Decomposition of a cycle as a product of transpositions
Can someone please explain the rules pertaining to different ways to write a cycle decomposition as products of 2-cycles, an example from textbook:
I understand this $$ (12345) = (54)(53)(52)(51) ,$$ but it also can be written
$$ (12345) = (54)(52)(21)(25)(23)(13) $$ which I don't understand how it can be.
abstract-algebra group-theory permutations
edited Oct 14 '13 at 12:47
user1729
17.2k64085
asked Oct 13 '13 at 13:21
Arief Anbiya
1,3601622
add a comment |
Can someone please explain the rules pertaining to different ways to write a cycle decomposition as products of 2-cycles, an example from textbook:
I understand this $$ (12345) = (54)(53)(52)(51) ,$$ but it also can be written
$$ (12345) = (54)(52)(21)(25)(23)(13) $$ which I don't understand how it can be.
abstract-algebra group-theory permutations
edited Oct 14 '13 at 12:47
user1729
17.2k64085
asked Oct 13 '13 at 13:21
Arief Anbiya
1,3601622
Have you tried writing out the right hand sides and multiplying them?
– Prahlad Vaidyanathan
Oct 13 '13 at 13:36
Now i understand. Thanks for the comment
– Arief Anbiya
Oct 13 '13 at 13:48
add a comment |
Can someone please explain the rules pertaining to different ways to write a cycle decomposition as products of 2-cycles, an example from textbook:
I understand this $$ (12345) = (54)(53)(52)(51) ,$$ but it also can be written
$$ (12345) = (54)(52)(21)(25)(23)(13) $$ which I don't understand how it can be.
abstract-algebra group-theory permutations
edited Oct 14 '13 at 12:47
user1729
17.2k64085
asked Oct 13 '13 at 13:21
Arief Anbiya
1,3601622
Can someone please explain the rules pertaining to different ways to write a cycle decomposition as products of 2-cycles, an example from textbook:
I understand this $$ (12345) = (54)(53)(52)(51) ,$$ but it also can be written
$$ (12345) = (54)(52)(21)(25)(23)(13) $$ which I don't understand how it can be.
abstract-algebra group-theory permutations
abstract-algebra group-theory permutations
edited Oct 14 '13 at 12:47
user1729
17.2k64085
asked Oct 13 '13 at 13:21
Arief Anbiya
1,3601622
edited Oct 14 '13 at 12:47
user1729
17.2k64085
asked Oct 13 '13 at 13:21
Arief Anbiya
1,3601622
edited Oct 14 '13 at 12:47
user1729
17.2k64085
edited Oct 14 '13 at 12:47
user1729
17.2k64085
edited Oct 14 '13 at 12:47
user1729
17.2k64085
17.2k64085
asked Oct 13 '13 at 13:21
Arief Anbiya
1,3601622
asked Oct 13 '13 at 13:21
Arief Anbiya
1,3601622
asked Oct 13 '13 at 13:21
Arief Anbiya
1,3601622
1,3601622
Have you tried writing out the right hand sides and multiplying them?
– Prahlad Vaidyanathan
Oct 13 '13 at 13:36
Now i understand. Thanks for the comment
– Arief Anbiya
Oct 13 '13 at 13:48
add a comment |
Have you tried writing out the right hand sides and multiplying them?
– Prahlad Vaidyanathan
Oct 13 '13 at 13:36
Now i understand. Thanks for the comment
– Arief Anbiya
Oct 13 '13 at 13:48
Have you tried writing out the right hand sides and multiplying them?
– Prahlad Vaidyanathan
Oct 13 '13 at 13:36
Have you tried writing out the right hand sides and multiplying them?
– Prahlad Vaidyanathan
Oct 13 '13 at 13:36
Now i understand. Thanks for the comment
– Arief Anbiya
Oct 13 '13 at 13:48
Now i understand. Thanks for the comment
– Arief Anbiya
Oct 13 '13 at 13:48
add a comment |
3 Answers
3
active
oldest
votes
Notice that $left(54right)left(53right)left(52right)left(51right)$
is actually a composite of permutations. Permution $left(153right)$ for instance sends $1$ to $5$, sends $5$ to $3$ and sends
$3$ to $1$. Here $2$ and $4$ (and eventually others) are untouched,
or equivalently are sent to themselves. Questioning where $1$ will
be sent to by $left(54right)left(53right)left(52right)left(51right)$ we find $left(54right)left(53right)left(52right)left(51right)left(1right)=left(54right)left(53right)left(52right)(5)=left(54right)left(53right)left(2right)=left(54right)left(2right)=2$.
Likewise we find that it sends $2$ to $3$, $3$ to $4$, $4$ to
$5$ and $5$ to $1$. So it can be recognized as permutation $left(12345right)$.
Composite $left(54right)left(52right)left(21right)left(25right)left(23right)left(13right)$
does exactly the same so can also be recognized as that permutation.
answered Oct 13 '13 at 14:05
drhab
98.1k544129
add a comment |
Usually by a "cycle decomposition" one means writing a permutation as a product of disjoint cycles. So neither of your products of transpositions is a cycle decomposition.
More formally, a "product of disjoint cycles" is a product of the form $sigma_1sigma_2cdotssigma_n$ for $nge 0$, where every $sigma_i$ is a cycle and $sigma_i$ is disjoint from $sigma_j$ unless $i=j$.
The case $n=0$ is the empty product which by definition is the identity element, so $e$.
In the case $n=1$ you have only one cycle, which is the product if itself and nothing else! This product automatically one of "disjoint cycles" in the vacuous sense that $i=j$ whenever $1le ile 1$ and $1le jle 1$.
You should have a theorem somewhere describing the uniqueness of cycle decomposition, so $(1,2,3,4,5)$ is itself its own cycle decomposition, and is its only cycle decomposition, except for other ways to write down the same cycle, such as $(3,4,5,1,2)$.
answered Oct 13 '13 at 13:37
Henning Makholm
238k16303538
add a comment |
$(13):(12345)->(32145)$
$(23)(13):(12345)->(23145)$
$(25)(23)(13):(12345)->(53142)$
$(21)(25)(23)(13):(12345)->(53241)$
$(52)(21)(25)(23)(13):(12345)->(23541)$
$(54)(52)(21)(25)(23)(13):(12345)->(23451)$
Since $(12345)$ is identical with $(23451)$,
$(54)(52)(21)(25)(23)(13)=(23451)$
answered Oct 13 '13 at 13:37
Math.StackExchange
2,347920
add a comment |
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3 Answers
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3 Answers
3
active
oldest
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active
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active
oldest
votes
Notice that $left(54right)left(53right)left(52right)left(51right)$
is actually a composite of permutations. Permution $left(153right)$ for instance sends $1$ to $5$, sends $5$ to $3$ and sends
$3$ to $1$. Here $2$ and $4$ (and eventually others) are untouched,
or equivalently are sent to themselves. Questioning where $1$ will
be sent to by $left(54right)left(53right)left(52right)left(51right)$ we find $left(54right)left(53right)left(52right)left(51right)left(1right)=left(54right)left(53right)left(52right)(5)=left(54right)left(53right)left(2right)=left(54right)left(2right)=2$.
Likewise we find that it sends $2$ to $3$, $3$ to $4$, $4$ to
$5$ and $5$ to $1$. So it can be recognized as permutation $left(12345right)$.
Composite $left(54right)left(52right)left(21right)left(25right)left(23right)left(13right)$
does exactly the same so can also be recognized as that permutation.
answered Oct 13 '13 at 14:05
drhab
98.1k544129
add a comment |
Notice that $left(54right)left(53right)left(52right)left(51right)$
is actually a composite of permutations. Permution $left(153right)$ for instance sends $1$ to $5$, sends $5$ to $3$ and sends
$3$ to $1$. Here $2$ and $4$ (and eventually others) are untouched,
or equivalently are sent to themselves. Questioning where $1$ will
be sent to by $left(54right)left(53right)left(52right)left(51right)$ we find $left(54right)left(53right)left(52right)left(51right)left(1right)=left(54right)left(53right)left(52right)(5)=left(54right)left(53right)left(2right)=left(54right)left(2right)=2$.
Likewise we find that it sends $2$ to $3$, $3$ to $4$, $4$ to
$5$ and $5$ to $1$. So it can be recognized as permutation $left(12345right)$.
Composite $left(54right)left(52right)left(21right)left(25right)left(23right)left(13right)$
does exactly the same so can also be recognized as that permutation.
answered Oct 13 '13 at 14:05
drhab
98.1k544129
add a comment |
Notice that $left(54right)left(53right)left(52right)left(51right)$
is actually a composite of permutations. Permution $left(153right)$ for instance sends $1$ to $5$, sends $5$ to $3$ and sends
$3$ to $1$. Here $2$ and $4$ (and eventually others) are untouched,
or equivalently are sent to themselves. Questioning where $1$ will
be sent to by $left(54right)left(53right)left(52right)left(51right)$ we find $left(54right)left(53right)left(52right)left(51right)left(1right)=left(54right)left(53right)left(52right)(5)=left(54right)left(53right)left(2right)=left(54right)left(2right)=2$.
Likewise we find that it sends $2$ to $3$, $3$ to $4$, $4$ to
$5$ and $5$ to $1$. So it can be recognized as permutation $left(12345right)$.
Composite $left(54right)left(52right)left(21right)left(25right)left(23right)left(13right)$
does exactly the same so can also be recognized as that permutation.
answered Oct 13 '13 at 14:05
drhab
98.1k544129
Notice that $left(54right)left(53right)left(52right)left(51right)$
is actually a composite of permutations. Permution $left(153right)$ for instance sends $1$ to $5$, sends $5$ to $3$ and sends
$3$ to $1$. Here $2$ and $4$ (and eventually others) are untouched,
or equivalently are sent to themselves. Questioning where $1$ will
be sent to by $left(54right)left(53right)left(52right)left(51right)$ we find $left(54right)left(53right)left(52right)left(51right)left(1right)=left(54right)left(53right)left(52right)(5)=left(54right)left(53right)left(2right)=left(54right)left(2right)=2$.
Likewise we find that it sends $2$ to $3$, $3$ to $4$, $4$ to
$5$ and $5$ to $1$. So it can be recognized as permutation $left(12345right)$.
Composite $left(54right)left(52right)left(21right)left(25right)left(23right)left(13right)$
does exactly the same so can also be recognized as that permutation.
answered Oct 13 '13 at 14:05
drhab
98.1k544129
answered Oct 13 '13 at 14:05
drhab
98.1k544129
answered Oct 13 '13 at 14:05
drhab
98.1k544129
answered Oct 13 '13 at 14:05
drhab
98.1k544129
98.1k544129
add a comment |
add a comment |
Usually by a "cycle decomposition" one means writing a permutation as a product of disjoint cycles. So neither of your products of transpositions is a cycle decomposition.
More formally, a "product of disjoint cycles" is a product of the form $sigma_1sigma_2cdotssigma_n$ for $nge 0$, where every $sigma_i$ is a cycle and $sigma_i$ is disjoint from $sigma_j$ unless $i=j$.
The case $n=0$ is the empty product which by definition is the identity element, so $e$.
In the case $n=1$ you have only one cycle, which is the product if itself and nothing else! This product automatically one of "disjoint cycles" in the vacuous sense that $i=j$ whenever $1le ile 1$ and $1le jle 1$.
You should have a theorem somewhere describing the uniqueness of cycle decomposition, so $(1,2,3,4,5)$ is itself its own cycle decomposition, and is its only cycle decomposition, except for other ways to write down the same cycle, such as $(3,4,5,1,2)$.
answered Oct 13 '13 at 13:37
Henning Makholm
238k16303538
add a comment |
Usually by a "cycle decomposition" one means writing a permutation as a product of disjoint cycles. So neither of your products of transpositions is a cycle decomposition.
More formally, a "product of disjoint cycles" is a product of the form $sigma_1sigma_2cdotssigma_n$ for $nge 0$, where every $sigma_i$ is a cycle and $sigma_i$ is disjoint from $sigma_j$ unless $i=j$.
The case $n=0$ is the empty product which by definition is the identity element, so $e$.
In the case $n=1$ you have only one cycle, which is the product if itself and nothing else! This product automatically one of "disjoint cycles" in the vacuous sense that $i=j$ whenever $1le ile 1$ and $1le jle 1$.
You should have a theorem somewhere describing the uniqueness of cycle decomposition, so $(1,2,3,4,5)$ is itself its own cycle decomposition, and is its only cycle decomposition, except for other ways to write down the same cycle, such as $(3,4,5,1,2)$.
answered Oct 13 '13 at 13:37
Henning Makholm
238k16303538
add a comment |
Usually by a "cycle decomposition" one means writing a permutation as a product of disjoint cycles. So neither of your products of transpositions is a cycle decomposition.
More formally, a "product of disjoint cycles" is a product of the form $sigma_1sigma_2cdotssigma_n$ for $nge 0$, where every $sigma_i$ is a cycle and $sigma_i$ is disjoint from $sigma_j$ unless $i=j$.
The case $n=0$ is the empty product which by definition is the identity element, so $e$.
In the case $n=1$ you have only one cycle, which is the product if itself and nothing else! This product automatically one of "disjoint cycles" in the vacuous sense that $i=j$ whenever $1le ile 1$ and $1le jle 1$.
You should have a theorem somewhere describing the uniqueness of cycle decomposition, so $(1,2,3,4,5)$ is itself its own cycle decomposition, and is its only cycle decomposition, except for other ways to write down the same cycle, such as $(3,4,5,1,2)$.
answered Oct 13 '13 at 13:37
Henning Makholm
238k16303538
Usually by a "cycle decomposition" one means writing a permutation as a product of disjoint cycles. So neither of your products of transpositions is a cycle decomposition.
More formally, a "product of disjoint cycles" is a product of the form $sigma_1sigma_2cdotssigma_n$ for $nge 0$, where every $sigma_i$ is a cycle and $sigma_i$ is disjoint from $sigma_j$ unless $i=j$.
The case $n=0$ is the empty product which by definition is the identity element, so $e$.
In the case $n=1$ you have only one cycle, which is the product if itself and nothing else! This product automatically one of "disjoint cycles" in the vacuous sense that $i=j$ whenever $1le ile 1$ and $1le jle 1$.
You should have a theorem somewhere describing the uniqueness of cycle decomposition, so $(1,2,3,4,5)$ is itself its own cycle decomposition, and is its only cycle decomposition, except for other ways to write down the same cycle, such as $(3,4,5,1,2)$.
answered Oct 13 '13 at 13:37
Henning Makholm
238k16303538
answered Oct 13 '13 at 13:37
Henning Makholm
238k16303538
answered Oct 13 '13 at 13:37
Henning Makholm
238k16303538
answered Oct 13 '13 at 13:37
Henning Makholm
238k16303538
238k16303538
add a comment |
add a comment |
$(13):(12345)->(32145)$
$(23)(13):(12345)->(23145)$
$(25)(23)(13):(12345)->(53142)$
$(21)(25)(23)(13):(12345)->(53241)$
$(52)(21)(25)(23)(13):(12345)->(23541)$
$(54)(52)(21)(25)(23)(13):(12345)->(23451)$
Since $(12345)$ is identical with $(23451)$,
$(54)(52)(21)(25)(23)(13)=(23451)$
answered Oct 13 '13 at 13:37
Math.StackExchange
2,347920
add a comment |
$(13):(12345)->(32145)$
$(23)(13):(12345)->(23145)$
$(25)(23)(13):(12345)->(53142)$
$(21)(25)(23)(13):(12345)->(53241)$
$(52)(21)(25)(23)(13):(12345)->(23541)$
$(54)(52)(21)(25)(23)(13):(12345)->(23451)$
Since $(12345)$ is identical with $(23451)$,
$(54)(52)(21)(25)(23)(13)=(23451)$
answered Oct 13 '13 at 13:37
Math.StackExchange
2,347920
add a comment |
$(13):(12345)->(32145)$
$(23)(13):(12345)->(23145)$
$(25)(23)(13):(12345)->(53142)$
$(21)(25)(23)(13):(12345)->(53241)$
$(52)(21)(25)(23)(13):(12345)->(23541)$
$(54)(52)(21)(25)(23)(13):(12345)->(23451)$
Since $(12345)$ is identical with $(23451)$,
$(54)(52)(21)(25)(23)(13)=(23451)$
answered Oct 13 '13 at 13:37
Math.StackExchange
2,347920
$(13):(12345)->(32145)$
$(23)(13):(12345)->(23145)$
$(25)(23)(13):(12345)->(53142)$
$(21)(25)(23)(13):(12345)->(53241)$
$(52)(21)(25)(23)(13):(12345)->(23541)$
$(54)(52)(21)(25)(23)(13):(12345)->(23451)$
Since $(12345)$ is identical with $(23451)$,
$(54)(52)(21)(25)(23)(13)=(23451)$
answered Oct 13 '13 at 13:37
Math.StackExchange
2,347920
answered Oct 13 '13 at 13:37
Math.StackExchange
2,347920
answered Oct 13 '13 at 13:37
Math.StackExchange
2,347920
answered Oct 13 '13 at 13:37
Math.StackExchange
2,347920
2,347920
add a comment |
add a comment |
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Have you tried writing out the right hand sides and multiplying them?
– Prahlad Vaidyanathan
Oct 13 '13 at 13:36
Now i understand. Thanks for the comment
– Arief Anbiya
Oct 13 '13 at 13:48