Determine the number of cycles of the form $(underline6)(underline2)(underline2)$ in $S_9$In $S_6$, write the result as a product of disjoint cycles and then in the 2-row form.Showing that $A_n$ is generated by the 3-cycles in $S_n$Finding the number of permutations in$S_9$ of the form $(a_1a_2)(a_3a_4)(a_5a_6)(a_7a_8a_9)$Count the permutations which are products of exactly two disjoint cycles.Show that the product of two transpositions can be expressed as a product of $3$-cyclesShow that $S_n$ can be generated from the two-cycles $(1,2),(1,3),dotsc,(1,n)$.Alternative proof that the 3-cycles generate the alternating group $A_n$Let $sigmain S_n$ be an $n$-cycle, and let $tauin S_n$ be a $2$-cycle.Then, $sigma$ and $tau$ need not generate $S_n$.Finding the number of elements of order 7 in $S_9$On the product of two cycles (and its conjugates)
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Determine the number of cycles of the form $(underline6)(underline2)(underline2)$ in $S_9$
In $S_6$, write the result as a product of disjoint cycles and then in the 2-row form.Showing that $A_n$ is generated by the 3-cycles in $S_n$Finding the number of permutations in$S_9$ of the form $(a_1a_2)(a_3a_4)(a_5a_6)(a_7a_8a_9)$Count the permutations which are products of exactly two disjoint cycles.Show that the product of two transpositions can be expressed as a product of $3$-cyclesShow that $S_n$ can be generated from the two-cycles $(1,2),(1,3),dotsc,(1,n)$.Alternative proof that the 3-cycles generate the alternating group $A_n$Let $sigmain S_n$ be an $n$-cycle, and let $tauin S_n$ be a $2$-cycle.Then, $sigma$ and $tau$ need not generate $S_n$.Finding the number of elements of order 7 in $S_9$On the product of two cycles (and its conjugates)
$begingroup$
Consider a product of cycles of the form $(underline6)(underline2)(underline2)$ in $S_9$ where $(underlinei)$ is a cycle with length $i$. I want to determine the number of such cycles.
However, in order for $(underline6)(underline2)(underline2)$ to even make sense, it must be the case that $(underline2)(underline2)$ is a product of transpositions. So if we fix $a = 1,dots,9$, it follows that such a product can be written as
$$(ax)(ay) = (ayx)$$
where $x,y = 1,2,dotsc,9$ and $xneq y$.
Therefore, the product $(underline6)(underline2)(underline2) = (underline6)(underline3)$, so it follows that there are
$$frac9!18 = 20,160$$
such elements of that form.
Is the above calculation correct? I feel as if I overcounted.
abstract-algebra group-theory permutation-cycles
edited Mar 21 at 20:32
Shaun
10.3k113686
asked Mar 21 at 19:52
Benedict VoltaireBenedict Voltaire
1,347929
$endgroup$
|
show 2 more comments
$begingroup$
Consider a product of cycles of the form $(underline6)(underline2)(underline2)$ in $S_9$ where $(underlinei)$ is a cycle with length $i$. I want to determine the number of such cycles.
However, in order for $(underline6)(underline2)(underline2)$ to even make sense, it must be the case that $(underline2)(underline2)$ is a product of transpositions. So if we fix $a = 1,dots,9$, it follows that such a product can be written as
$$(ax)(ay) = (ayx)$$
where $x,y = 1,2,dotsc,9$ and $xneq y$.
Therefore, the product $(underline6)(underline2)(underline2) = (underline6)(underline3)$, so it follows that there are
$$frac9!18 = 20,160$$
such elements of that form.
Is the above calculation correct? I feel as if I overcounted.
abstract-algebra group-theory permutation-cycles
edited Mar 21 at 20:32
Shaun
10.3k113686
asked Mar 21 at 19:52
Benedict VoltaireBenedict Voltaire
1,347929
$endgroup$
2
$begingroup$
If you're allowing overlap - why not allow the 6-cycle to overlap with the 2-cycles?
$endgroup$
– jmerry
Mar 21 at 19:55
1
$begingroup$
What throws a spanner in the works is that it may not be the case that permutations in that form have a unique representation in that form; there may be different $6$ and $2$ cycles that produce the same permutation. (I don't have a counterexample, so there may not be as well!)
$endgroup$
– Theo Bendit
Mar 21 at 20:03
1
$begingroup$
I suspect something has gone horribly wrong in the formulation of this question. As it happens, the number of elements in $S_9$ which can be written in the form $abc$ where $a$ is a $6$-cycle and $b$ and $c$ are $2$-cycles turns out to be $160740 = 2^2 cdot 3^2 cdot 5 cdot 19 cdot 47$. But that's almost certainly not what the intended question is. That's approximately 88.59% of all elements in $S_9$ which are not in $A_9$ (certainly all the terms are odd so the product is not in $A_9$).
$endgroup$
– user655377
Mar 22 at 0:43
1
$begingroup$
@TheoBendit It is easy to come up with examples like that. One way is to use the fact that the product $sigma(ij)$ either splits a cycle containing both $i$ and $j$ in two, or fuses together the two cycles containing $i$ and $j$. All according to whether $i$ and $j$ were in separate cycles or not to begin with. So for example $$(123456)(35)=(1236)(45),$$ splitting the 6-cycle containing both $3$ and $5$, and then $$(1236)(45)(34)=(123546),$$ fusing as $3$ and $4$ were in separate cycles of $(1236)(45)$.
$endgroup$
– Jyrki Lahtonen
Mar 22 at 7:06
1
$begingroup$
(cont'd) Anyway, we get non-uniqueness like $$(123456)(35)(34)=(123546)(78)(78).$$ Other kind of non-uniqueness also comes from different "fusions", such as $$(123456)(78)(69)=(1234569)(78)=(234569)(78)(91)$$
$endgroup$
– Jyrki Lahtonen
Mar 22 at 7:10
|
show 2 more comments
$begingroup$
Consider a product of cycles of the form $(underline6)(underline2)(underline2)$ in $S_9$ where $(underlinei)$ is a cycle with length $i$. I want to determine the number of such cycles.
However, in order for $(underline6)(underline2)(underline2)$ to even make sense, it must be the case that $(underline2)(underline2)$ is a product of transpositions. So if we fix $a = 1,dots,9$, it follows that such a product can be written as
$$(ax)(ay) = (ayx)$$
where $x,y = 1,2,dotsc,9$ and $xneq y$.
Therefore, the product $(underline6)(underline2)(underline2) = (underline6)(underline3)$, so it follows that there are
$$frac9!18 = 20,160$$
such elements of that form.
Is the above calculation correct? I feel as if I overcounted.
abstract-algebra group-theory permutation-cycles
edited Mar 21 at 20:32
Shaun
10.3k113686
asked Mar 21 at 19:52
Benedict VoltaireBenedict Voltaire
1,347929
$endgroup$
Consider a product of cycles of the form $(underline6)(underline2)(underline2)$ in $S_9$ where $(underlinei)$ is a cycle with length $i$. I want to determine the number of such cycles.
However, in order for $(underline6)(underline2)(underline2)$ to even make sense, it must be the case that $(underline2)(underline2)$ is a product of transpositions. So if we fix $a = 1,dots,9$, it follows that such a product can be written as
$$(ax)(ay) = (ayx)$$
where $x,y = 1,2,dotsc,9$ and $xneq y$.
Therefore, the product $(underline6)(underline2)(underline2) = (underline6)(underline3)$, so it follows that there are
$$frac9!18 = 20,160$$
such elements of that form.
Is the above calculation correct? I feel as if I overcounted.
abstract-algebra group-theory permutation-cycles
abstract-algebra group-theory permutation-cycles
edited Mar 21 at 20:32
Shaun
10.3k113686
asked Mar 21 at 19:52
Benedict VoltaireBenedict Voltaire
1,347929
edited Mar 21 at 20:32
Shaun
10.3k113686
asked Mar 21 at 19:52
Benedict VoltaireBenedict Voltaire
1,347929
edited Mar 21 at 20:32
Shaun
10.3k113686
edited Mar 21 at 20:32
Shaun
10.3k113686
edited Mar 21 at 20:32
Shaun
10.3k113686
10.3k113686
asked Mar 21 at 19:52
Benedict VoltaireBenedict Voltaire
1,347929
asked Mar 21 at 19:52
Benedict VoltaireBenedict Voltaire
1,347929
asked Mar 21 at 19:52
Benedict VoltaireBenedict Voltaire
1,347929
1,347929
2
$begingroup$
If you're allowing overlap - why not allow the 6-cycle to overlap with the 2-cycles?
$endgroup$
– jmerry
Mar 21 at 19:55
1
$begingroup$
What throws a spanner in the works is that it may not be the case that permutations in that form have a unique representation in that form; there may be different $6$ and $2$ cycles that produce the same permutation. (I don't have a counterexample, so there may not be as well!)
$endgroup$
– Theo Bendit
Mar 21 at 20:03
1
$begingroup$
I suspect something has gone horribly wrong in the formulation of this question. As it happens, the number of elements in $S_9$ which can be written in the form $abc$ where $a$ is a $6$-cycle and $b$ and $c$ are $2$-cycles turns out to be $160740 = 2^2 cdot 3^2 cdot 5 cdot 19 cdot 47$. But that's almost certainly not what the intended question is. That's approximately 88.59% of all elements in $S_9$ which are not in $A_9$ (certainly all the terms are odd so the product is not in $A_9$).
$endgroup$
– user655377
Mar 22 at 0:43
1
$begingroup$
@TheoBendit It is easy to come up with examples like that. One way is to use the fact that the product $sigma(ij)$ either splits a cycle containing both $i$ and $j$ in two, or fuses together the two cycles containing $i$ and $j$. All according to whether $i$ and $j$ were in separate cycles or not to begin with. So for example $$(123456)(35)=(1236)(45),$$ splitting the 6-cycle containing both $3$ and $5$, and then $$(1236)(45)(34)=(123546),$$ fusing as $3$ and $4$ were in separate cycles of $(1236)(45)$.
$endgroup$
– Jyrki Lahtonen
Mar 22 at 7:06
1
$begingroup$
(cont'd) Anyway, we get non-uniqueness like $$(123456)(35)(34)=(123546)(78)(78).$$ Other kind of non-uniqueness also comes from different "fusions", such as $$(123456)(78)(69)=(1234569)(78)=(234569)(78)(91)$$
$endgroup$
– Jyrki Lahtonen
Mar 22 at 7:10
|
show 2 more comments
2
$begingroup$
If you're allowing overlap - why not allow the 6-cycle to overlap with the 2-cycles?
$endgroup$
– jmerry
Mar 21 at 19:55
1
$begingroup$
What throws a spanner in the works is that it may not be the case that permutations in that form have a unique representation in that form; there may be different $6$ and $2$ cycles that produce the same permutation. (I don't have a counterexample, so there may not be as well!)
$endgroup$
– Theo Bendit
Mar 21 at 20:03
1
$begingroup$
I suspect something has gone horribly wrong in the formulation of this question. As it happens, the number of elements in $S_9$ which can be written in the form $abc$ where $a$ is a $6$-cycle and $b$ and $c$ are $2$-cycles turns out to be $160740 = 2^2 cdot 3^2 cdot 5 cdot 19 cdot 47$. But that's almost certainly not what the intended question is. That's approximately 88.59% of all elements in $S_9$ which are not in $A_9$ (certainly all the terms are odd so the product is not in $A_9$).
$endgroup$
– user655377
Mar 22 at 0:43
1
$begingroup$
@TheoBendit It is easy to come up with examples like that. One way is to use the fact that the product $sigma(ij)$ either splits a cycle containing both $i$ and $j$ in two, or fuses together the two cycles containing $i$ and $j$. All according to whether $i$ and $j$ were in separate cycles or not to begin with. So for example $$(123456)(35)=(1236)(45),$$ splitting the 6-cycle containing both $3$ and $5$, and then $$(1236)(45)(34)=(123546),$$ fusing as $3$ and $4$ were in separate cycles of $(1236)(45)$.
$endgroup$
– Jyrki Lahtonen
Mar 22 at 7:06
1
$begingroup$
(cont'd) Anyway, we get non-uniqueness like $$(123456)(35)(34)=(123546)(78)(78).$$ Other kind of non-uniqueness also comes from different "fusions", such as $$(123456)(78)(69)=(1234569)(78)=(234569)(78)(91)$$
$endgroup$
– Jyrki Lahtonen
Mar 22 at 7:10
2
2
$begingroup$
If you're allowing overlap - why not allow the 6-cycle to overlap with the 2-cycles?
$endgroup$
– jmerry
Mar 21 at 19:55
$begingroup$
If you're allowing overlap - why not allow the 6-cycle to overlap with the 2-cycles?
$endgroup$
– jmerry
Mar 21 at 19:55
1
1
$begingroup$
What throws a spanner in the works is that it may not be the case that permutations in that form have a unique representation in that form; there may be different $6$ and $2$ cycles that produce the same permutation. (I don't have a counterexample, so there may not be as well!)
$endgroup$
– Theo Bendit
Mar 21 at 20:03
$begingroup$
What throws a spanner in the works is that it may not be the case that permutations in that form have a unique representation in that form; there may be different $6$ and $2$ cycles that produce the same permutation. (I don't have a counterexample, so there may not be as well!)
$endgroup$
– Theo Bendit
Mar 21 at 20:03
1
1
$begingroup$
I suspect something has gone horribly wrong in the formulation of this question. As it happens, the number of elements in $S_9$ which can be written in the form $abc$ where $a$ is a $6$-cycle and $b$ and $c$ are $2$-cycles turns out to be $160740 = 2^2 cdot 3^2 cdot 5 cdot 19 cdot 47$. But that's almost certainly not what the intended question is. That's approximately 88.59% of all elements in $S_9$ which are not in $A_9$ (certainly all the terms are odd so the product is not in $A_9$).
$endgroup$
– user655377
Mar 22 at 0:43
$begingroup$
I suspect something has gone horribly wrong in the formulation of this question. As it happens, the number of elements in $S_9$ which can be written in the form $abc$ where $a$ is a $6$-cycle and $b$ and $c$ are $2$-cycles turns out to be $160740 = 2^2 cdot 3^2 cdot 5 cdot 19 cdot 47$. But that's almost certainly not what the intended question is. That's approximately 88.59% of all elements in $S_9$ which are not in $A_9$ (certainly all the terms are odd so the product is not in $A_9$).
$endgroup$
– user655377
Mar 22 at 0:43
1
1
$begingroup$
@TheoBendit It is easy to come up with examples like that. One way is to use the fact that the product $sigma(ij)$ either splits a cycle containing both $i$ and $j$ in two, or fuses together the two cycles containing $i$ and $j$. All according to whether $i$ and $j$ were in separate cycles or not to begin with. So for example $$(123456)(35)=(1236)(45),$$ splitting the 6-cycle containing both $3$ and $5$, and then $$(1236)(45)(34)=(123546),$$ fusing as $3$ and $4$ were in separate cycles of $(1236)(45)$.
$endgroup$
– Jyrki Lahtonen
Mar 22 at 7:06
$begingroup$
@TheoBendit It is easy to come up with examples like that. One way is to use the fact that the product $sigma(ij)$ either splits a cycle containing both $i$ and $j$ in two, or fuses together the two cycles containing $i$ and $j$. All according to whether $i$ and $j$ were in separate cycles or not to begin with. So for example $$(123456)(35)=(1236)(45),$$ splitting the 6-cycle containing both $3$ and $5$, and then $$(1236)(45)(34)=(123546),$$ fusing as $3$ and $4$ were in separate cycles of $(1236)(45)$.
$endgroup$
– Jyrki Lahtonen
Mar 22 at 7:06
1
1
$begingroup$
(cont'd) Anyway, we get non-uniqueness like $$(123456)(35)(34)=(123546)(78)(78).$$ Other kind of non-uniqueness also comes from different "fusions", such as $$(123456)(78)(69)=(1234569)(78)=(234569)(78)(91)$$
$endgroup$
– Jyrki Lahtonen
Mar 22 at 7:10
$begingroup$
(cont'd) Anyway, we get non-uniqueness like $$(123456)(35)(34)=(123546)(78)(78).$$ Other kind of non-uniqueness also comes from different "fusions", such as $$(123456)(78)(69)=(1234569)(78)=(234569)(78)(91)$$
$endgroup$
– Jyrki Lahtonen
Mar 22 at 7:10
|
show 2 more comments
0
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oldest
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2
$begingroup$
If you're allowing overlap - why not allow the 6-cycle to overlap with the 2-cycles?
$endgroup$
– jmerry
Mar 21 at 19:55
1
$begingroup$
What throws a spanner in the works is that it may not be the case that permutations in that form have a unique representation in that form; there may be different $6$ and $2$ cycles that produce the same permutation. (I don't have a counterexample, so there may not be as well!)
$endgroup$
– Theo Bendit
Mar 21 at 20:03
1
$begingroup$
I suspect something has gone horribly wrong in the formulation of this question. As it happens, the number of elements in $S_9$ which can be written in the form $abc$ where $a$ is a $6$-cycle and $b$ and $c$ are $2$-cycles turns out to be $160740 = 2^2 cdot 3^2 cdot 5 cdot 19 cdot 47$. But that's almost certainly not what the intended question is. That's approximately 88.59% of all elements in $S_9$ which are not in $A_9$ (certainly all the terms are odd so the product is not in $A_9$).
$endgroup$
– user655377
Mar 22 at 0:43
1
$begingroup$
@TheoBendit It is easy to come up with examples like that. One way is to use the fact that the product $sigma(ij)$ either splits a cycle containing both $i$ and $j$ in two, or fuses together the two cycles containing $i$ and $j$. All according to whether $i$ and $j$ were in separate cycles or not to begin with. So for example $$(123456)(35)=(1236)(45),$$ splitting the 6-cycle containing both $3$ and $5$, and then $$(1236)(45)(34)=(123546),$$ fusing as $3$ and $4$ were in separate cycles of $(1236)(45)$.
$endgroup$
– Jyrki Lahtonen
Mar 22 at 7:06
1
$begingroup$
(cont'd) Anyway, we get non-uniqueness like $$(123456)(35)(34)=(123546)(78)(78).$$ Other kind of non-uniqueness also comes from different "fusions", such as $$(123456)(78)(69)=(1234569)(78)=(234569)(78)(91)$$
$endgroup$
– Jyrki Lahtonen
Mar 22 at 7:10