Find $f^−100g^146f^301$ (permutations to high powers) The Next CEO of Stack OverflowCycle notation questionBasic question on permutationsLet $f in 8^8$ where the permutation is given in two line form:Questions on trace invariance of product of square matrices under cyclic permutationsPreservation of cycle lengthorder of permutation $(2,9,6,3,7)(4,8)$ in $S_9$Is this permutation even or odd?Disjoint cycle notation of $(a_1, a_2, . . . , a_n)^−1$.With a given permutation $π$, how many functions $f: N_10 → N_10$ are $f(π(i)) = πf(i)$, where all $i ∈ N_10$?Find all solutions in an equation with permutations in $S_10$
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Find $f^−100g^146f^301$ (permutations to high powers)
The Next CEO of Stack OverflowCycle notation questionBasic question on permutationsLet $f in 8^8$ where the permutation is given in two line form:Questions on trace invariance of product of square matrices under cyclic permutationsPreservation of cycle lengthorder of permutation $(2,9,6,3,7)(4,8)$ in $S_9$Is this permutation even or odd?Disjoint cycle notation of $(a_1, a_2, . . . , a_n)^−1$.With a given permutation $π$, how many functions $f: N_10 → N_10$ are $f(π(i)) = πf(i)$, where all $i ∈ N_10$?Find all solutions in an equation with permutations in $S_10$
$begingroup$
Find $f^-100g^146f^301$ where
$$f = beginpmatrix
1 & 2 & 3& 4 & 5 & 6 & 7 \
3 & 1 & 5 & 7 & 2 & 6 & 4endpmatrix, \
g = beginpmatrix
1 & 2 & 3 & 4 & 5 & 6 & 7 \
3 & 1 & 7 & 6 & 4 & 5 & 2endpmatrix.
$$
can someone please help with a step by step guide for this as I can't find any examples anywhere and Im really confused. I can put it in cycle notation and find order but don't know how to use that information to find answer.
Thanks
Edit-
cycle notation for f = (1352)(47) and order is lcm(4,2)=4
cycle notation for g = (1372)(465) and order is lcm(4,3)=12
proof-writing permutations exponentiation permutation-cycles
asked yesterday
AnonymousAnonymous
11
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Find $f^-100g^146f^301$ where
$$f = beginpmatrix
1 & 2 & 3& 4 & 5 & 6 & 7 \
3 & 1 & 5 & 7 & 2 & 6 & 4endpmatrix, \
g = beginpmatrix
1 & 2 & 3 & 4 & 5 & 6 & 7 \
3 & 1 & 7 & 6 & 4 & 5 & 2endpmatrix.
$$
can someone please help with a step by step guide for this as I can't find any examples anywhere and Im really confused. I can put it in cycle notation and find order but don't know how to use that information to find answer.
Thanks
Edit-
cycle notation for f = (1352)(47) and order is lcm(4,2)=4
cycle notation for g = (1372)(465) and order is lcm(4,3)=12
proof-writing permutations exponentiation permutation-cycles
asked yesterday
AnonymousAnonymous
11
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
You should edit your answer to show your working so far ("I can put it in cycle notation and find order").
$endgroup$
– TonyK
yesterday
$begingroup$
Good! So $f^4$ is the identity permutation. That means that $f^100=$... what?
$endgroup$
– TonyK
yesterday
$begingroup$
before I do that I have a question - do you do the inverse first then you have f^4 using the inverse or do you find it at the end
$endgroup$
– Anonymous
yesterday
$begingroup$
Either way, whichever is easiest. That's because $(f^n)^-1=(f^-1)^n$ for all $n$.
$endgroup$
– TonyK
yesterday
add a comment |
$begingroup$
Find $f^-100g^146f^301$ where
$$f = beginpmatrix
1 & 2 & 3& 4 & 5 & 6 & 7 \
3 & 1 & 5 & 7 & 2 & 6 & 4endpmatrix, \
g = beginpmatrix
1 & 2 & 3 & 4 & 5 & 6 & 7 \
3 & 1 & 7 & 6 & 4 & 5 & 2endpmatrix.
$$
can someone please help with a step by step guide for this as I can't find any examples anywhere and Im really confused. I can put it in cycle notation and find order but don't know how to use that information to find answer.
Thanks
Edit-
cycle notation for f = (1352)(47) and order is lcm(4,2)=4
cycle notation for g = (1372)(465) and order is lcm(4,3)=12
proof-writing permutations exponentiation permutation-cycles
asked yesterday
AnonymousAnonymous
11
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Find $f^-100g^146f^301$ where
$$f = beginpmatrix
1 & 2 & 3& 4 & 5 & 6 & 7 \
3 & 1 & 5 & 7 & 2 & 6 & 4endpmatrix, \
g = beginpmatrix
1 & 2 & 3 & 4 & 5 & 6 & 7 \
3 & 1 & 7 & 6 & 4 & 5 & 2endpmatrix.
$$
can someone please help with a step by step guide for this as I can't find any examples anywhere and Im really confused. I can put it in cycle notation and find order but don't know how to use that information to find answer.
Thanks
Edit-
cycle notation for f = (1352)(47) and order is lcm(4,2)=4
cycle notation for g = (1372)(465) and order is lcm(4,3)=12
proof-writing permutations exponentiation permutation-cycles
proof-writing permutations exponentiation permutation-cycles
asked yesterday
AnonymousAnonymous
11
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
AnonymousAnonymous
11
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Anonymous
asked yesterday
AnonymousAnonymous
11
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
AnonymousAnonymous
11
asked yesterday
AnonymousAnonymous
11
11
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
You should edit your answer to show your working so far ("I can put it in cycle notation and find order").
$endgroup$
– TonyK
yesterday
$begingroup$
Good! So $f^4$ is the identity permutation. That means that $f^100=$... what?
$endgroup$
– TonyK
yesterday
$begingroup$
before I do that I have a question - do you do the inverse first then you have f^4 using the inverse or do you find it at the end
$endgroup$
– Anonymous
yesterday
$begingroup$
Either way, whichever is easiest. That's because $(f^n)^-1=(f^-1)^n$ for all $n$.
$endgroup$
– TonyK
yesterday
add a comment |
$begingroup$
You should edit your answer to show your working so far ("I can put it in cycle notation and find order").
$endgroup$
– TonyK
yesterday
$begingroup$
Good! So $f^4$ is the identity permutation. That means that $f^100=$... what?
$endgroup$
– TonyK
yesterday
$begingroup$
before I do that I have a question - do you do the inverse first then you have f^4 using the inverse or do you find it at the end
$endgroup$
– Anonymous
yesterday
$begingroup$
Either way, whichever is easiest. That's because $(f^n)^-1=(f^-1)^n$ for all $n$.
$endgroup$
– TonyK
yesterday
$begingroup$
You should edit your answer to show your working so far ("I can put it in cycle notation and find order").
$endgroup$
– TonyK
yesterday
$begingroup$
You should edit your answer to show your working so far ("I can put it in cycle notation and find order").
$endgroup$
– TonyK
yesterday
$begingroup$
Good! So $f^4$ is the identity permutation. That means that $f^100=$... what?
$endgroup$
– TonyK
yesterday
$begingroup$
Good! So $f^4$ is the identity permutation. That means that $f^100=$... what?
$endgroup$
– TonyK
yesterday
$begingroup$
before I do that I have a question - do you do the inverse first then you have f^4 using the inverse or do you find it at the end
$endgroup$
– Anonymous
yesterday
$begingroup$
before I do that I have a question - do you do the inverse first then you have f^4 using the inverse or do you find it at the end
$endgroup$
– Anonymous
yesterday
$begingroup$
Either way, whichever is easiest. That's because $(f^n)^-1=(f^-1)^n$ for all $n$.
$endgroup$
– TonyK
yesterday
$begingroup$
Either way, whichever is easiest. That's because $(f^n)^-1=(f^-1)^n$ for all $n$.
$endgroup$
– TonyK
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
To calculate easily powers oa permutation, decompose it first as a product of disjoint cycles.
E.g., $;f=(1,3,5,2)(4,7)$.
Now a cycle has order its length, and a product of disjoint cycles has order the l.c.m. of the orders of their orders, i.e. the l.c.m. of thelengths of each cycle. Thus, $f$ has order $operatornamelcm(4,2)=4$.
On the other hand, $f^2$ is easily obtained by the rule than an element in the cycle maps onto the second next, as if these elements were distributed clockwise on a circle: so that $;f^2=(1,5)(2,3)$. Similarly, for $f^3$, an element maps onto the third next: $;f^3=(1,2,5,3)$. You can observe it is the same as $f^-1$, which is obtained mapping each element onto the next o,e counterclockwise.
Last hint: as $f^4=1$, $; f^301=f^{301bmod 4$f^1$.
Can you perform all the calculations with these elements?
answered yesterday
BernardBernard
123k741117
$endgroup$
add a comment |
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votes
$begingroup$
Hint:
To calculate easily powers oa permutation, decompose it first as a product of disjoint cycles.
E.g., $;f=(1,3,5,2)(4,7)$.
Now a cycle has order its length, and a product of disjoint cycles has order the l.c.m. of the orders of their orders, i.e. the l.c.m. of thelengths of each cycle. Thus, $f$ has order $operatornamelcm(4,2)=4$.
On the other hand, $f^2$ is easily obtained by the rule than an element in the cycle maps onto the second next, as if these elements were distributed clockwise on a circle: so that $;f^2=(1,5)(2,3)$. Similarly, for $f^3$, an element maps onto the third next: $;f^3=(1,2,5,3)$. You can observe it is the same as $f^-1$, which is obtained mapping each element onto the next o,e counterclockwise.
Last hint: as $f^4=1$, $; f^301=f^{301bmod 4$f^1$.
Can you perform all the calculations with these elements?
answered yesterday
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Hint:
To calculate easily powers oa permutation, decompose it first as a product of disjoint cycles.
E.g., $;f=(1,3,5,2)(4,7)$.
Now a cycle has order its length, and a product of disjoint cycles has order the l.c.m. of the orders of their orders, i.e. the l.c.m. of thelengths of each cycle. Thus, $f$ has order $operatornamelcm(4,2)=4$.
On the other hand, $f^2$ is easily obtained by the rule than an element in the cycle maps onto the second next, as if these elements were distributed clockwise on a circle: so that $;f^2=(1,5)(2,3)$. Similarly, for $f^3$, an element maps onto the third next: $;f^3=(1,2,5,3)$. You can observe it is the same as $f^-1$, which is obtained mapping each element onto the next o,e counterclockwise.
Last hint: as $f^4=1$, $; f^301=f^{301bmod 4$f^1$.
Can you perform all the calculations with these elements?
answered yesterday
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Hint:
To calculate easily powers oa permutation, decompose it first as a product of disjoint cycles.
E.g., $;f=(1,3,5,2)(4,7)$.
Now a cycle has order its length, and a product of disjoint cycles has order the l.c.m. of the orders of their orders, i.e. the l.c.m. of thelengths of each cycle. Thus, $f$ has order $operatornamelcm(4,2)=4$.
On the other hand, $f^2$ is easily obtained by the rule than an element in the cycle maps onto the second next, as if these elements were distributed clockwise on a circle: so that $;f^2=(1,5)(2,3)$. Similarly, for $f^3$, an element maps onto the third next: $;f^3=(1,2,5,3)$. You can observe it is the same as $f^-1$, which is obtained mapping each element onto the next o,e counterclockwise.
Last hint: as $f^4=1$, $; f^301=f^{301bmod 4$f^1$.
Can you perform all the calculations with these elements?
answered yesterday
BernardBernard
123k741117
$endgroup$
Hint:
To calculate easily powers oa permutation, decompose it first as a product of disjoint cycles.
E.g., $;f=(1,3,5,2)(4,7)$.
Now a cycle has order its length, and a product of disjoint cycles has order the l.c.m. of the orders of their orders, i.e. the l.c.m. of thelengths of each cycle. Thus, $f$ has order $operatornamelcm(4,2)=4$.
On the other hand, $f^2$ is easily obtained by the rule than an element in the cycle maps onto the second next, as if these elements were distributed clockwise on a circle: so that $;f^2=(1,5)(2,3)$. Similarly, for $f^3$, an element maps onto the third next: $;f^3=(1,2,5,3)$. You can observe it is the same as $f^-1$, which is obtained mapping each element onto the next o,e counterclockwise.
Last hint: as $f^4=1$, $; f^301=f^{301bmod 4$f^1$.
Can you perform all the calculations with these elements?
answered yesterday
BernardBernard
123k741117
answered yesterday
BernardBernard
123k741117
answered yesterday
BernardBernard
123k741117
answered yesterday
BernardBernard
123k741117
123k741117
add a comment |
add a comment |
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$begingroup$
You should edit your answer to show your working so far ("I can put it in cycle notation and find order").
$endgroup$
– TonyK
yesterday
$begingroup$
Good! So $f^4$ is the identity permutation. That means that $f^100=$... what?
$endgroup$
– TonyK
yesterday
$begingroup$
before I do that I have a question - do you do the inverse first then you have f^4 using the inverse or do you find it at the end
$endgroup$
– Anonymous
yesterday
$begingroup$
Either way, whichever is easiest. That's because $(f^n)^-1=(f^-1)^n$ for all $n$.
$endgroup$
– TonyK
yesterday