Prove: two elements in the group of permutations satisfy A^2 = B^2 then A = B.
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How does one prove when two cycles in the group of permutations satisfy A^2 = B^2 that A = B?
I think the fact the inverses A_*A = e and A_ = A^(l-1) and A = A^(l+1) is useful. l is the length of the cycles A and B, l is odd. Because V^2 is a cycle if and only if the length is odd.
I have a tendency to assume proofs have to be more complicated then they need to be.
abstract-algebra permutation-cycles
asked Nov 14 at 5:04
Jennifer Clarke
6
add a comment |
up vote
-1
down vote
favorite
How does one prove when two cycles in the group of permutations satisfy A^2 = B^2 that A = B?
I think the fact the inverses A_*A = e and A_ = A^(l-1) and A = A^(l+1) is useful. l is the length of the cycles A and B, l is odd. Because V^2 is a cycle if and only if the length is odd.
I have a tendency to assume proofs have to be more complicated then they need to be.
abstract-algebra permutation-cycles
asked Nov 14 at 5:04
Jennifer Clarke
6
1
transpositions?
– user10354138
Nov 14 at 5:07
1
In the initial statement, odd length wasn't specified. I think that restriction is necessary for the statement to be true.
– coffeemath
Nov 14 at 5:11
@coffeemath I didn't specify that A, B, A^2, and B^2 were all cycles either. I have to be more careful in the future.
– Jennifer Clarke
Nov 14 at 16:59
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
How does one prove when two cycles in the group of permutations satisfy A^2 = B^2 that A = B?
I think the fact the inverses A_*A = e and A_ = A^(l-1) and A = A^(l+1) is useful. l is the length of the cycles A and B, l is odd. Because V^2 is a cycle if and only if the length is odd.
I have a tendency to assume proofs have to be more complicated then they need to be.
abstract-algebra permutation-cycles
asked Nov 14 at 5:04
Jennifer Clarke
6
How does one prove when two cycles in the group of permutations satisfy A^2 = B^2 that A = B?
I think the fact the inverses A_*A = e and A_ = A^(l-1) and A = A^(l+1) is useful. l is the length of the cycles A and B, l is odd. Because V^2 is a cycle if and only if the length is odd.
I have a tendency to assume proofs have to be more complicated then they need to be.
abstract-algebra permutation-cycles
abstract-algebra permutation-cycles
asked Nov 14 at 5:04
Jennifer Clarke
6
asked Nov 14 at 5:04
Jennifer Clarke
6
asked Nov 14 at 5:04
Jennifer Clarke
6
asked Nov 14 at 5:04
Jennifer Clarke
6
asked Nov 14 at 5:04
Jennifer Clarke
6
6
1
transpositions?
– user10354138
Nov 14 at 5:07
1
In the initial statement, odd length wasn't specified. I think that restriction is necessary for the statement to be true.
– coffeemath
Nov 14 at 5:11
@coffeemath I didn't specify that A, B, A^2, and B^2 were all cycles either. I have to be more careful in the future.
– Jennifer Clarke
Nov 14 at 16:59
add a comment |
1
transpositions?
– user10354138
Nov 14 at 5:07
1
In the initial statement, odd length wasn't specified. I think that restriction is necessary for the statement to be true.
– coffeemath
Nov 14 at 5:11
@coffeemath I didn't specify that A, B, A^2, and B^2 were all cycles either. I have to be more careful in the future.
– Jennifer Clarke
Nov 14 at 16:59
1
1
transpositions?
– user10354138
Nov 14 at 5:07
transpositions?
– user10354138
Nov 14 at 5:07
1
1
In the initial statement, odd length wasn't specified. I think that restriction is necessary for the statement to be true.
– coffeemath
Nov 14 at 5:11
In the initial statement, odd length wasn't specified. I think that restriction is necessary for the statement to be true.
– coffeemath
Nov 14 at 5:11
@coffeemath I didn't specify that A, B, A^2, and B^2 were all cycles either. I have to be more careful in the future.
– Jennifer Clarke
Nov 14 at 16:59
@coffeemath I didn't specify that A, B, A^2, and B^2 were all cycles either. I have to be more careful in the future.
– Jennifer Clarke
Nov 14 at 16:59
add a comment |
1 Answer
1
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up vote
1
down vote
We need to assume cycle lengths of $A,B$ are odd. Otherwise $(1,2,3,4)$ and $(1,4,3,2)$ have the same squares. Maybe you want all of $A,B,A^2,B^2$ to be cycles, then the lengths must be odd. [But that wasn't said in question]
Anyway first show if $A^2=B^2$ the $A,B$ have the same length, say $2n-1.$ Then since $A^{2n-1}=1$ [because cycle to length of cycle is 1] you have $A^{2n}=A.$ Similarly $B^{2n}=B.$ But then
$$A=A^{2n}=[A^2]^n=[B^2]^n=B^{2n}=B,$$
where in the middle we used assumption $A^2=B^2.$
answered Nov 14 at 15:46
coffeemath
1,8951313
But isn't this simply saying the identity function satisfies A^2 = B^2? Since a cycle of one element can only be the identity (I think).
– Jennifer Clarke
Nov 14 at 16:41
@JenniferClarke I see your issue-- but the problem itself says to assume $A^2=B^2$ and then show $A=B$ follows. So the conclusion of the problem is the two permutations are equal, i.e. one can say the identity function gives the relation between $A$ and $B.$ That relation was not assumed in the problem, but had to be derived.Also no assumption that the cycle was of only $1$ element was used.
– coffeemath
Nov 14 at 16:46
How about A = A^(l + 1), where l is odd. (and l is the length of the cycle). So A = A^(2n + 1 + 1) = A^(2n + 2) = A^(2(n+1)) = [A^2]^(n+1) = [B^2]^(n+1) = B^(2(n+1)) = B^(2n + 1 + 1) = B^(l + 1) = B. Now I'm wondering if I can assume the B has l length like A, because I like making things complicated.
– Jennifer Clarke
Nov 14 at 16:54
@JenniferClarke Your argument in comment is just a shifted version of mine (notation for cycle length different). About assuming B has same length as A: If $A=(x_1,x_2,cdot x_l)$ when squared gives $(x_1,x_3,cdots,x_l,x_2,x_4,cdots x_{l-1})$ so same length as A. When we assume squares equal, we therefore assume lengths equal also.
– coffeemath
Nov 14 at 18:51
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
We need to assume cycle lengths of $A,B$ are odd. Otherwise $(1,2,3,4)$ and $(1,4,3,2)$ have the same squares. Maybe you want all of $A,B,A^2,B^2$ to be cycles, then the lengths must be odd. [But that wasn't said in question]
Anyway first show if $A^2=B^2$ the $A,B$ have the same length, say $2n-1.$ Then since $A^{2n-1}=1$ [because cycle to length of cycle is 1] you have $A^{2n}=A.$ Similarly $B^{2n}=B.$ But then
$$A=A^{2n}=[A^2]^n=[B^2]^n=B^{2n}=B,$$
where in the middle we used assumption $A^2=B^2.$
answered Nov 14 at 15:46
coffeemath
1,8951313
But isn't this simply saying the identity function satisfies A^2 = B^2? Since a cycle of one element can only be the identity (I think).
– Jennifer Clarke
Nov 14 at 16:41
@JenniferClarke I see your issue-- but the problem itself says to assume $A^2=B^2$ and then show $A=B$ follows. So the conclusion of the problem is the two permutations are equal, i.e. one can say the identity function gives the relation between $A$ and $B.$ That relation was not assumed in the problem, but had to be derived.Also no assumption that the cycle was of only $1$ element was used.
– coffeemath
Nov 14 at 16:46
How about A = A^(l + 1), where l is odd. (and l is the length of the cycle). So A = A^(2n + 1 + 1) = A^(2n + 2) = A^(2(n+1)) = [A^2]^(n+1) = [B^2]^(n+1) = B^(2(n+1)) = B^(2n + 1 + 1) = B^(l + 1) = B. Now I'm wondering if I can assume the B has l length like A, because I like making things complicated.
– Jennifer Clarke
Nov 14 at 16:54
@JenniferClarke Your argument in comment is just a shifted version of mine (notation for cycle length different). About assuming B has same length as A: If $A=(x_1,x_2,cdot x_l)$ when squared gives $(x_1,x_3,cdots,x_l,x_2,x_4,cdots x_{l-1})$ so same length as A. When we assume squares equal, we therefore assume lengths equal also.
– coffeemath
Nov 14 at 18:51
add a comment |
up vote
1
down vote
We need to assume cycle lengths of $A,B$ are odd. Otherwise $(1,2,3,4)$ and $(1,4,3,2)$ have the same squares. Maybe you want all of $A,B,A^2,B^2$ to be cycles, then the lengths must be odd. [But that wasn't said in question]
Anyway first show if $A^2=B^2$ the $A,B$ have the same length, say $2n-1.$ Then since $A^{2n-1}=1$ [because cycle to length of cycle is 1] you have $A^{2n}=A.$ Similarly $B^{2n}=B.$ But then
$$A=A^{2n}=[A^2]^n=[B^2]^n=B^{2n}=B,$$
where in the middle we used assumption $A^2=B^2.$
answered Nov 14 at 15:46
coffeemath
1,8951313
But isn't this simply saying the identity function satisfies A^2 = B^2? Since a cycle of one element can only be the identity (I think).
– Jennifer Clarke
Nov 14 at 16:41
@JenniferClarke I see your issue-- but the problem itself says to assume $A^2=B^2$ and then show $A=B$ follows. So the conclusion of the problem is the two permutations are equal, i.e. one can say the identity function gives the relation between $A$ and $B.$ That relation was not assumed in the problem, but had to be derived.Also no assumption that the cycle was of only $1$ element was used.
– coffeemath
Nov 14 at 16:46
How about A = A^(l + 1), where l is odd. (and l is the length of the cycle). So A = A^(2n + 1 + 1) = A^(2n + 2) = A^(2(n+1)) = [A^2]^(n+1) = [B^2]^(n+1) = B^(2(n+1)) = B^(2n + 1 + 1) = B^(l + 1) = B. Now I'm wondering if I can assume the B has l length like A, because I like making things complicated.
– Jennifer Clarke
Nov 14 at 16:54
@JenniferClarke Your argument in comment is just a shifted version of mine (notation for cycle length different). About assuming B has same length as A: If $A=(x_1,x_2,cdot x_l)$ when squared gives $(x_1,x_3,cdots,x_l,x_2,x_4,cdots x_{l-1})$ so same length as A. When we assume squares equal, we therefore assume lengths equal also.
– coffeemath
Nov 14 at 18:51
add a comment |
up vote
1
down vote
up vote
1
down vote
We need to assume cycle lengths of $A,B$ are odd. Otherwise $(1,2,3,4)$ and $(1,4,3,2)$ have the same squares. Maybe you want all of $A,B,A^2,B^2$ to be cycles, then the lengths must be odd. [But that wasn't said in question]
Anyway first show if $A^2=B^2$ the $A,B$ have the same length, say $2n-1.$ Then since $A^{2n-1}=1$ [because cycle to length of cycle is 1] you have $A^{2n}=A.$ Similarly $B^{2n}=B.$ But then
$$A=A^{2n}=[A^2]^n=[B^2]^n=B^{2n}=B,$$
where in the middle we used assumption $A^2=B^2.$
answered Nov 14 at 15:46
coffeemath
1,8951313
We need to assume cycle lengths of $A,B$ are odd. Otherwise $(1,2,3,4)$ and $(1,4,3,2)$ have the same squares. Maybe you want all of $A,B,A^2,B^2$ to be cycles, then the lengths must be odd. [But that wasn't said in question]
Anyway first show if $A^2=B^2$ the $A,B$ have the same length, say $2n-1.$ Then since $A^{2n-1}=1$ [because cycle to length of cycle is 1] you have $A^{2n}=A.$ Similarly $B^{2n}=B.$ But then
$$A=A^{2n}=[A^2]^n=[B^2]^n=B^{2n}=B,$$
where in the middle we used assumption $A^2=B^2.$
answered Nov 14 at 15:46
coffeemath
1,8951313
answered Nov 14 at 15:46
coffeemath
1,8951313
answered Nov 14 at 15:46
coffeemath
1,8951313
answered Nov 14 at 15:46
coffeemath
1,8951313
1,8951313
But isn't this simply saying the identity function satisfies A^2 = B^2? Since a cycle of one element can only be the identity (I think).
– Jennifer Clarke
Nov 14 at 16:41
@JenniferClarke I see your issue-- but the problem itself says to assume $A^2=B^2$ and then show $A=B$ follows. So the conclusion of the problem is the two permutations are equal, i.e. one can say the identity function gives the relation between $A$ and $B.$ That relation was not assumed in the problem, but had to be derived.Also no assumption that the cycle was of only $1$ element was used.
– coffeemath
Nov 14 at 16:46
How about A = A^(l + 1), where l is odd. (and l is the length of the cycle). So A = A^(2n + 1 + 1) = A^(2n + 2) = A^(2(n+1)) = [A^2]^(n+1) = [B^2]^(n+1) = B^(2(n+1)) = B^(2n + 1 + 1) = B^(l + 1) = B. Now I'm wondering if I can assume the B has l length like A, because I like making things complicated.
– Jennifer Clarke
Nov 14 at 16:54
@JenniferClarke Your argument in comment is just a shifted version of mine (notation for cycle length different). About assuming B has same length as A: If $A=(x_1,x_2,cdot x_l)$ when squared gives $(x_1,x_3,cdots,x_l,x_2,x_4,cdots x_{l-1})$ so same length as A. When we assume squares equal, we therefore assume lengths equal also.
– coffeemath
Nov 14 at 18:51
add a comment |
But isn't this simply saying the identity function satisfies A^2 = B^2? Since a cycle of one element can only be the identity (I think).
– Jennifer Clarke
Nov 14 at 16:41
@JenniferClarke I see your issue-- but the problem itself says to assume $A^2=B^2$ and then show $A=B$ follows. So the conclusion of the problem is the two permutations are equal, i.e. one can say the identity function gives the relation between $A$ and $B.$ That relation was not assumed in the problem, but had to be derived.Also no assumption that the cycle was of only $1$ element was used.
– coffeemath
Nov 14 at 16:46
How about A = A^(l + 1), where l is odd. (and l is the length of the cycle). So A = A^(2n + 1 + 1) = A^(2n + 2) = A^(2(n+1)) = [A^2]^(n+1) = [B^2]^(n+1) = B^(2(n+1)) = B^(2n + 1 + 1) = B^(l + 1) = B. Now I'm wondering if I can assume the B has l length like A, because I like making things complicated.
– Jennifer Clarke
Nov 14 at 16:54
@JenniferClarke Your argument in comment is just a shifted version of mine (notation for cycle length different). About assuming B has same length as A: If $A=(x_1,x_2,cdot x_l)$ when squared gives $(x_1,x_3,cdots,x_l,x_2,x_4,cdots x_{l-1})$ so same length as A. When we assume squares equal, we therefore assume lengths equal also.
– coffeemath
Nov 14 at 18:51
But isn't this simply saying the identity function satisfies A^2 = B^2? Since a cycle of one element can only be the identity (I think).
– Jennifer Clarke
Nov 14 at 16:41
But isn't this simply saying the identity function satisfies A^2 = B^2? Since a cycle of one element can only be the identity (I think).
– Jennifer Clarke
Nov 14 at 16:41
@JenniferClarke I see your issue-- but the problem itself says to assume $A^2=B^2$ and then show $A=B$ follows. So the conclusion of the problem is the two permutations are equal, i.e. one can say the identity function gives the relation between $A$ and $B.$ That relation was not assumed in the problem, but had to be derived.Also no assumption that the cycle was of only $1$ element was used.
– coffeemath
Nov 14 at 16:46
@JenniferClarke I see your issue-- but the problem itself says to assume $A^2=B^2$ and then show $A=B$ follows. So the conclusion of the problem is the two permutations are equal, i.e. one can say the identity function gives the relation between $A$ and $B.$ That relation was not assumed in the problem, but had to be derived.Also no assumption that the cycle was of only $1$ element was used.
– coffeemath
Nov 14 at 16:46
How about A = A^(l + 1), where l is odd. (and l is the length of the cycle). So A = A^(2n + 1 + 1) = A^(2n + 2) = A^(2(n+1)) = [A^2]^(n+1) = [B^2]^(n+1) = B^(2(n+1)) = B^(2n + 1 + 1) = B^(l + 1) = B. Now I'm wondering if I can assume the B has l length like A, because I like making things complicated.
– Jennifer Clarke
Nov 14 at 16:54
How about A = A^(l + 1), where l is odd. (and l is the length of the cycle). So A = A^(2n + 1 + 1) = A^(2n + 2) = A^(2(n+1)) = [A^2]^(n+1) = [B^2]^(n+1) = B^(2(n+1)) = B^(2n + 1 + 1) = B^(l + 1) = B. Now I'm wondering if I can assume the B has l length like A, because I like making things complicated.
– Jennifer Clarke
Nov 14 at 16:54
@JenniferClarke Your argument in comment is just a shifted version of mine (notation for cycle length different). About assuming B has same length as A: If $A=(x_1,x_2,cdot x_l)$ when squared gives $(x_1,x_3,cdots,x_l,x_2,x_4,cdots x_{l-1})$ so same length as A. When we assume squares equal, we therefore assume lengths equal also.
– coffeemath
Nov 14 at 18:51
@JenniferClarke Your argument in comment is just a shifted version of mine (notation for cycle length different). About assuming B has same length as A: If $A=(x_1,x_2,cdot x_l)$ when squared gives $(x_1,x_3,cdots,x_l,x_2,x_4,cdots x_{l-1})$ so same length as A. When we assume squares equal, we therefore assume lengths equal also.
– coffeemath
Nov 14 at 18:51
add a comment |
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transpositions?
– user10354138
Nov 14 at 5:07
1
In the initial statement, odd length wasn't specified. I think that restriction is necessary for the statement to be true.
– coffeemath
Nov 14 at 5:11
@coffeemath I didn't specify that A, B, A^2, and B^2 were all cycles either. I have to be more careful in the future.
– Jennifer Clarke
Nov 14 at 16:59