Mixing
Basics on group operation
$begingroup$
Let's $G$ be a group and $H le G$. $H$ brings a partition of $G$, $lambda_H(G)=(H,complement_GH$), where $complement_GH:=G setminus H$. Firstly, I notice that, as $g in G Leftrightarrow g^{-1} in G$, then also $g in complement_GH Leftrightarrow g^{-1} in complement_GH$.
Now, I claim that $g in H, g' in complement_GH Rightarrow gg' in complement_GH$. In fact, $gg' in H Rightarrow g'=g^{-1}(gg') in H$: contradiction. Then, $gg' in complement_GH$.
So, denoted by $f colon G times G rightarrow G$ the group operation, we get:
- $f(G times G) subseteq G$
- $f(H times H) subseteq H$
- $f(H times complement_GH) subseteq complement_GH$
- $f(complement_GH times H) subseteq complement_GH$
Finally, I claim that $f(complement_GH times complement_GH) cap H ne emptyset$. In fact, differently, we'd have that $f(complement_GH times complement_GH) subseteq complement_GH$ and then $complement_GH le G$: contradiction, because $e notin complement_GH$.
Is this whole correct?
abstract-algebra group-theory
asked Jan 3 at 8:49
LucaLuca
1288
$endgroup$
add a comment |
$begingroup$
Let's $G$ be a group and $H le G$. $H$ brings a partition of $G$, $lambda_H(G)=(H,complement_GH$), where $complement_GH:=G setminus H$. Firstly, I notice that, as $g in G Leftrightarrow g^{-1} in G$, then also $g in complement_GH Leftrightarrow g^{-1} in complement_GH$.
Now, I claim that $g in H, g' in complement_GH Rightarrow gg' in complement_GH$. In fact, $gg' in H Rightarrow g'=g^{-1}(gg') in H$: contradiction. Then, $gg' in complement_GH$.
So, denoted by $f colon G times G rightarrow G$ the group operation, we get:
- $f(G times G) subseteq G$
- $f(H times H) subseteq H$
- $f(H times complement_GH) subseteq complement_GH$
- $f(complement_GH times H) subseteq complement_GH$
Finally, I claim that $f(complement_GH times complement_GH) cap H ne emptyset$. In fact, differently, we'd have that $f(complement_GH times complement_GH) subseteq complement_GH$ and then $complement_GH le G$: contradiction, because $e notin complement_GH$.
Is this whole correct?
abstract-algebra group-theory
asked Jan 3 at 8:49
LucaLuca
1288
$endgroup$
2
$begingroup$
Yes, this all looks fine, though it is somewhat more complicated than it needs to be. $G$ is the union of cosets of $H$ and acts on these with $H$ being the stabilizer of itself, so $H$ acts on the remaining cosets, which is your complement.
$endgroup$
– Tobias Kildetoft
Jan 3 at 9:05
$begingroup$
Thank you Tobias. I'll come back to your comment when I'll be more acquainted with concepts of higher-level than current mine, like action, stabilizer, etc.
$endgroup$
– Luca
Jan 3 at 9:11
add a comment |
$begingroup$
Let's $G$ be a group and $H le G$. $H$ brings a partition of $G$, $lambda_H(G)=(H,complement_GH$), where $complement_GH:=G setminus H$. Firstly, I notice that, as $g in G Leftrightarrow g^{-1} in G$, then also $g in complement_GH Leftrightarrow g^{-1} in complement_GH$.
Now, I claim that $g in H, g' in complement_GH Rightarrow gg' in complement_GH$. In fact, $gg' in H Rightarrow g'=g^{-1}(gg') in H$: contradiction. Then, $gg' in complement_GH$.
So, denoted by $f colon G times G rightarrow G$ the group operation, we get:
- $f(G times G) subseteq G$
- $f(H times H) subseteq H$
- $f(H times complement_GH) subseteq complement_GH$
- $f(complement_GH times H) subseteq complement_GH$
Finally, I claim that $f(complement_GH times complement_GH) cap H ne emptyset$. In fact, differently, we'd have that $f(complement_GH times complement_GH) subseteq complement_GH$ and then $complement_GH le G$: contradiction, because $e notin complement_GH$.
Is this whole correct?
abstract-algebra group-theory
asked Jan 3 at 8:49
LucaLuca
1288
$endgroup$
Let's $G$ be a group and $H le G$. $H$ brings a partition of $G$, $lambda_H(G)=(H,complement_GH$), where $complement_GH:=G setminus H$. Firstly, I notice that, as $g in G Leftrightarrow g^{-1} in G$, then also $g in complement_GH Leftrightarrow g^{-1} in complement_GH$.
Now, I claim that $g in H, g' in complement_GH Rightarrow gg' in complement_GH$. In fact, $gg' in H Rightarrow g'=g^{-1}(gg') in H$: contradiction. Then, $gg' in complement_GH$.
So, denoted by $f colon G times G rightarrow G$ the group operation, we get:
- $f(G times G) subseteq G$
- $f(H times H) subseteq H$
- $f(H times complement_GH) subseteq complement_GH$
- $f(complement_GH times H) subseteq complement_GH$
Finally, I claim that $f(complement_GH times complement_GH) cap H ne emptyset$. In fact, differently, we'd have that $f(complement_GH times complement_GH) subseteq complement_GH$ and then $complement_GH le G$: contradiction, because $e notin complement_GH$.
Is this whole correct?
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 3 at 8:49
LucaLuca
1288
asked Jan 3 at 8:49
LucaLuca
1288
asked Jan 3 at 8:49
LucaLuca
1288
asked Jan 3 at 8:49
LucaLuca
1288
asked Jan 3 at 8:49
LucaLuca
1288
1288
2
$begingroup$
Yes, this all looks fine, though it is somewhat more complicated than it needs to be. $G$ is the union of cosets of $H$ and acts on these with $H$ being the stabilizer of itself, so $H$ acts on the remaining cosets, which is your complement.
$endgroup$
– Tobias Kildetoft
Jan 3 at 9:05
$begingroup$
Thank you Tobias. I'll come back to your comment when I'll be more acquainted with concepts of higher-level than current mine, like action, stabilizer, etc.
$endgroup$
– Luca
Jan 3 at 9:11
add a comment |
2
$begingroup$
Yes, this all looks fine, though it is somewhat more complicated than it needs to be. $G$ is the union of cosets of $H$ and acts on these with $H$ being the stabilizer of itself, so $H$ acts on the remaining cosets, which is your complement.
$endgroup$
– Tobias Kildetoft
Jan 3 at 9:05
$begingroup$
Thank you Tobias. I'll come back to your comment when I'll be more acquainted with concepts of higher-level than current mine, like action, stabilizer, etc.
$endgroup$
– Luca
Jan 3 at 9:11
2
2
$begingroup$
Yes, this all looks fine, though it is somewhat more complicated than it needs to be. $G$ is the union of cosets of $H$ and acts on these with $H$ being the stabilizer of itself, so $H$ acts on the remaining cosets, which is your complement.
$endgroup$
– Tobias Kildetoft
Jan 3 at 9:05
$begingroup$
Yes, this all looks fine, though it is somewhat more complicated than it needs to be. $G$ is the union of cosets of $H$ and acts on these with $H$ being the stabilizer of itself, so $H$ acts on the remaining cosets, which is your complement.
$endgroup$
– Tobias Kildetoft
Jan 3 at 9:05
$begingroup$
Thank you Tobias. I'll come back to your comment when I'll be more acquainted with concepts of higher-level than current mine, like action, stabilizer, etc.
$endgroup$
– Luca
Jan 3 at 9:11
$begingroup$
Thank you Tobias. I'll come back to your comment when I'll be more acquainted with concepts of higher-level than current mine, like action, stabilizer, etc.
$endgroup$
– Luca
Jan 3 at 9:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is not really clear what you are asking about. Though, your reasoning is correct.
But, note that, you cannot say that $H$ partitions the group $G$. It would be saying that it is a set's property to partition any other sets/groups/categories etc. in which it is included. You see, having a complement is not what we call partitioning, even if it seems similar. With complements, it is already too trivial! For example, you could omit all your proof between second line and the before last one: As $H$ is a subgroup of $G$, thus the identity element of $G$ is included in $H$, thus no way it can be for the complement of $H$; so that we conclude the complement can never be a subgroup of $G$.
But the properties you showed would still remain valid.
Except that:
When you tried to proceed by contradiction at the last line, you began with a logically false claim. The proof holds on itself. But it is not how it is done. We prefer(!) beginning with a logically valid statement which doesn't hold with the given propositional phrase, then we try to arrive either to a logically false conclusion or any conclusion which does not hold with predicate to prove a contradiction.
I surely think that $ f( bar{H} times bar{H}) subseteq bar{H} $ cannot imply that $bar{H}$ is a subgroup of $G$ too. (where $bar{H}$ is the complement). It only shows an inclusion; and nothing about a property at all.
answered Jan 3 at 14:51
freehumoristfreehumorist
174112
$endgroup$
$begingroup$
My aim was to prove that $f(complement_GH times complement_GH) subseteq complement_GH$ is false; along the proof, I used the fact that $complement_GH$ is not a subgroup of $G$, as can be straightforwardly obtained as you said. Since I have pretty no basis in logics, I'd really appreaciate a logically flawless version of my proof. Thanks
$endgroup$
– Luca
Jan 3 at 15:23
$begingroup$
Well, and actually -given an $H le G$- I wished to make an "inventory" of $f(dots times dots)$ in terms of $H$ and $complement_GH$
$endgroup$
– Luca
Jan 3 at 15:30
$begingroup$
You should, on this forum, mention explicitly your goal. Otherwise the answers tend to vary. Said you that you wanted a detailed "logical argumentation" as if exercise, I'd do it already.
$endgroup$
– freehumorist
Jan 3 at 16:43
add a comment |
$begingroup$
The inclusion relation is what we call an ordering. It is, in fact, a linear ordering in this case. A linear ordering is a partial ordering in which the comparaison quality is well defined for every element pairwise distinct. Now, I will initiate you to a higher theorem:
("higher" in the sense that, here, we don't make use of it within its usual definition for limits; but for a different metric, ie. topological space with different ordering, but which holds well because they are isomorphic, ie. homoemorphic.)
So, you can use here the Sandwich Theorem. You see, the image of complements is ordered by usual set inclusion between the complement of $G$ and the $H$ itself. Though we know that we cannot have any element outside $G$. But we cannot have any in $H$ neither: Otherwise we would have images of $H$ and its complement intertwining. And as $f$ is a homomorphism of groups, it cannot be, thus contradiction.
You see, with the reasoning I told above, adapting it to different cases is only up to your redaction and comprehension. It becomes free of your choice for underlying set (ie. $Dom(f)$ here). You just need: 1. the linear ordering of inclusion (which you surely have here -unless in higher mathematics), 2. the function $f$ to be homomorphism of groups.
answered Jan 3 at 16:46
freehumoristfreehumorist
174112
$endgroup$
$begingroup$
I have a very hard time following this answer. For example, $f$ is certainly not going to be a group homomorphism in most cases.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:00
$begingroup$
@TobiasKildetoft the trouble with your comprehension is the same reason I supposed $f$ homomorphic here. You are right to say not every $f$ is a morphism, but we have a stranger case here: $f$ is the internal law. So in fact, the question would be nothing but empty as it is. Luca , can you review your question? if $f$ is a function defined on $(G,times)$ than we have interest in the case only when it is homomorphism. otherwise no conclusion.
$endgroup$
– freehumorist
Jan 3 at 17:19
$begingroup$
The $f$ defined in the question is completely fine and makes sense. It defines an action of $G$ on itself. The fact that it will only be a homomorphism when the group is abelian is irrelevant to the actual question.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:22
$begingroup$
who said about abelian? My reasoning holds for $G$ monoid
$endgroup$
– freehumorist
Jan 3 at 17:24
$begingroup$
I have no idea what your "reasoning" is. The map $f$ in question is a homomorphism iff $G$ is abelian, and you seemed to take it for granted that it was a homomorphism.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:26
|
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is not really clear what you are asking about. Though, your reasoning is correct.
But, note that, you cannot say that $H$ partitions the group $G$. It would be saying that it is a set's property to partition any other sets/groups/categories etc. in which it is included. You see, having a complement is not what we call partitioning, even if it seems similar. With complements, it is already too trivial! For example, you could omit all your proof between second line and the before last one: As $H$ is a subgroup of $G$, thus the identity element of $G$ is included in $H$, thus no way it can be for the complement of $H$; so that we conclude the complement can never be a subgroup of $G$.
But the properties you showed would still remain valid.
Except that:
When you tried to proceed by contradiction at the last line, you began with a logically false claim. The proof holds on itself. But it is not how it is done. We prefer(!) beginning with a logically valid statement which doesn't hold with the given propositional phrase, then we try to arrive either to a logically false conclusion or any conclusion which does not hold with predicate to prove a contradiction.
I surely think that $ f( bar{H} times bar{H}) subseteq bar{H} $ cannot imply that $bar{H}$ is a subgroup of $G$ too. (where $bar{H}$ is the complement). It only shows an inclusion; and nothing about a property at all.
answered Jan 3 at 14:51
freehumoristfreehumorist
174112
$endgroup$
$begingroup$
My aim was to prove that $f(complement_GH times complement_GH) subseteq complement_GH$ is false; along the proof, I used the fact that $complement_GH$ is not a subgroup of $G$, as can be straightforwardly obtained as you said. Since I have pretty no basis in logics, I'd really appreaciate a logically flawless version of my proof. Thanks
$endgroup$
– Luca
Jan 3 at 15:23
$begingroup$
Well, and actually -given an $H le G$- I wished to make an "inventory" of $f(dots times dots)$ in terms of $H$ and $complement_GH$
$endgroup$
– Luca
Jan 3 at 15:30
$begingroup$
You should, on this forum, mention explicitly your goal. Otherwise the answers tend to vary. Said you that you wanted a detailed "logical argumentation" as if exercise, I'd do it already.
$endgroup$
– freehumorist
Jan 3 at 16:43
add a comment |
$begingroup$
It is not really clear what you are asking about. Though, your reasoning is correct.
But, note that, you cannot say that $H$ partitions the group $G$. It would be saying that it is a set's property to partition any other sets/groups/categories etc. in which it is included. You see, having a complement is not what we call partitioning, even if it seems similar. With complements, it is already too trivial! For example, you could omit all your proof between second line and the before last one: As $H$ is a subgroup of $G$, thus the identity element of $G$ is included in $H$, thus no way it can be for the complement of $H$; so that we conclude the complement can never be a subgroup of $G$.
But the properties you showed would still remain valid.
Except that:
When you tried to proceed by contradiction at the last line, you began with a logically false claim. The proof holds on itself. But it is not how it is done. We prefer(!) beginning with a logically valid statement which doesn't hold with the given propositional phrase, then we try to arrive either to a logically false conclusion or any conclusion which does not hold with predicate to prove a contradiction.
I surely think that $ f( bar{H} times bar{H}) subseteq bar{H} $ cannot imply that $bar{H}$ is a subgroup of $G$ too. (where $bar{H}$ is the complement). It only shows an inclusion; and nothing about a property at all.
answered Jan 3 at 14:51
freehumoristfreehumorist
174112
$endgroup$
$begingroup$
My aim was to prove that $f(complement_GH times complement_GH) subseteq complement_GH$ is false; along the proof, I used the fact that $complement_GH$ is not a subgroup of $G$, as can be straightforwardly obtained as you said. Since I have pretty no basis in logics, I'd really appreaciate a logically flawless version of my proof. Thanks
$endgroup$
– Luca
Jan 3 at 15:23
$begingroup$
Well, and actually -given an $H le G$- I wished to make an "inventory" of $f(dots times dots)$ in terms of $H$ and $complement_GH$
$endgroup$
– Luca
Jan 3 at 15:30
$begingroup$
You should, on this forum, mention explicitly your goal. Otherwise the answers tend to vary. Said you that you wanted a detailed "logical argumentation" as if exercise, I'd do it already.
$endgroup$
– freehumorist
Jan 3 at 16:43
add a comment |
$begingroup$
It is not really clear what you are asking about. Though, your reasoning is correct.
But, note that, you cannot say that $H$ partitions the group $G$. It would be saying that it is a set's property to partition any other sets/groups/categories etc. in which it is included. You see, having a complement is not what we call partitioning, even if it seems similar. With complements, it is already too trivial! For example, you could omit all your proof between second line and the before last one: As $H$ is a subgroup of $G$, thus the identity element of $G$ is included in $H$, thus no way it can be for the complement of $H$; so that we conclude the complement can never be a subgroup of $G$.
But the properties you showed would still remain valid.
Except that:
When you tried to proceed by contradiction at the last line, you began with a logically false claim. The proof holds on itself. But it is not how it is done. We prefer(!) beginning with a logically valid statement which doesn't hold with the given propositional phrase, then we try to arrive either to a logically false conclusion or any conclusion which does not hold with predicate to prove a contradiction.
I surely think that $ f( bar{H} times bar{H}) subseteq bar{H} $ cannot imply that $bar{H}$ is a subgroup of $G$ too. (where $bar{H}$ is the complement). It only shows an inclusion; and nothing about a property at all.
answered Jan 3 at 14:51
freehumoristfreehumorist
174112
$endgroup$
It is not really clear what you are asking about. Though, your reasoning is correct.
But, note that, you cannot say that $H$ partitions the group $G$. It would be saying that it is a set's property to partition any other sets/groups/categories etc. in which it is included. You see, having a complement is not what we call partitioning, even if it seems similar. With complements, it is already too trivial! For example, you could omit all your proof between second line and the before last one: As $H$ is a subgroup of $G$, thus the identity element of $G$ is included in $H$, thus no way it can be for the complement of $H$; so that we conclude the complement can never be a subgroup of $G$.
But the properties you showed would still remain valid.
Except that:
When you tried to proceed by contradiction at the last line, you began with a logically false claim. The proof holds on itself. But it is not how it is done. We prefer(!) beginning with a logically valid statement which doesn't hold with the given propositional phrase, then we try to arrive either to a logically false conclusion or any conclusion which does not hold with predicate to prove a contradiction.
I surely think that $ f( bar{H} times bar{H}) subseteq bar{H} $ cannot imply that $bar{H}$ is a subgroup of $G$ too. (where $bar{H}$ is the complement). It only shows an inclusion; and nothing about a property at all.
answered Jan 3 at 14:51
freehumoristfreehumorist
174112
answered Jan 3 at 14:51
freehumoristfreehumorist
174112
answered Jan 3 at 14:51
freehumoristfreehumorist
174112
answered Jan 3 at 14:51
freehumoristfreehumorist
174112
174112
$begingroup$
My aim was to prove that $f(complement_GH times complement_GH) subseteq complement_GH$ is false; along the proof, I used the fact that $complement_GH$ is not a subgroup of $G$, as can be straightforwardly obtained as you said. Since I have pretty no basis in logics, I'd really appreaciate a logically flawless version of my proof. Thanks
$endgroup$
– Luca
Jan 3 at 15:23
$begingroup$
Well, and actually -given an $H le G$- I wished to make an "inventory" of $f(dots times dots)$ in terms of $H$ and $complement_GH$
$endgroup$
– Luca
Jan 3 at 15:30
$begingroup$
You should, on this forum, mention explicitly your goal. Otherwise the answers tend to vary. Said you that you wanted a detailed "logical argumentation" as if exercise, I'd do it already.
$endgroup$
– freehumorist
Jan 3 at 16:43
add a comment |
$begingroup$
My aim was to prove that $f(complement_GH times complement_GH) subseteq complement_GH$ is false; along the proof, I used the fact that $complement_GH$ is not a subgroup of $G$, as can be straightforwardly obtained as you said. Since I have pretty no basis in logics, I'd really appreaciate a logically flawless version of my proof. Thanks
$endgroup$
– Luca
Jan 3 at 15:23
$begingroup$
Well, and actually -given an $H le G$- I wished to make an "inventory" of $f(dots times dots)$ in terms of $H$ and $complement_GH$
$endgroup$
– Luca
Jan 3 at 15:30
$begingroup$
You should, on this forum, mention explicitly your goal. Otherwise the answers tend to vary. Said you that you wanted a detailed "logical argumentation" as if exercise, I'd do it already.
$endgroup$
– freehumorist
Jan 3 at 16:43
$begingroup$
My aim was to prove that $f(complement_GH times complement_GH) subseteq complement_GH$ is false; along the proof, I used the fact that $complement_GH$ is not a subgroup of $G$, as can be straightforwardly obtained as you said. Since I have pretty no basis in logics, I'd really appreaciate a logically flawless version of my proof. Thanks
$endgroup$
– Luca
Jan 3 at 15:23
$begingroup$
My aim was to prove that $f(complement_GH times complement_GH) subseteq complement_GH$ is false; along the proof, I used the fact that $complement_GH$ is not a subgroup of $G$, as can be straightforwardly obtained as you said. Since I have pretty no basis in logics, I'd really appreaciate a logically flawless version of my proof. Thanks
$endgroup$
– Luca
Jan 3 at 15:23
$begingroup$
Well, and actually -given an $H le G$- I wished to make an "inventory" of $f(dots times dots)$ in terms of $H$ and $complement_GH$
$endgroup$
– Luca
Jan 3 at 15:30
$begingroup$
Well, and actually -given an $H le G$- I wished to make an "inventory" of $f(dots times dots)$ in terms of $H$ and $complement_GH$
$endgroup$
– Luca
Jan 3 at 15:30
$begingroup$
You should, on this forum, mention explicitly your goal. Otherwise the answers tend to vary. Said you that you wanted a detailed "logical argumentation" as if exercise, I'd do it already.
$endgroup$
– freehumorist
Jan 3 at 16:43
$begingroup$
You should, on this forum, mention explicitly your goal. Otherwise the answers tend to vary. Said you that you wanted a detailed "logical argumentation" as if exercise, I'd do it already.
$endgroup$
– freehumorist
Jan 3 at 16:43
add a comment |
$begingroup$
The inclusion relation is what we call an ordering. It is, in fact, a linear ordering in this case. A linear ordering is a partial ordering in which the comparaison quality is well defined for every element pairwise distinct. Now, I will initiate you to a higher theorem:
("higher" in the sense that, here, we don't make use of it within its usual definition for limits; but for a different metric, ie. topological space with different ordering, but which holds well because they are isomorphic, ie. homoemorphic.)
So, you can use here the Sandwich Theorem. You see, the image of complements is ordered by usual set inclusion between the complement of $G$ and the $H$ itself. Though we know that we cannot have any element outside $G$. But we cannot have any in $H$ neither: Otherwise we would have images of $H$ and its complement intertwining. And as $f$ is a homomorphism of groups, it cannot be, thus contradiction.
You see, with the reasoning I told above, adapting it to different cases is only up to your redaction and comprehension. It becomes free of your choice for underlying set (ie. $Dom(f)$ here). You just need: 1. the linear ordering of inclusion (which you surely have here -unless in higher mathematics), 2. the function $f$ to be homomorphism of groups.
answered Jan 3 at 16:46
freehumoristfreehumorist
174112
$endgroup$
$begingroup$
I have a very hard time following this answer. For example, $f$ is certainly not going to be a group homomorphism in most cases.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:00
$begingroup$
@TobiasKildetoft the trouble with your comprehension is the same reason I supposed $f$ homomorphic here. You are right to say not every $f$ is a morphism, but we have a stranger case here: $f$ is the internal law. So in fact, the question would be nothing but empty as it is. Luca , can you review your question? if $f$ is a function defined on $(G,times)$ than we have interest in the case only when it is homomorphism. otherwise no conclusion.
$endgroup$
– freehumorist
Jan 3 at 17:19
$begingroup$
The $f$ defined in the question is completely fine and makes sense. It defines an action of $G$ on itself. The fact that it will only be a homomorphism when the group is abelian is irrelevant to the actual question.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:22
$begingroup$
who said about abelian? My reasoning holds for $G$ monoid
$endgroup$
– freehumorist
Jan 3 at 17:24
$begingroup$
I have no idea what your "reasoning" is. The map $f$ in question is a homomorphism iff $G$ is abelian, and you seemed to take it for granted that it was a homomorphism.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:26
|
show 1 more comment
$begingroup$
The inclusion relation is what we call an ordering. It is, in fact, a linear ordering in this case. A linear ordering is a partial ordering in which the comparaison quality is well defined for every element pairwise distinct. Now, I will initiate you to a higher theorem:
("higher" in the sense that, here, we don't make use of it within its usual definition for limits; but for a different metric, ie. topological space with different ordering, but which holds well because they are isomorphic, ie. homoemorphic.)
So, you can use here the Sandwich Theorem. You see, the image of complements is ordered by usual set inclusion between the complement of $G$ and the $H$ itself. Though we know that we cannot have any element outside $G$. But we cannot have any in $H$ neither: Otherwise we would have images of $H$ and its complement intertwining. And as $f$ is a homomorphism of groups, it cannot be, thus contradiction.
You see, with the reasoning I told above, adapting it to different cases is only up to your redaction and comprehension. It becomes free of your choice for underlying set (ie. $Dom(f)$ here). You just need: 1. the linear ordering of inclusion (which you surely have here -unless in higher mathematics), 2. the function $f$ to be homomorphism of groups.
answered Jan 3 at 16:46
freehumoristfreehumorist
174112
$endgroup$
$begingroup$
I have a very hard time following this answer. For example, $f$ is certainly not going to be a group homomorphism in most cases.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:00
$begingroup$
@TobiasKildetoft the trouble with your comprehension is the same reason I supposed $f$ homomorphic here. You are right to say not every $f$ is a morphism, but we have a stranger case here: $f$ is the internal law. So in fact, the question would be nothing but empty as it is. Luca , can you review your question? if $f$ is a function defined on $(G,times)$ than we have interest in the case only when it is homomorphism. otherwise no conclusion.
$endgroup$
– freehumorist
Jan 3 at 17:19
$begingroup$
The $f$ defined in the question is completely fine and makes sense. It defines an action of $G$ on itself. The fact that it will only be a homomorphism when the group is abelian is irrelevant to the actual question.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:22
$begingroup$
who said about abelian? My reasoning holds for $G$ monoid
$endgroup$
– freehumorist
Jan 3 at 17:24
$begingroup$
I have no idea what your "reasoning" is. The map $f$ in question is a homomorphism iff $G$ is abelian, and you seemed to take it for granted that it was a homomorphism.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:26
|
show 1 more comment
$begingroup$
The inclusion relation is what we call an ordering. It is, in fact, a linear ordering in this case. A linear ordering is a partial ordering in which the comparaison quality is well defined for every element pairwise distinct. Now, I will initiate you to a higher theorem:
("higher" in the sense that, here, we don't make use of it within its usual definition for limits; but for a different metric, ie. topological space with different ordering, but which holds well because they are isomorphic, ie. homoemorphic.)
So, you can use here the Sandwich Theorem. You see, the image of complements is ordered by usual set inclusion between the complement of $G$ and the $H$ itself. Though we know that we cannot have any element outside $G$. But we cannot have any in $H$ neither: Otherwise we would have images of $H$ and its complement intertwining. And as $f$ is a homomorphism of groups, it cannot be, thus contradiction.
You see, with the reasoning I told above, adapting it to different cases is only up to your redaction and comprehension. It becomes free of your choice for underlying set (ie. $Dom(f)$ here). You just need: 1. the linear ordering of inclusion (which you surely have here -unless in higher mathematics), 2. the function $f$ to be homomorphism of groups.
answered Jan 3 at 16:46
freehumoristfreehumorist
174112
$endgroup$
The inclusion relation is what we call an ordering. It is, in fact, a linear ordering in this case. A linear ordering is a partial ordering in which the comparaison quality is well defined for every element pairwise distinct. Now, I will initiate you to a higher theorem:
("higher" in the sense that, here, we don't make use of it within its usual definition for limits; but for a different metric, ie. topological space with different ordering, but which holds well because they are isomorphic, ie. homoemorphic.)
So, you can use here the Sandwich Theorem. You see, the image of complements is ordered by usual set inclusion between the complement of $G$ and the $H$ itself. Though we know that we cannot have any element outside $G$. But we cannot have any in $H$ neither: Otherwise we would have images of $H$ and its complement intertwining. And as $f$ is a homomorphism of groups, it cannot be, thus contradiction.
You see, with the reasoning I told above, adapting it to different cases is only up to your redaction and comprehension. It becomes free of your choice for underlying set (ie. $Dom(f)$ here). You just need: 1. the linear ordering of inclusion (which you surely have here -unless in higher mathematics), 2. the function $f$ to be homomorphism of groups.
answered Jan 3 at 16:46
freehumoristfreehumorist
174112
edited Jan 3 at 16:51
answered Jan 3 at 16:46
freehumoristfreehumorist
174112
answered Jan 3 at 16:46
freehumoristfreehumorist
174112
answered Jan 3 at 16:46
freehumoristfreehumorist
174112
174112
$begingroup$
I have a very hard time following this answer. For example, $f$ is certainly not going to be a group homomorphism in most cases.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:00
$begingroup$
@TobiasKildetoft the trouble with your comprehension is the same reason I supposed $f$ homomorphic here. You are right to say not every $f$ is a morphism, but we have a stranger case here: $f$ is the internal law. So in fact, the question would be nothing but empty as it is. Luca , can you review your question? if $f$ is a function defined on $(G,times)$ than we have interest in the case only when it is homomorphism. otherwise no conclusion.
$endgroup$
– freehumorist
Jan 3 at 17:19
$begingroup$
The $f$ defined in the question is completely fine and makes sense. It defines an action of $G$ on itself. The fact that it will only be a homomorphism when the group is abelian is irrelevant to the actual question.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:22
$begingroup$
who said about abelian? My reasoning holds for $G$ monoid
$endgroup$
– freehumorist
Jan 3 at 17:24
$begingroup$
I have no idea what your "reasoning" is. The map $f$ in question is a homomorphism iff $G$ is abelian, and you seemed to take it for granted that it was a homomorphism.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:26
|
show 1 more comment
$begingroup$
I have a very hard time following this answer. For example, $f$ is certainly not going to be a group homomorphism in most cases.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:00
$begingroup$
@TobiasKildetoft the trouble with your comprehension is the same reason I supposed $f$ homomorphic here. You are right to say not every $f$ is a morphism, but we have a stranger case here: $f$ is the internal law. So in fact, the question would be nothing but empty as it is. Luca , can you review your question? if $f$ is a function defined on $(G,times)$ than we have interest in the case only when it is homomorphism. otherwise no conclusion.
$endgroup$
– freehumorist
Jan 3 at 17:19
$begingroup$
The $f$ defined in the question is completely fine and makes sense. It defines an action of $G$ on itself. The fact that it will only be a homomorphism when the group is abelian is irrelevant to the actual question.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:22
$begingroup$
who said about abelian? My reasoning holds for $G$ monoid
$endgroup$
– freehumorist
Jan 3 at 17:24
$begingroup$
I have no idea what your "reasoning" is. The map $f$ in question is a homomorphism iff $G$ is abelian, and you seemed to take it for granted that it was a homomorphism.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:26
$begingroup$
I have a very hard time following this answer. For example, $f$ is certainly not going to be a group homomorphism in most cases.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:00
$begingroup$
I have a very hard time following this answer. For example, $f$ is certainly not going to be a group homomorphism in most cases.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:00
$begingroup$
@TobiasKildetoft the trouble with your comprehension is the same reason I supposed $f$ homomorphic here. You are right to say not every $f$ is a morphism, but we have a stranger case here: $f$ is the internal law. So in fact, the question would be nothing but empty as it is. Luca , can you review your question? if $f$ is a function defined on $(G,times)$ than we have interest in the case only when it is homomorphism. otherwise no conclusion.
$endgroup$
– freehumorist
Jan 3 at 17:19
$begingroup$
@TobiasKildetoft the trouble with your comprehension is the same reason I supposed $f$ homomorphic here. You are right to say not every $f$ is a morphism, but we have a stranger case here: $f$ is the internal law. So in fact, the question would be nothing but empty as it is. Luca , can you review your question? if $f$ is a function defined on $(G,times)$ than we have interest in the case only when it is homomorphism. otherwise no conclusion.
$endgroup$
– freehumorist
Jan 3 at 17:19
$begingroup$
The $f$ defined in the question is completely fine and makes sense. It defines an action of $G$ on itself. The fact that it will only be a homomorphism when the group is abelian is irrelevant to the actual question.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:22
$begingroup$
The $f$ defined in the question is completely fine and makes sense. It defines an action of $G$ on itself. The fact that it will only be a homomorphism when the group is abelian is irrelevant to the actual question.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:22
$begingroup$
who said about abelian? My reasoning holds for $G$ monoid
$endgroup$
– freehumorist
Jan 3 at 17:24
$begingroup$
who said about abelian? My reasoning holds for $G$ monoid
$endgroup$
– freehumorist
Jan 3 at 17:24
$begingroup$
I have no idea what your "reasoning" is. The map $f$ in question is a homomorphism iff $G$ is abelian, and you seemed to take it for granted that it was a homomorphism.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:26
$begingroup$
I have no idea what your "reasoning" is. The map $f$ in question is a homomorphism iff $G$ is abelian, and you seemed to take it for granted that it was a homomorphism.
$endgroup$
– Tobias Kildetoft
Jan 3 at 17:26
|
show 1 more comment
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$begingroup$
Yes, this all looks fine, though it is somewhat more complicated than it needs to be. $G$ is the union of cosets of $H$ and acts on these with $H$ being the stabilizer of itself, so $H$ acts on the remaining cosets, which is your complement.
$endgroup$
– Tobias Kildetoft
Jan 3 at 9:05
$begingroup$
Thank you Tobias. I'll come back to your comment when I'll be more acquainted with concepts of higher-level than current mine, like action, stabilizer, etc.
$endgroup$
– Luca
Jan 3 at 9:11