume I have found minimal generating sets $A$ and $B$ for a the two groups respectively (of equal size) and additionally (see comments) I know at least one decomposition into generating elements of all $g in G$.
I now want to find out if the two groups are isomorphic give this extra information of the minimal generating sets.
Is there an approach along these lines:
Define a bijective function $f: Ato B$. Extend the function to all of $G$ in the following manner:
For $g in G$ find a decomposition of $g = a_1 + cdots + a_m$ where $a_i in A$ and define $f(g) = sum_{i=1}^m f(a_i)$. If this extension fulfills the homomorphism property $f(g_1 + g_2) = f(g_1) + f(g_2)$ for all $g_1, g_2 in G$ then it is an isomorphism.
Question 1. Is the above statement correct? Do I need to check $f(G) = H$ or is this already implicitly true?
Question 2. Here I need to expand the function for all $g in G$ and check the homomorphism property for all pairs of elements from $G$.
Given the information of two minimal generating sets, can I reduce the amount of checking I have to do?
(Checking only the pairs of generators is obviously not enough since the extension fulfills the homomorphism property for elements from $A$ by construction)
group-theory finite-groups group-isomorphism
edited Jan 29 at 13:50
Shaun
9,869113684
asked Jan 29 at 11:32
elfeckelfeck
1428
$endgroup$
$begingroup$
Are you assuming your groups are abelian? How do you know that that the extension of $f$ that you define is independent of the choice of decomposition of $g$? If, say, $a_1$ has order $3$ and $f(a_1)=b_1$ has order $2$ then $f(a_1^2)=b_1^2=e_H$ so your function isn't even a set theoretic bijection.
$endgroup$
– lulu
Jan 29 at 11:38
$begingroup$
No I don't assume commutativity. Good point, I had not thought about that. Is there a way to rescue this?
$endgroup$
– elfeck
Jan 29 at 11:41
$begingroup$
Well, I don't know. It's hard to say much about a group just knowing a generating set.
$endgroup$
– lulu
Jan 29 at 11:43
$begingroup$
If the extension of $f$ were an isomorphism, then it must also be independent of the decomposition of any $g$. So perhaps it is not necessary to require (a-priori) that the extension of $f$ is bijective
$endgroup$
– elfeck
Jan 29 at 11:45
$begingroup$
You haven't even defined an extension of $f$, your definition depends on that independence.
$endgroup$
– lulu
Jan 29 at 11:46
|
show 6 more comments
$begingroup$
Assume I have two finite groups $G$ and $H$ of equal order. Further assume I have found minimal generating sets $A$ and $B$ for a the two groups respectively (of equal size) and additionally (see comments) I know at least one decomposition into generating elements of all $g in G$.
I now want to find out if the two groups are isomorphic give this extra information of the minimal generating sets.
Is there an approach along these lines:
Define a bijective function $f: Ato B$. Extend the function to all of $G$ in the following manner:
For $g in G$ find a decomposition of $g = a_1 + cdots + a_m$ where $a_i in A$ and define $f(g) = sum_{i=1}^m f(a_i)$. If this extension fulfills the homomorphism property $f(g_1 + g_2) = f(g_1) + f(g_2)$ for all $g_1, g_2 in G$ then it is an isomorphism.
Question 1. Is the above statement correct? Do I need to check $f(G) = H$ or is this already implicitly true?
Question 2. Here I need to expand the function for all $g in G$ and check the homomorphism property for all pairs of elements from $G$.
Given the information of two minimal generating sets, can I reduce the amount of checking I have to do?
(Checking only the pairs of generators is obviously not enough since the extension fulfills the homomorphism property for elements from $A$ by construction)
group-theory finite-groups group-isomorphism
edited Jan 29 at 13:50
Shaun
9,869113684
asked Jan 29 at 11:32
elfeckelfeck
1428
$endgroup$
Assume I have two finite groups $G$ and $H$ of equal order. Further assume I have found minimal generating sets $A$ and $B$ for a the two groups respectively (of equal size) and additionally (see comments) I know at least one decomposition into generating elements of all $g in G$.
I now want to find out if the two groups are isomorphic give this extra information of the minimal generating sets.
Is there an approach along these lines:
Define a bijective function $f: Ato B$. Extend the function to all of $G$ in the following manner:
For $g in G$ find a decomposition of $g = a_1 + cdots + a_m$ where $a_i in A$ and define $f(g) = sum_{i=1}^m f(a_i)$. If this extension fulfills the homomorphism property $f(g_1 + g_2) = f(g_1) + f(g_2)$ for all $g_1, g_2 in G$ then it is an isomorphism.
Question 1. Is the above statement correct? Do I need to check $f(G) = H$ or is this already implicitly true?
Question 2. Here I need to expand the function for all $g in G$ and check the homomorphism property for all pairs of elements from $G$.
Given the information of two minimal generating sets, can I reduce the amount of checking I have to do?
(Checking only the pairs of generators is obviously not enough since the extension fulfills the homomorphism property for elements from $A$ by construction)
group-theory finite-groups group-isomorphism
group-theory finite-groups group-isomorphism
edited Jan 29 at 13:50
Shaun
9,869113684
asked Jan 29 at 11:32
elfeckelfeck
1428
edited Jan 29 at 13:50
Shaun
9,869113684
asked Jan 29 at 11:32
elfeckelfeck
1428
edited Jan 29 at 13:50
Shaun
9,869113684
edited Jan 29 at 13:50
Shaun
9,869113684
edited Jan 29 at 13:50
Shaun
9,869113684
9,869113684
asked Jan 29 at 11:32
elfeckelfeck
1428
asked Jan 29 at 11:32
elfeckelfeck
1428
asked Jan 29 at 11:32
elfeckelfeck
1428
1428
$begingroup$
Are you assuming your groups are abelian? How do you know that that the extension of $f$ that you define is independent of the choice of decomposition of $g$? If, say, $a_1$ has order $3$ and $f(a_1)=b_1$ has order $2$ then $f(a_1^2)=b_1^2=e_H$ so your function isn't even a set theoretic bijection.
$endgroup$
– lulu
Jan 29 at 11:38
$begingroup$
No I don't assume commutativity. Good point, I had not thought about that. Is there a way to rescue this?
$endgroup$
– elfeck
Jan 29 at 11:41
$begingroup$
Well, I don't know. It's hard to say much about a group just knowing a generating set.
$endgroup$
– lulu
Jan 29 at 11:43
$begingroup$
If the extension of $f$ were an isomorphism, then it must also be independent of the decomposition of any $g$. So perhaps it is not necessary to require (a-priori) that the extension of $f$ is bijective
$endgroup$
– elfeck
Jan 29 at 11:45
$begingroup$
You haven't even defined an extension of $f$, your definition depends on that independence.
$endgroup$
– lulu
Jan 29 at 11:46
|
show 6 more comments
$begingroup$
Are you assuming your groups are abelian? How do you know that that the extension of $f$ that you define is independent of the choice of decomposition of $g$? If, say, $a_1$ has order $3$ and $f(a_1)=b_1$ has order $2$ then $f(a_1^2)=b_1^2=e_H$ so your function isn't even a set theoretic bijection.
$endgroup$
– lulu
Jan 29 at 11:38
$begingroup$
No I don't assume commutativity. Good point, I had not thought about that. Is there a way to rescue this?
$endgroup$
– elfeck
Jan 29 at 11:41
$begingroup$
Well, I don't know. It's hard to say much about a group just knowing a generating set.
$endgroup$
– lulu
Jan 29 at 11:43
$begingroup$
If the extension of $f$ were an isomorphism, then it must also be independent of the decomposition of any $g$. So perhaps it is not necessary to require (a-priori) that the extension of $f$ is bijective
$endgroup$
– elfeck
Jan 29 at 11:45
$begingroup$
You haven't even defined an extension of $f$, your definition depends on that independence.
$endgroup$
– lulu
Jan 29 at 11:46
$begingroup$
Are you assuming your groups are abelian? How do you know that that the extension of $f$ that you define is independent of the choice of decomposition of $g$? If, say, $a_1$ has order $3$ and $f(a_1)=b_1$ has order $2$ then $f(a_1^2)=b_1^2=e_H$ so your function isn't even a set theoretic bijection.
$endgroup$
– lulu
Jan 29 at 11:38
$begingroup$
Are you assuming your groups are abelian? How do you know that that the extension of $f$ that you define is independent of the choice of decomposition of $g$? If, say, $a_1$ has order $3$ and $f(a_1)=b_1$ has order $2$ then $f(a_1^2)=b_1^2=e_H$ so your function isn't even a set theoretic bijection.
$endgroup$
– lulu
Jan 29 at 11:38
$begingroup$
No I don't assume commutativity. Good point, I had not thought about that. Is there a way to rescue this?
$endgroup$
– elfeck
Jan 29 at 11:41
$begingroup$
No I don't assume commutativity. Good point, I had not thought about that. Is there a way to rescue this?
$endgroup$
– elfeck
Jan 29 at 11:41
$begingroup$
Well, I don't know. It's hard to say much about a group just knowing a generating set.
$endgroup$
– lulu
Jan 29 at 11:43
$begingroup$
Well, I don't know. It's hard to say much about a group just knowing a generating set.
$endgroup$
– lulu
Jan 29 at 11:43
$begingroup$
If the extension of $f$ were an isomorphism, then it must also be independent of the decomposition of any $g$. So perhaps it is not necessary to require (a-priori) that the extension of $f$ is bijective
$endgroup$
– elfeck
Jan 29 at 11:45
$begingroup$
If the extension of $f$ were an isomorphism, then it must also be independent of the decomposition of any $g$. So perhaps it is not necessary to require (a-priori) that the extension of $f$ is bijective
$endgroup$
– elfeck
Jan 29 at 11:45
$begingroup$
You haven't even defined an extension of $f$, your definition depends on that independence.
$endgroup$
– lulu
Jan 29 at 11:46
$begingroup$
You haven't even defined an extension of $f$, your definition depends on that independence.
$endgroup$
– lulu
Jan 29 at 11:46
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
(I should state at the start that the ideas in the question are close to pretty standard ideas. All I am doing here is trying to explain the link to these standard ideas.)
There are two things you seem to be missing.
- If $f$ is a homomorphism then its extension to the whole group is not necessarily a bijection. You need to verify that it is injective or surjective (or both if your group is infinite).
- The image group $H$ may have lots of minimal generating sets. So you need to find all minimal generating sets $B_i$ (for speed you can take them up to conjugacy, or up to automorphism). For example, consider $mathbb{Z}_3timesmathbb{Z}_{18}$. This is isomorphic to $mathbb{Z}_6timesmathbb{Z}_9$. One group is generated by an element of order $3$ and an element of order $18$, while the other is generated by an element of order $6$ and an element of order $9$.
The following is then true:
Theorem. Let $A$ be a minimal generating set for $G$, and let ${B_i}$ be the set of all minimal generating sets of $H$ which have cardinality $|A|$. Then $Gcong H$ if and only if there exists a bijection $f_{i, j}: Arightarrow B_i$ which extends to an isomorphism $Grightarrow H$ in the way you describe.
Note that you have to check that the extension is bijective and is a homomorphism. These are non-trivial tasks*. Also, the subscript $j$ is because there are lots of bijections between $A$ and $B_i$, but not all of these will extend to be a homomorphism.
*If $G$ is given by a presentation $langle mathbf{x}midmathbf{r}rangle$ (alternatively this presentation can be computed, but not necessarily quickly!, as your group is finite) then determining if $f$ is a homomorphism is relatively easy, as you just need to verify that $f(R)=_H1$ for all $Rin mathbf{r}$.
answered Jan 29 at 11:58
user1729user1729
17.5k64193
$endgroup$
$begingroup$
Thank you for your answer. So what you are implying that is not even sufficient to fix two generating sets $A$ and $B$ of G and H and check all extensions of all bijections $f: A to B$ for bijectivity and homomorphism? Edit: So there might not even exists a bijective homomorphisms $G to H$ that maps $A$ to $B$.
$endgroup$
– elfeck
Jan 29 at 12:05
1
$begingroup$
Nope! For example, consider $mathbb{Z}_3timesmathbb{Z}_{18}$. This is isomorphic to $mathbb{Z}_6timesmathbb{Z}_9$. One group is generated by an element of order $3$ and an element of order $18$, while the other is generated by an element of order $6$ and an element of order $9$.
$endgroup$
– user1729
Jan 29 at 12:08
$begingroup$
(I've added the above comment into the answer.)
$endgroup$
– user1729
Jan 29 at 12:10
$begingroup$
Aiaiai, okay, I see. I assume restricting the generating sets to contain elements of the same order is not sufficient either, although coming up with a counter example might be a bit tricky?
$endgroup$
– elfeck
Jan 29 at 12:10
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1 Answer
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1 Answer
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votes
$begingroup$
(I should state at the start that the ideas in the question are close to pretty standard ideas. All I am doing here is trying to explain the link to these standard ideas.)
There are two things you seem to be missing.
- If $f$ is a homomorphism then its extension to the whole group is not necessarily a bijection. You need to verify that it is injective or surjective (or both if your group is infinite).
- The image group $H$ may have lots of minimal generating sets. So you need to find all minimal generating sets $B_i$ (for speed you can take them up to conjugacy, or up to automorphism). For example, consider $mathbb{Z}_3timesmathbb{Z}_{18}$. This is isomorphic to $mathbb{Z}_6timesmathbb{Z}_9$. One group is generated by an element of order $3$ and an element of order $18$, while the other is generated by an element of order $6$ and an element of order $9$.
The following is then true:
Theorem. Let $A$ be a minimal generating set for $G$, and let ${B_i}$ be the set of all minimal generating sets of $H$ which have cardinality $|A|$. Then $Gcong H$ if and only if there exists a bijection $f_{i, j}: Arightarrow B_i$ which extends to an isomorphism $Grightarrow H$ in the way you describe.
Note that you have to check that the extension is bijective and is a homomorphism. These are non-trivial tasks*. Also, the subscript $j$ is because there are lots of bijections between $A$ and $B_i$, but not all of these will extend to be a homomorphism.
*If $G$ is given by a presentation $langle mathbf{x}midmathbf{r}rangle$ (alternatively this presentation can be computed, but not necessarily quickly!, as your group is finite) then determining if $f$ is a homomorphism is relatively easy, as you just need to verify that $f(R)=_H1$ for all $Rin mathbf{r}$.
answered Jan 29 at 11:58
user1729user1729
17.5k64193
$endgroup$
$begingroup$
Thank you for your answer. So what you are implying that is not even sufficient to fix two generating sets $A$ and $B$ of G and H and check all extensions of all bijections $f: A to B$ for bijectivity and homomorphism? Edit: So there might not even exists a bijective homomorphisms $G to H$ that maps $A$ to $B$.
$endgroup$
– elfeck
Jan 29 at 12:05
1
$begingroup$
Nope! For example, consider $mathbb{Z}_3timesmathbb{Z}_{18}$. This is isomorphic to $mathbb{Z}_6timesmathbb{Z}_9$. One group is generated by an element of order $3$ and an element of order $18$, while the other is generated by an element of order $6$ and an element of order $9$.
$endgroup$
– user1729
Jan 29 at 12:08
$begingroup$
(I've added the above comment into the answer.)
$endgroup$
– user1729
Jan 29 at 12:10
$begingroup$
Aiaiai, okay, I see. I assume restricting the generating sets to contain elements of the same order is not sufficient either, although coming up with a counter example might be a bit tricky?
$endgroup$
– elfeck
Jan 29 at 12:10
add a comment |
$begingroup$
(I should state at the start that the ideas in the question are close to pretty standard ideas. All I am doing here is trying to explain the link to these standard ideas.)
There are two things you seem to be missing.
- If $f$ is a homomorphism then its extension to the whole group is not necessarily a bijection. You need to verify that it is injective or surjective (or both if your group is infinite).
- The image group $H$ may have lots of minimal generating sets. So you need to find all minimal generating sets $B_i$ (for speed you can take them up to conjugacy, or up to automorphism). For example, consider $mathbb{Z}_3timesmathbb{Z}_{18}$. This is isomorphic to $mathbb{Z}_6timesmathbb{Z}_9$. One group is generated by an element of order $3$ and an element of order $18$, while the other is generated by an element of order $6$ and an element of order $9$.
The following is then true:
Theorem. Let $A$ be a minimal generating set for $G$, and let ${B_i}$ be the set of all minimal generating sets of $H$ which have cardinality $|A|$. Then $Gcong H$ if and only if there exists a bijection $f_{i, j}: Arightarrow B_i$ which extends to an isomorphism $Grightarrow H$ in the way you describe.
Note that you have to check that the extension is bijective and is a homomorphism. These are non-trivial tasks*. Also, the subscript $j$ is because there are lots of bijections between $A$ and $B_i$, but not all of these will extend to be a homomorphism.
*If $G$ is given by a presentation $langle mathbf{x}midmathbf{r}rangle$ (alternatively this presentation can be computed, but not necessarily quickly!, as your group is finite) then determining if $f$ is a homomorphism is relatively easy, as you just need to verify that $f(R)=_H1$ for all $Rin mathbf{r}$.
answered Jan 29 at 11:58
user1729user1729
17.5k64193
$endgroup$
$begingroup$
Thank you for your answer. So what you are implying that is not even sufficient to fix two generating sets $A$ and $B$ of G and H and check all extensions of all bijections $f: A to B$ for bijectivity and homomorphism? Edit: So there might not even exists a bijective homomorphisms $G to H$ that maps $A$ to $B$.
$endgroup$
– elfeck
Jan 29 at 12:05
1
$begingroup$
Nope! For example, consider $mathbb{Z}_3timesmathbb{Z}_{18}$. This is isomorphic to $mathbb{Z}_6timesmathbb{Z}_9$. One group is generated by an element of order $3$ and an element of order $18$, while the other is generated by an element of order $6$ and an element of order $9$.
$endgroup$
– user1729
Jan 29 at 12:08
$begingroup$
(I've added the above comment into the answer.)
$endgroup$
– user1729
Jan 29 at 12:10
$begingroup$
Aiaiai, okay, I see. I assume restricting the generating sets to contain elements of the same order is not sufficient either, although coming up with a counter example might be a bit tricky?
$endgroup$
– elfeck
Jan 29 at 12:10
add a comment |
$begingroup$
(I should state at the start that the ideas in the question are close to pretty standard ideas. All I am doing here is trying to explain the link to these standard ideas.)
There are two things you seem to be missing.
- If $f$ is a homomorphism then its extension to the whole group is not necessarily a bijection. You need to verify that it is injective or surjective (or both if your group is infinite).
- The image group $H$ may have lots of minimal generating sets. So you need to find all minimal generating sets $B_i$ (for speed you can take them up to conjugacy, or up to automorphism). For example, consider $mathbb{Z}_3timesmathbb{Z}_{18}$. This is isomorphic to $mathbb{Z}_6timesmathbb{Z}_9$. One group is generated by an element of order $3$ and an element of order $18$, while the other is generated by an element of order $6$ and an element of order $9$.
The following is then true:
Theorem. Let $A$ be a minimal generating set for $G$, and let ${B_i}$ be the set of all minimal generating sets of $H$ which have cardinality $|A|$. Then $Gcong H$ if and only if there exists a bijection $f_{i, j}: Arightarrow B_i$ which extends to an isomorphism $Grightarrow H$ in the way you describe.
Note that you have to check that the extension is bijective and is a homomorphism. These are non-trivial tasks*. Also, the subscript $j$ is because there are lots of bijections between $A$ and $B_i$, but not all of these will extend to be a homomorphism.
*If $G$ is given by a presentation $langle mathbf{x}midmathbf{r}rangle$ (alternatively this presentation can be computed, but not necessarily quickly!, as your group is finite) then determining if $f$ is a homomorphism is relatively easy, as you just need to verify that $f(R)=_H1$ for all $Rin mathbf{r}$.
answered Jan 29 at 11:58
user1729user1729
17.5k64193
$endgroup$
(I should state at the start that the ideas in the question are close to pretty standard ideas. All I am doing here is trying to explain the link to these standard ideas.)
There are two things you seem to be missing.
- If $f$ is a homomorphism then its extension to the whole group is not necessarily a bijection. You need to verify that it is injective or surjective (or both if your group is infinite).
- The image group $H$ may have lots of minimal generating sets. So you need to find all minimal generating sets $B_i$ (for speed you can take them up to conjugacy, or up to automorphism). For example, consider $mathbb{Z}_3timesmathbb{Z}_{18}$. This is isomorphic to $mathbb{Z}_6timesmathbb{Z}_9$. One group is generated by an element of order $3$ and an element of order $18$, while the other is generated by an element of order $6$ and an element of order $9$.
The following is then true:
Theorem. Let $A$ be a minimal generating set for $G$, and let ${B_i}$ be the set of all minimal generating sets of $H$ which have cardinality $|A|$. Then $Gcong H$ if and only if there exists a bijection $f_{i, j}: Arightarrow B_i$ which extends to an isomorphism $Grightarrow H$ in the way you describe.
Note that you have to check that the extension is bijective and is a homomorphism. These are non-trivial tasks*. Also, the subscript $j$ is because there are lots of bijections between $A$ and $B_i$, but not all of these will extend to be a homomorphism.
*If $G$ is given by a presentation $langle mathbf{x}midmathbf{r}rangle$ (alternatively this presentation can be computed, but not necessarily quickly!, as your group is finite) then determining if $f$ is a homomorphism is relatively easy, as you just need to verify that $f(R)=_H1$ for all $Rin mathbf{r}$.
answered Jan 29 at 11:58
user1729user1729
17.5k64193
edited Jan 29 at 12:14
answered Jan 29 at 11:58
user1729user1729
17.5k64193
answered Jan 29 at 11:58
user1729user1729
17.5k64193
answered Jan 29 at 11:58
user1729user1729
17.5k64193
17.5k64193
$begingroup$
Thank you for your answer. So what you are implying that is not even sufficient to fix two generating sets $A$ and $B$ of G and H and check all extensions of all bijections $f: A to B$ for bijectivity and homomorphism? Edit: So there might not even exists a bijective homomorphisms $G to H$ that maps $A$ to $B$.
$endgroup$
– elfeck
Jan 29 at 12:05
1
$begingroup$
Nope! For example, consider $mathbb{Z}_3timesmathbb{Z}_{18}$. This is isomorphic to $mathbb{Z}_6timesmathbb{Z}_9$. One group is generated by an element of order $3$ and an element of order $18$, while the other is generated by an element of order $6$ and an element of order $9$.
$endgroup$
– user1729
Jan 29 at 12:08
$begingroup$
(I've added the above comment into the answer.)
$endgroup$
– user1729
Jan 29 at 12:10
$begingroup$
Aiaiai, okay, I see. I assume restricting the generating sets to contain elements of the same order is not sufficient either, although coming up with a counter example might be a bit tricky?
$endgroup$
– elfeck
Jan 29 at 12:10
add a comment |
$begingroup$
Thank you for your answer. So what you are implying that is not even sufficient to fix two generating sets $A$ and $B$ of G and H and check all extensions of all bijections $f: A to B$ for bijectivity and homomorphism? Edit: So there might not even exists a bijective homomorphisms $G to H$ that maps $A$ to $B$.
$endgroup$
– elfeck
Jan 29 at 12:05
1
$begingroup$
Nope! For example, consider $mathbb{Z}_3timesmathbb{Z}_{18}$. This is isomorphic to $mathbb{Z}_6timesmathbb{Z}_9$. One group is generated by an element of order $3$ and an element of order $18$, while the other is generated by an element of order $6$ and an element of order $9$.
$endgroup$
– user1729
Jan 29 at 12:08
$begingroup$
(I've added the above comment into the answer.)
$endgroup$
– user1729
Jan 29 at 12:10
$begingroup$
Aiaiai, okay, I see. I assume restricting the generating sets to contain elements of the same order is not sufficient either, although coming up with a counter example might be a bit tricky?
$endgroup$
– elfeck
Jan 29 at 12:10
$begingroup$
Thank you for your answer. So what you are implying that is not even sufficient to fix two generating sets $A$ and $B$ of G and H and check all extensions of all bijections $f: A to B$ for bijectivity and homomorphism? Edit: So there might not even exists a bijective homomorphisms $G to H$ that maps $A$ to $B$.
$endgroup$
– elfeck
Jan 29 at 12:05
$begingroup$
Thank you for your answer. So what you are implying that is not even sufficient to fix two generating sets $A$ and $B$ of G and H and check all extensions of all bijections $f: A to B$ for bijectivity and homomorphism? Edit: So there might not even exists a bijective homomorphisms $G to H$ that maps $A$ to $B$.
$endgroup$
– elfeck
Jan 29 at 12:05
1
1
$begingroup$
Nope! For example, consider $mathbb{Z}_3timesmathbb{Z}_{18}$. This is isomorphic to $mathbb{Z}_6timesmathbb{Z}_9$. One group is generated by an element of order $3$ and an element of order $18$, while the other is generated by an element of order $6$ and an element of order $9$.
$endgroup$
– user1729
Jan 29 at 12:08
$begingroup$
Nope! For example, consider $mathbb{Z}_3timesmathbb{Z}_{18}$. This is isomorphic to $mathbb{Z}_6timesmathbb{Z}_9$. One group is generated by an element of order $3$ and an element of order $18$, while the other is generated by an element of order $6$ and an element of order $9$.
$endgroup$
– user1729
Jan 29 at 12:08
$begingroup$
(I've added the above comment into the answer.)
$endgroup$
– user1729
Jan 29 at 12:10
$begingroup$
(I've added the above comment into the answer.)
$endgroup$
– user1729
Jan 29 at 12:10
$begingroup$
Aiaiai, okay, I see. I assume restricting the generating sets to contain elements of the same order is not sufficient either, although coming up with a counter example might be a bit tricky?
$endgroup$
– elfeck
Jan 29 at 12:10
$begingroup$
Aiaiai, okay, I see. I assume restricting the generating sets to contain elements of the same order is not sufficient either, although coming up with a counter example might be a bit tricky?
$endgroup$
– elfeck
Jan 29 at 12:10
add a comment |
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$begingroup$
Are you assuming your groups are abelian? How do you know that that the extension of $f$ that you define is independent of the choice of decomposition of $g$? If, say, $a_1$ has order $3$ and $f(a_1)=b_1$ has order $2$ then $f(a_1^2)=b_1^2=e_H$ so your function isn't even a set theoretic bijection.
$endgroup$
– lulu
Jan 29 at 11:38
$begingroup$
No I don't assume commutativity. Good point, I had not thought about that. Is there a way to rescue this?
$endgroup$
– elfeck
Jan 29 at 11:41
$begingroup$
Well, I don't know. It's hard to say much about a group just knowing a generating set.
$endgroup$
– lulu
Jan 29 at 11:43
$begingroup$
If the extension of $f$ were an isomorphism, then it must also be independent of the decomposition of any $g$. So perhaps it is not necessary to require (a-priori) that the extension of $f$ is bijective
$endgroup$
– elfeck
Jan 29 at 11:45
$begingroup$
You haven't even defined an extension of $f$, your definition depends on that independence.
$endgroup$
– lulu
Jan 29 at 11:46