Cycles and symmetric groups [closed]
$begingroup$
Let $n in N$ and $S_n$ be the symmetric group on $n$.
(A)
Let $pi in S_n$ and $z$ be the number of disjoint cycles of $pi$ (here the 1-cycles are counted). Show that then
${rm sgn} pi = (-1)^{n-z}$
(B)
Show that the subset
To: = {π∈Sn|sgn (π) = 1} ⊆Sn
is a subset of Sn.
(C)
Determine the number of elements $|A_n|$ the subgroup $A_n$ from part (b).
group-theory symmetric-groups
edited Dec 21 '18 at 21:39
Shaun
9,221113684
asked Dec 20 '18 at 12:50
ireallydontknowitireallydontknowit
11
$endgroup$
closed as off-topic by amWhy, Mike Pierce, user10354138, Paul Frost, Jyrki Lahtonen Dec 20 '18 at 19:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Mike Pierce, user10354138, Paul Frost, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $n in N$ and $S_n$ be the symmetric group on $n$.
(A)
Let $pi in S_n$ and $z$ be the number of disjoint cycles of $pi$ (here the 1-cycles are counted). Show that then
${rm sgn} pi = (-1)^{n-z}$
(B)
Show that the subset
To: = {π∈Sn|sgn (π) = 1} ⊆Sn
is a subset of Sn.
(C)
Determine the number of elements $|A_n|$ the subgroup $A_n$ from part (b).
group-theory symmetric-groups
edited Dec 21 '18 at 21:39
Shaun
9,221113684
asked Dec 20 '18 at 12:50
ireallydontknowitireallydontknowit
11
$endgroup$
closed as off-topic by amWhy, Mike Pierce, user10354138, Paul Frost, Jyrki Lahtonen Dec 20 '18 at 19:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Mike Pierce, user10354138, Paul Frost, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
For part (A) I'm partial to this argument. In other words, this question is also a duplicate. Askers are recommended to spend a while searching the site. Chances are that a standard question like this has already been covered.
$endgroup$
– Jyrki Lahtonen
Dec 20 '18 at 19:12
add a comment |
$begingroup$
Let $n in N$ and $S_n$ be the symmetric group on $n$.
(A)
Let $pi in S_n$ and $z$ be the number of disjoint cycles of $pi$ (here the 1-cycles are counted). Show that then
${rm sgn} pi = (-1)^{n-z}$
(B)
Show that the subset
To: = {π∈Sn|sgn (π) = 1} ⊆Sn
is a subset of Sn.
(C)
Determine the number of elements $|A_n|$ the subgroup $A_n$ from part (b).
group-theory symmetric-groups
edited Dec 21 '18 at 21:39
Shaun
9,221113684
asked Dec 20 '18 at 12:50
ireallydontknowitireallydontknowit
11
$endgroup$
Let $n in N$ and $S_n$ be the symmetric group on $n$.
(A)
Let $pi in S_n$ and $z$ be the number of disjoint cycles of $pi$ (here the 1-cycles are counted). Show that then
${rm sgn} pi = (-1)^{n-z}$
(B)
Show that the subset
To: = {π∈Sn|sgn (π) = 1} ⊆Sn
is a subset of Sn.
(C)
Determine the number of elements $|A_n|$ the subgroup $A_n$ from part (b).
group-theory symmetric-groups
group-theory symmetric-groups
edited Dec 21 '18 at 21:39
Shaun
9,221113684
asked Dec 20 '18 at 12:50
ireallydontknowitireallydontknowit
11
edited Dec 21 '18 at 21:39
Shaun
9,221113684
asked Dec 20 '18 at 12:50
ireallydontknowitireallydontknowit
11
edited Dec 21 '18 at 21:39
Shaun
9,221113684
edited Dec 21 '18 at 21:39
Shaun
9,221113684
edited Dec 21 '18 at 21:39
Shaun
9,221113684
9,221113684
asked Dec 20 '18 at 12:50
ireallydontknowitireallydontknowit
11
asked Dec 20 '18 at 12:50
ireallydontknowitireallydontknowit
11
asked Dec 20 '18 at 12:50
ireallydontknowitireallydontknowit
11
11
closed as off-topic by amWhy, Mike Pierce, user10354138, Paul Frost, Jyrki Lahtonen Dec 20 '18 at 19:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Mike Pierce, user10354138, Paul Frost, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Mike Pierce, user10354138, Paul Frost, Jyrki Lahtonen Dec 20 '18 at 19:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Mike Pierce, user10354138, Paul Frost, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
For part (A) I'm partial to this argument. In other words, this question is also a duplicate. Askers are recommended to spend a while searching the site. Chances are that a standard question like this has already been covered.
$endgroup$
– Jyrki Lahtonen
Dec 20 '18 at 19:12
add a comment |
$begingroup$
For part (A) I'm partial to this argument. In other words, this question is also a duplicate. Askers are recommended to spend a while searching the site. Chances are that a standard question like this has already been covered.
$endgroup$
– Jyrki Lahtonen
Dec 20 '18 at 19:12
$begingroup$
For part (A) I'm partial to this argument. In other words, this question is also a duplicate. Askers are recommended to spend a while searching the site. Chances are that a standard question like this has already been covered.
$endgroup$
– Jyrki Lahtonen
Dec 20 '18 at 19:12
$begingroup$
For part (A) I'm partial to this argument. In other words, this question is also a duplicate. Askers are recommended to spend a while searching the site. Chances are that a standard question like this has already been covered.
$endgroup$
– Jyrki Lahtonen
Dec 20 '18 at 19:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For some $piin S_n$, $mathrm{sgn}(pi) = (-1)^r$ where $pi = tau_1tau_2ldotstau_r$ as a product of transpositions. This is well defined (although this is non-trivial to check).
(A) A cycle of length $k$ can be written as a product of $k-1$ transpositions - e.g. $(1 2 3 ldots k) = (1 2)(2 3)ldots(k-1, k)$. If $pi$ is a product of $z$ cycles of length $k_1,ldots,k_z$ respectively then $pi$ is a product of $$sum_{i = 1}^z(k_i-1) = left(sum_1^zk_iright) - z = n-z$$ transpositions and hence $textrm{sgn}(pi) = (-1)^{n-z}$.
(B) Check that $sigma : S_n rightarrow {pm 1}$ where $sigma(pi) = textrm{sgn}(pi)$ is a well defined homomorphism. Then $A_n = {piin S_n : textrm{sgn}(pi) = 1}$ is precisely the kernel of $sigma$ and hence is a subgroup of $S_n$.
(C) Use the $sigma$ from (B) and the first ismorphism theorem to show that $|A_n| = frac{|S_n|}{2}$.
answered Dec 20 '18 at 13:22
ODFODF
1,486510
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For some $piin S_n$, $mathrm{sgn}(pi) = (-1)^r$ where $pi = tau_1tau_2ldotstau_r$ as a product of transpositions. This is well defined (although this is non-trivial to check).
(A) A cycle of length $k$ can be written as a product of $k-1$ transpositions - e.g. $(1 2 3 ldots k) = (1 2)(2 3)ldots(k-1, k)$. If $pi$ is a product of $z$ cycles of length $k_1,ldots,k_z$ respectively then $pi$ is a product of $$sum_{i = 1}^z(k_i-1) = left(sum_1^zk_iright) - z = n-z$$ transpositions and hence $textrm{sgn}(pi) = (-1)^{n-z}$.
(B) Check that $sigma : S_n rightarrow {pm 1}$ where $sigma(pi) = textrm{sgn}(pi)$ is a well defined homomorphism. Then $A_n = {piin S_n : textrm{sgn}(pi) = 1}$ is precisely the kernel of $sigma$ and hence is a subgroup of $S_n$.
(C) Use the $sigma$ from (B) and the first ismorphism theorem to show that $|A_n| = frac{|S_n|}{2}$.
answered Dec 20 '18 at 13:22
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
For some $piin S_n$, $mathrm{sgn}(pi) = (-1)^r$ where $pi = tau_1tau_2ldotstau_r$ as a product of transpositions. This is well defined (although this is non-trivial to check).
(A) A cycle of length $k$ can be written as a product of $k-1$ transpositions - e.g. $(1 2 3 ldots k) = (1 2)(2 3)ldots(k-1, k)$. If $pi$ is a product of $z$ cycles of length $k_1,ldots,k_z$ respectively then $pi$ is a product of $$sum_{i = 1}^z(k_i-1) = left(sum_1^zk_iright) - z = n-z$$ transpositions and hence $textrm{sgn}(pi) = (-1)^{n-z}$.
(B) Check that $sigma : S_n rightarrow {pm 1}$ where $sigma(pi) = textrm{sgn}(pi)$ is a well defined homomorphism. Then $A_n = {piin S_n : textrm{sgn}(pi) = 1}$ is precisely the kernel of $sigma$ and hence is a subgroup of $S_n$.
(C) Use the $sigma$ from (B) and the first ismorphism theorem to show that $|A_n| = frac{|S_n|}{2}$.
answered Dec 20 '18 at 13:22
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
For some $piin S_n$, $mathrm{sgn}(pi) = (-1)^r$ where $pi = tau_1tau_2ldotstau_r$ as a product of transpositions. This is well defined (although this is non-trivial to check).
(A) A cycle of length $k$ can be written as a product of $k-1$ transpositions - e.g. $(1 2 3 ldots k) = (1 2)(2 3)ldots(k-1, k)$. If $pi$ is a product of $z$ cycles of length $k_1,ldots,k_z$ respectively then $pi$ is a product of $$sum_{i = 1}^z(k_i-1) = left(sum_1^zk_iright) - z = n-z$$ transpositions and hence $textrm{sgn}(pi) = (-1)^{n-z}$.
(B) Check that $sigma : S_n rightarrow {pm 1}$ where $sigma(pi) = textrm{sgn}(pi)$ is a well defined homomorphism. Then $A_n = {piin S_n : textrm{sgn}(pi) = 1}$ is precisely the kernel of $sigma$ and hence is a subgroup of $S_n$.
(C) Use the $sigma$ from (B) and the first ismorphism theorem to show that $|A_n| = frac{|S_n|}{2}$.
answered Dec 20 '18 at 13:22
ODFODF
1,486510
$endgroup$
For some $piin S_n$, $mathrm{sgn}(pi) = (-1)^r$ where $pi = tau_1tau_2ldotstau_r$ as a product of transpositions. This is well defined (although this is non-trivial to check).
(A) A cycle of length $k$ can be written as a product of $k-1$ transpositions - e.g. $(1 2 3 ldots k) = (1 2)(2 3)ldots(k-1, k)$. If $pi$ is a product of $z$ cycles of length $k_1,ldots,k_z$ respectively then $pi$ is a product of $$sum_{i = 1}^z(k_i-1) = left(sum_1^zk_iright) - z = n-z$$ transpositions and hence $textrm{sgn}(pi) = (-1)^{n-z}$.
(B) Check that $sigma : S_n rightarrow {pm 1}$ where $sigma(pi) = textrm{sgn}(pi)$ is a well defined homomorphism. Then $A_n = {piin S_n : textrm{sgn}(pi) = 1}$ is precisely the kernel of $sigma$ and hence is a subgroup of $S_n$.
(C) Use the $sigma$ from (B) and the first ismorphism theorem to show that $|A_n| = frac{|S_n|}{2}$.
answered Dec 20 '18 at 13:22
ODFODF
1,486510
answered Dec 20 '18 at 13:22
ODFODF
1,486510
answered Dec 20 '18 at 13:22
ODFODF
1,486510
answered Dec 20 '18 at 13:22
ODFODF
1,486510
1,486510
add a comment |
add a comment |
$begingroup$
For part (A) I'm partial to this argument. In other words, this question is also a duplicate. Askers are recommended to spend a while searching the site. Chances are that a standard question like this has already been covered.
$endgroup$
– Jyrki Lahtonen
Dec 20 '18 at 19:12