Proving Order of Double Coset [duplicate]
This question already has an answer here:
Order of Double Coset
2 answers
Problem: Prove that if $G$ is a finite group, then the number of elements in the double coset $AxB$ is $frac{|A| * |B|}{|A cap xBx^{-1}|}$.
So here's what I know, and I don't know if these pieces of information will be useful. $|AB| = frac{|A| * |B|}{|A cap B|}$ and $[G:B] = [G:A][A:B]$.
I am especially caught up on the $xBx^{-1}$ part. I understand that $xBx^{-1} = B$ if the left and right cosets are equal (that's how the professor explained it anyway) and I know that the left and right cosets are equal if the subgroup is normal. I don't even know if that is useful for this problem though.
I really don't have any level of understanding when it comes to cosets beyond the definitions as my professor has really not taught this and the textbook does not seem very clear. There was a preceding problem and the link will be provided in case that is necessary to prove this problem.
Equivalence class of double coset
abstract-algebra
edited Apr 13 '17 at 12:21
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asked Jan 29 '17 at 18:15
Kaptain
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marked as duplicate by darij grinberg, Lord Shark the Unknown abstract-algebra
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Nov 26 at 19:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Order of Double Coset
2 answers
Problem: Prove that if $G$ is a finite group, then the number of elements in the double coset $AxB$ is $frac{|A| * |B|}{|A cap xBx^{-1}|}$.
So here's what I know, and I don't know if these pieces of information will be useful. $|AB| = frac{|A| * |B|}{|A cap B|}$ and $[G:B] = [G:A][A:B]$.
I am especially caught up on the $xBx^{-1}$ part. I understand that $xBx^{-1} = B$ if the left and right cosets are equal (that's how the professor explained it anyway) and I know that the left and right cosets are equal if the subgroup is normal. I don't even know if that is useful for this problem though.
I really don't have any level of understanding when it comes to cosets beyond the definitions as my professor has really not taught this and the textbook does not seem very clear. There was a preceding problem and the link will be provided in case that is necessary to prove this problem.
Equivalence class of double coset
abstract-algebra
edited Apr 13 '17 at 12:21
Community♦
1
asked Jan 29 '17 at 18:15
Kaptain
426
marked as duplicate by darij grinberg, Lord Shark the Unknown abstract-algebra
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Nov 26 at 19:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Hint: $|AxB|=big|Aleft(xBx^{-1}right)big|$.
– Batominovski
Jan 29 '17 at 18:54
Hint #2: even if $B$ is not normal, $xBx^{-1}$ is still a subgroup, call it $C$ (you should prove this to your own satisfaction). Show that $b mapsto xbx^{-1}$ is a bijection $B to C$. You already know what $|AC|$ is.
– David Wheeler
Jan 29 '17 at 19:12
So would that then lead to something like this? $|A(xBx^{-1})|=frac{|A|*|xBx^{-1}|}{|Acap xBx^{-1}|}$. Or am I doing something wrong here?
– Kaptain
Jan 29 '17 at 19:14
Yes, that's the right idea, and you just need to show that $|B| = |xBx^{-1}|$, which shouldn't be too hard (use the same idea you used to show all left (or right) cosets have the same size).
– David Wheeler
Jan 29 '17 at 19:18
Oh, so it's not showing that $B = xBx^{-1}$, it's showing that $|B| = |xBx^{-1}|$. Thanks for the assist.
– Kaptain
Jan 29 '17 at 19:25
add a comment |
This question already has an answer here:
Order of Double Coset
2 answers
Problem: Prove that if $G$ is a finite group, then the number of elements in the double coset $AxB$ is $frac{|A| * |B|}{|A cap xBx^{-1}|}$.
So here's what I know, and I don't know if these pieces of information will be useful. $|AB| = frac{|A| * |B|}{|A cap B|}$ and $[G:B] = [G:A][A:B]$.
I am especially caught up on the $xBx^{-1}$ part. I understand that $xBx^{-1} = B$ if the left and right cosets are equal (that's how the professor explained it anyway) and I know that the left and right cosets are equal if the subgroup is normal. I don't even know if that is useful for this problem though.
I really don't have any level of understanding when it comes to cosets beyond the definitions as my professor has really not taught this and the textbook does not seem very clear. There was a preceding problem and the link will be provided in case that is necessary to prove this problem.
Equivalence class of double coset
abstract-algebra
edited Apr 13 '17 at 12:21
Community♦
1
asked Jan 29 '17 at 18:15
Kaptain
426
This question already has an answer here:
Order of Double Coset
2 answers
Problem: Prove that if $G$ is a finite group, then the number of elements in the double coset $AxB$ is $frac{|A| * |B|}{|A cap xBx^{-1}|}$.
So here's what I know, and I don't know if these pieces of information will be useful. $|AB| = frac{|A| * |B|}{|A cap B|}$ and $[G:B] = [G:A][A:B]$.
I am especially caught up on the $xBx^{-1}$ part. I understand that $xBx^{-1} = B$ if the left and right cosets are equal (that's how the professor explained it anyway) and I know that the left and right cosets are equal if the subgroup is normal. I don't even know if that is useful for this problem though.
I really don't have any level of understanding when it comes to cosets beyond the definitions as my professor has really not taught this and the textbook does not seem very clear. There was a preceding problem and the link will be provided in case that is necessary to prove this problem.
Equivalence class of double coset
This question already has an answer here:
Order of Double Coset
2 answers
abstract-algebra
abstract-algebra
edited Apr 13 '17 at 12:21
Community♦
1
asked Jan 29 '17 at 18:15
Kaptain
426
edited Apr 13 '17 at 12:21
Community♦
1
asked Jan 29 '17 at 18:15
Kaptain
426
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
asked Jan 29 '17 at 18:15
Kaptain
426
asked Jan 29 '17 at 18:15
Kaptain
426
asked Jan 29 '17 at 18:15
Kaptain
426
426
marked as duplicate by darij grinberg, Lord Shark the Unknown abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
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Nov 26 at 19:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Nov 26 at 19:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Hint: $|AxB|=big|Aleft(xBx^{-1}right)big|$.
– Batominovski
Jan 29 '17 at 18:54
Hint #2: even if $B$ is not normal, $xBx^{-1}$ is still a subgroup, call it $C$ (you should prove this to your own satisfaction). Show that $b mapsto xbx^{-1}$ is a bijection $B to C$. You already know what $|AC|$ is.
– David Wheeler
Jan 29 '17 at 19:12
So would that then lead to something like this? $|A(xBx^{-1})|=frac{|A|*|xBx^{-1}|}{|Acap xBx^{-1}|}$. Or am I doing something wrong here?
– Kaptain
Jan 29 '17 at 19:14
Yes, that's the right idea, and you just need to show that $|B| = |xBx^{-1}|$, which shouldn't be too hard (use the same idea you used to show all left (or right) cosets have the same size).
– David Wheeler
Jan 29 '17 at 19:18
Oh, so it's not showing that $B = xBx^{-1}$, it's showing that $|B| = |xBx^{-1}|$. Thanks for the assist.
– Kaptain
Jan 29 '17 at 19:25
add a comment |
1
Hint: $|AxB|=big|Aleft(xBx^{-1}right)big|$.
– Batominovski
Jan 29 '17 at 18:54
Hint #2: even if $B$ is not normal, $xBx^{-1}$ is still a subgroup, call it $C$ (you should prove this to your own satisfaction). Show that $b mapsto xbx^{-1}$ is a bijection $B to C$. You already know what $|AC|$ is.
– David Wheeler
Jan 29 '17 at 19:12
So would that then lead to something like this? $|A(xBx^{-1})|=frac{|A|*|xBx^{-1}|}{|Acap xBx^{-1}|}$. Or am I doing something wrong here?
– Kaptain
Jan 29 '17 at 19:14
Yes, that's the right idea, and you just need to show that $|B| = |xBx^{-1}|$, which shouldn't be too hard (use the same idea you used to show all left (or right) cosets have the same size).
– David Wheeler
Jan 29 '17 at 19:18
Oh, so it's not showing that $B = xBx^{-1}$, it's showing that $|B| = |xBx^{-1}|$. Thanks for the assist.
– Kaptain
Jan 29 '17 at 19:25
1
1
Hint: $|AxB|=big|Aleft(xBx^{-1}right)big|$.
– Batominovski
Jan 29 '17 at 18:54
Hint: $|AxB|=big|Aleft(xBx^{-1}right)big|$.
– Batominovski
Jan 29 '17 at 18:54
Hint #2: even if $B$ is not normal, $xBx^{-1}$ is still a subgroup, call it $C$ (you should prove this to your own satisfaction). Show that $b mapsto xbx^{-1}$ is a bijection $B to C$. You already know what $|AC|$ is.
– David Wheeler
Jan 29 '17 at 19:12
Hint #2: even if $B$ is not normal, $xBx^{-1}$ is still a subgroup, call it $C$ (you should prove this to your own satisfaction). Show that $b mapsto xbx^{-1}$ is a bijection $B to C$. You already know what $|AC|$ is.
– David Wheeler
Jan 29 '17 at 19:12
So would that then lead to something like this? $|A(xBx^{-1})|=frac{|A|*|xBx^{-1}|}{|Acap xBx^{-1}|}$. Or am I doing something wrong here?
– Kaptain
Jan 29 '17 at 19:14
So would that then lead to something like this? $|A(xBx^{-1})|=frac{|A|*|xBx^{-1}|}{|Acap xBx^{-1}|}$. Or am I doing something wrong here?
– Kaptain
Jan 29 '17 at 19:14
Yes, that's the right idea, and you just need to show that $|B| = |xBx^{-1}|$, which shouldn't be too hard (use the same idea you used to show all left (or right) cosets have the same size).
– David Wheeler
Jan 29 '17 at 19:18
Yes, that's the right idea, and you just need to show that $|B| = |xBx^{-1}|$, which shouldn't be too hard (use the same idea you used to show all left (or right) cosets have the same size).
– David Wheeler
Jan 29 '17 at 19:18
Oh, so it's not showing that $B = xBx^{-1}$, it's showing that $|B| = |xBx^{-1}|$. Thanks for the assist.
– Kaptain
Jan 29 '17 at 19:25
Oh, so it's not showing that $B = xBx^{-1}$, it's showing that $|B| = |xBx^{-1}|$. Thanks for the assist.
– Kaptain
Jan 29 '17 at 19:25
add a comment |
1 Answer
1
active
oldest
votes
Decompose $AxB = amalg_{1leq i leq n} Axb_i$ , and $|AxB|=|A|times n$.
Noticing $B= amalg_{1leq i leq n} (B cap x^{-1}Ax)b_i$ , thus $n = frac{|B|}{|B cap x^{-1}Ax|}$, $|AxB|=frac{|A|times |B|}{|B cap x^{-1}Ax|}$.
Order of Double Coset
answered Nov 26 at 19:23
Andrews
374317
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Decompose $AxB = amalg_{1leq i leq n} Axb_i$ , and $|AxB|=|A|times n$.
Noticing $B= amalg_{1leq i leq n} (B cap x^{-1}Ax)b_i$ , thus $n = frac{|B|}{|B cap x^{-1}Ax|}$, $|AxB|=frac{|A|times |B|}{|B cap x^{-1}Ax|}$.
Order of Double Coset
answered Nov 26 at 19:23
Andrews
374317
add a comment |
Decompose $AxB = amalg_{1leq i leq n} Axb_i$ , and $|AxB|=|A|times n$.
Noticing $B= amalg_{1leq i leq n} (B cap x^{-1}Ax)b_i$ , thus $n = frac{|B|}{|B cap x^{-1}Ax|}$, $|AxB|=frac{|A|times |B|}{|B cap x^{-1}Ax|}$.
Order of Double Coset
answered Nov 26 at 19:23
Andrews
374317
add a comment |
Decompose $AxB = amalg_{1leq i leq n} Axb_i$ , and $|AxB|=|A|times n$.
Noticing $B= amalg_{1leq i leq n} (B cap x^{-1}Ax)b_i$ , thus $n = frac{|B|}{|B cap x^{-1}Ax|}$, $|AxB|=frac{|A|times |B|}{|B cap x^{-1}Ax|}$.
Order of Double Coset
answered Nov 26 at 19:23
Andrews
374317
Decompose $AxB = amalg_{1leq i leq n} Axb_i$ , and $|AxB|=|A|times n$.
Noticing $B= amalg_{1leq i leq n} (B cap x^{-1}Ax)b_i$ , thus $n = frac{|B|}{|B cap x^{-1}Ax|}$, $|AxB|=frac{|A|times |B|}{|B cap x^{-1}Ax|}$.
Order of Double Coset
answered Nov 26 at 19:23
Andrews
374317
edited Nov 26 at 19:32
answered Nov 26 at 19:23
Andrews
374317
answered Nov 26 at 19:23
Andrews
374317
answered Nov 26 at 19:23
Andrews
374317
374317
add a comment |
add a comment |
1
Hint: $|AxB|=big|Aleft(xBx^{-1}right)big|$.
– Batominovski
Jan 29 '17 at 18:54
Hint #2: even if $B$ is not normal, $xBx^{-1}$ is still a subgroup, call it $C$ (you should prove this to your own satisfaction). Show that $b mapsto xbx^{-1}$ is a bijection $B to C$. You already know what $|AC|$ is.
– David Wheeler
Jan 29 '17 at 19:12
So would that then lead to something like this? $|A(xBx^{-1})|=frac{|A|*|xBx^{-1}|}{|Acap xBx^{-1}|}$. Or am I doing something wrong here?
– Kaptain
Jan 29 '17 at 19:14
Yes, that's the right idea, and you just need to show that $|B| = |xBx^{-1}|$, which shouldn't be too hard (use the same idea you used to show all left (or right) cosets have the same size).
– David Wheeler
Jan 29 '17 at 19:18
Oh, so it's not showing that $B = xBx^{-1}$, it's showing that $|B| = |xBx^{-1}|$. Thanks for the assist.
– Kaptain
Jan 29 '17 at 19:25