Mixing
Determine whether the following elements of $S_6$ are of the form $sigma^2$.
$begingroup$
The Problem:
Let $S_6$ be the symmetric group on six letters. Determine whether the following elements of $S_6$ are squares (i.e., of the form $sigma^2$).
(a) $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4)$.
(b) $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5)$.
(c) $(1 hspace{1mm} 2 hspace{1mm} 3)(4 hspace{1mm} 5)$.
My Progress:
(a) is clear. $sigma^2$ must be an even permutation, and $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4)$ is an odd permutation. Therefore there can be no $sigma in S_6$ such that $sigma^2 = (1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4)$.
(b) is not so clear to me. $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5)$ is, indeed, an even permutation, so the parity doesn't help me here. I've pretty much exhausted all of my tools. I'm basically at the trial-and-error point now (e.g., trying to write $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5)$ as the square of different permutations) which is getting me nowhere fast. Basically, I'm not sure what else can be said about an element of the form $sigma^2$ other than the fact that it's even.
(c) is giving me the same issues as (b). [EDIT: Nevermind. $(1 hspace{1mm} 2 hspace{1mm} 3)(4 hspace{1mm} 5)$ is odd since it has an odd number of cycles of even length; so it cannot be of the form $sigma^2$.]
abstract-algebra group-theory symmetric-groups
asked Jan 10 at 1:36
thisisourconcerndudethisisourconcerndude
1,1271022
$endgroup$
|
show 1 more comment
$begingroup$
The Problem:
Let $S_6$ be the symmetric group on six letters. Determine whether the following elements of $S_6$ are squares (i.e., of the form $sigma^2$).
(a) $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4)$.
(b) $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5)$.
(c) $(1 hspace{1mm} 2 hspace{1mm} 3)(4 hspace{1mm} 5)$.
My Progress:
(a) is clear. $sigma^2$ must be an even permutation, and $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4)$ is an odd permutation. Therefore there can be no $sigma in S_6$ such that $sigma^2 = (1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4)$.
(b) is not so clear to me. $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5)$ is, indeed, an even permutation, so the parity doesn't help me here. I've pretty much exhausted all of my tools. I'm basically at the trial-and-error point now (e.g., trying to write $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5)$ as the square of different permutations) which is getting me nowhere fast. Basically, I'm not sure what else can be said about an element of the form $sigma^2$ other than the fact that it's even.
(c) is giving me the same issues as (b). [EDIT: Nevermind. $(1 hspace{1mm} 2 hspace{1mm} 3)(4 hspace{1mm} 5)$ is odd since it has an odd number of cycles of even length; so it cannot be of the form $sigma^2$.]
abstract-algebra group-theory symmetric-groups
asked Jan 10 at 1:36
thisisourconcerndudethisisourconcerndude
1,1271022
$endgroup$
1
$begingroup$
(c) is odd, not even.
$endgroup$
– Theo Bendit
Jan 10 at 1:55
$begingroup$
@TheoBendit Oh, yes. You're right. I'm not sure how I made that mistake. I'll edit to reflect this.
$endgroup$
– thisisourconcerndude
Jan 10 at 1:57
3
$begingroup$
For (b), notice that the permutation is essentially adding $1$ modulo $5$. Maybe look at adding $1/2$ (by which I mean, the multiplicative inverse of $2$) modulo $5$, and see if that becomes a valid square root?
$endgroup$
– Theo Bendit
Jan 10 at 1:59
1
$begingroup$
Any permutation of odd order is a square.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 4:14
1
$begingroup$
Hint: if $tau^5=id$ then $tau=tau^6=(tau^3)^2$.
$endgroup$
– bof
Jan 10 at 13:59
|
show 1 more comment
$begingroup$
The Problem:
Let $S_6$ be the symmetric group on six letters. Determine whether the following elements of $S_6$ are squares (i.e., of the form $sigma^2$).
(a) $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4)$.
(b) $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5)$.
(c) $(1 hspace{1mm} 2 hspace{1mm} 3)(4 hspace{1mm} 5)$.
My Progress:
(a) is clear. $sigma^2$ must be an even permutation, and $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4)$ is an odd permutation. Therefore there can be no $sigma in S_6$ such that $sigma^2 = (1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4)$.
(b) is not so clear to me. $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5)$ is, indeed, an even permutation, so the parity doesn't help me here. I've pretty much exhausted all of my tools. I'm basically at the trial-and-error point now (e.g., trying to write $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5)$ as the square of different permutations) which is getting me nowhere fast. Basically, I'm not sure what else can be said about an element of the form $sigma^2$ other than the fact that it's even.
(c) is giving me the same issues as (b). [EDIT: Nevermind. $(1 hspace{1mm} 2 hspace{1mm} 3)(4 hspace{1mm} 5)$ is odd since it has an odd number of cycles of even length; so it cannot be of the form $sigma^2$.]
abstract-algebra group-theory symmetric-groups
asked Jan 10 at 1:36
thisisourconcerndudethisisourconcerndude
1,1271022
$endgroup$
The Problem:
Let $S_6$ be the symmetric group on six letters. Determine whether the following elements of $S_6$ are squares (i.e., of the form $sigma^2$).
(a) $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4)$.
(b) $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5)$.
(c) $(1 hspace{1mm} 2 hspace{1mm} 3)(4 hspace{1mm} 5)$.
My Progress:
(a) is clear. $sigma^2$ must be an even permutation, and $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4)$ is an odd permutation. Therefore there can be no $sigma in S_6$ such that $sigma^2 = (1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4)$.
(b) is not so clear to me. $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5)$ is, indeed, an even permutation, so the parity doesn't help me here. I've pretty much exhausted all of my tools. I'm basically at the trial-and-error point now (e.g., trying to write $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5)$ as the square of different permutations) which is getting me nowhere fast. Basically, I'm not sure what else can be said about an element of the form $sigma^2$ other than the fact that it's even.
(c) is giving me the same issues as (b). [EDIT: Nevermind. $(1 hspace{1mm} 2 hspace{1mm} 3)(4 hspace{1mm} 5)$ is odd since it has an odd number of cycles of even length; so it cannot be of the form $sigma^2$.]
abstract-algebra group-theory symmetric-groups
abstract-algebra group-theory symmetric-groups
asked Jan 10 at 1:36
thisisourconcerndudethisisourconcerndude
1,1271022
asked Jan 10 at 1:36
thisisourconcerndudethisisourconcerndude
1,1271022
edited Jan 10 at 1:59
thisisourconcerndude
asked Jan 10 at 1:36
thisisourconcerndudethisisourconcerndude
1,1271022
asked Jan 10 at 1:36
thisisourconcerndudethisisourconcerndude
1,1271022
asked Jan 10 at 1:36
thisisourconcerndudethisisourconcerndude
1,1271022
1,1271022
1
$begingroup$
(c) is odd, not even.
$endgroup$
– Theo Bendit
Jan 10 at 1:55
$begingroup$
@TheoBendit Oh, yes. You're right. I'm not sure how I made that mistake. I'll edit to reflect this.
$endgroup$
– thisisourconcerndude
Jan 10 at 1:57
3
$begingroup$
For (b), notice that the permutation is essentially adding $1$ modulo $5$. Maybe look at adding $1/2$ (by which I mean, the multiplicative inverse of $2$) modulo $5$, and see if that becomes a valid square root?
$endgroup$
– Theo Bendit
Jan 10 at 1:59
1
$begingroup$
Any permutation of odd order is a square.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 4:14
1
$begingroup$
Hint: if $tau^5=id$ then $tau=tau^6=(tau^3)^2$.
$endgroup$
– bof
Jan 10 at 13:59
|
show 1 more comment
1
$begingroup$
(c) is odd, not even.
$endgroup$
– Theo Bendit
Jan 10 at 1:55
$begingroup$
@TheoBendit Oh, yes. You're right. I'm not sure how I made that mistake. I'll edit to reflect this.
$endgroup$
– thisisourconcerndude
Jan 10 at 1:57
3
$begingroup$
For (b), notice that the permutation is essentially adding $1$ modulo $5$. Maybe look at adding $1/2$ (by which I mean, the multiplicative inverse of $2$) modulo $5$, and see if that becomes a valid square root?
$endgroup$
– Theo Bendit
Jan 10 at 1:59
1
$begingroup$
Any permutation of odd order is a square.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 4:14
1
$begingroup$
Hint: if $tau^5=id$ then $tau=tau^6=(tau^3)^2$.
$endgroup$
– bof
Jan 10 at 13:59
1
1
$begingroup$
(c) is odd, not even.
$endgroup$
– Theo Bendit
Jan 10 at 1:55
$begingroup$
(c) is odd, not even.
$endgroup$
– Theo Bendit
Jan 10 at 1:55
$begingroup$
@TheoBendit Oh, yes. You're right. I'm not sure how I made that mistake. I'll edit to reflect this.
$endgroup$
– thisisourconcerndude
Jan 10 at 1:57
$begingroup$
@TheoBendit Oh, yes. You're right. I'm not sure how I made that mistake. I'll edit to reflect this.
$endgroup$
– thisisourconcerndude
Jan 10 at 1:57
3
3
$begingroup$
For (b), notice that the permutation is essentially adding $1$ modulo $5$. Maybe look at adding $1/2$ (by which I mean, the multiplicative inverse of $2$) modulo $5$, and see if that becomes a valid square root?
$endgroup$
– Theo Bendit
Jan 10 at 1:59
$begingroup$
For (b), notice that the permutation is essentially adding $1$ modulo $5$. Maybe look at adding $1/2$ (by which I mean, the multiplicative inverse of $2$) modulo $5$, and see if that becomes a valid square root?
$endgroup$
– Theo Bendit
Jan 10 at 1:59
1
1
$begingroup$
Any permutation of odd order is a square.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 4:14
$begingroup$
Any permutation of odd order is a square.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 4:14
1
1
$begingroup$
Hint: if $tau^5=id$ then $tau=tau^6=(tau^3)^2$.
$endgroup$
– bof
Jan 10 at 13:59
$begingroup$
Hint: if $tau^5=id$ then $tau=tau^6=(tau^3)^2$.
$endgroup$
– bof
Jan 10 at 13:59
|
show 1 more comment
2 Answers
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$begingroup$
Hint: If $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5) = sigma^2$ , then $sigma^{10}$ is the identity, so its order must divide 10. Also, $sigma$ can be written as a product of disjoint cycles. What would its order be as a function of the orders of those cycles? What would the orders of those cycles have to be? Is it possible for a product of disjoint cycles of those orders to have its square equal to $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5)$ ?
Edit:
In light of the observation of Lord shark the unknown's in his first comment above (which follows from the fact that if $p^{2n+1} = I$ then $p = p^{2n+2} = left(p^{n+1}right)^2$ ), trying to answer the questions I posed above turns out to be a rather clumsy way of tackling the problem.
answered Jan 10 at 3:46
lonza leggieralonza leggiera
5416
$endgroup$
add a comment |
$begingroup$
$(12345)$ generates a cyclic subgroup $H$ of order $5$. Since the squares of an abelian group are a subgroup, the squares $S$ in $H$ are a subgroup of $H$. Since $(12345)^2 in S$, we have $S neq {id}$. Since $5$ is prime, it follows by lagrange’s theorem that $|S|=5$, so $H=S$, and we are done.
answered Jan 10 at 14:50
M. VanM. Van
2,645311
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Hint: If $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5) = sigma^2$ , then $sigma^{10}$ is the identity, so its order must divide 10. Also, $sigma$ can be written as a product of disjoint cycles. What would its order be as a function of the orders of those cycles? What would the orders of those cycles have to be? Is it possible for a product of disjoint cycles of those orders to have its square equal to $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5)$ ?
Edit:
In light of the observation of Lord shark the unknown's in his first comment above (which follows from the fact that if $p^{2n+1} = I$ then $p = p^{2n+2} = left(p^{n+1}right)^2$ ), trying to answer the questions I posed above turns out to be a rather clumsy way of tackling the problem.
answered Jan 10 at 3:46
lonza leggieralonza leggiera
5416
$endgroup$
add a comment |
$begingroup$
Hint: If $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5) = sigma^2$ , then $sigma^{10}$ is the identity, so its order must divide 10. Also, $sigma$ can be written as a product of disjoint cycles. What would its order be as a function of the orders of those cycles? What would the orders of those cycles have to be? Is it possible for a product of disjoint cycles of those orders to have its square equal to $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5)$ ?
Edit:
In light of the observation of Lord shark the unknown's in his first comment above (which follows from the fact that if $p^{2n+1} = I$ then $p = p^{2n+2} = left(p^{n+1}right)^2$ ), trying to answer the questions I posed above turns out to be a rather clumsy way of tackling the problem.
answered Jan 10 at 3:46
lonza leggieralonza leggiera
5416
$endgroup$
add a comment |
$begingroup$
Hint: If $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5) = sigma^2$ , then $sigma^{10}$ is the identity, so its order must divide 10. Also, $sigma$ can be written as a product of disjoint cycles. What would its order be as a function of the orders of those cycles? What would the orders of those cycles have to be? Is it possible for a product of disjoint cycles of those orders to have its square equal to $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5)$ ?
Edit:
In light of the observation of Lord shark the unknown's in his first comment above (which follows from the fact that if $p^{2n+1} = I$ then $p = p^{2n+2} = left(p^{n+1}right)^2$ ), trying to answer the questions I posed above turns out to be a rather clumsy way of tackling the problem.
answered Jan 10 at 3:46
lonza leggieralonza leggiera
5416
$endgroup$
Hint: If $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5) = sigma^2$ , then $sigma^{10}$ is the identity, so its order must divide 10. Also, $sigma$ can be written as a product of disjoint cycles. What would its order be as a function of the orders of those cycles? What would the orders of those cycles have to be? Is it possible for a product of disjoint cycles of those orders to have its square equal to $(1 hspace{1mm} 2 hspace{1mm} 3 hspace{1mm} 4 hspace{1mm} 5)$ ?
Edit:
In light of the observation of Lord shark the unknown's in his first comment above (which follows from the fact that if $p^{2n+1} = I$ then $p = p^{2n+2} = left(p^{n+1}right)^2$ ), trying to answer the questions I posed above turns out to be a rather clumsy way of tackling the problem.
answered Jan 10 at 3:46
lonza leggieralonza leggiera
5416
edited Jan 10 at 13:55
answered Jan 10 at 3:46
lonza leggieralonza leggiera
5416
answered Jan 10 at 3:46
lonza leggieralonza leggiera
5416
answered Jan 10 at 3:46
lonza leggieralonza leggiera
5416
5416
add a comment |
add a comment |
$begingroup$
$(12345)$ generates a cyclic subgroup $H$ of order $5$. Since the squares of an abelian group are a subgroup, the squares $S$ in $H$ are a subgroup of $H$. Since $(12345)^2 in S$, we have $S neq {id}$. Since $5$ is prime, it follows by lagrange’s theorem that $|S|=5$, so $H=S$, and we are done.
answered Jan 10 at 14:50
M. VanM. Van
2,645311
$endgroup$
add a comment |
$begingroup$
$(12345)$ generates a cyclic subgroup $H$ of order $5$. Since the squares of an abelian group are a subgroup, the squares $S$ in $H$ are a subgroup of $H$. Since $(12345)^2 in S$, we have $S neq {id}$. Since $5$ is prime, it follows by lagrange’s theorem that $|S|=5$, so $H=S$, and we are done.
answered Jan 10 at 14:50
M. VanM. Van
2,645311
$endgroup$
add a comment |
$begingroup$
$(12345)$ generates a cyclic subgroup $H$ of order $5$. Since the squares of an abelian group are a subgroup, the squares $S$ in $H$ are a subgroup of $H$. Since $(12345)^2 in S$, we have $S neq {id}$. Since $5$ is prime, it follows by lagrange’s theorem that $|S|=5$, so $H=S$, and we are done.
answered Jan 10 at 14:50
M. VanM. Van
2,645311
$endgroup$
$(12345)$ generates a cyclic subgroup $H$ of order $5$. Since the squares of an abelian group are a subgroup, the squares $S$ in $H$ are a subgroup of $H$. Since $(12345)^2 in S$, we have $S neq {id}$. Since $5$ is prime, it follows by lagrange’s theorem that $|S|=5$, so $H=S$, and we are done.
answered Jan 10 at 14:50
M. VanM. Van
2,645311
answered Jan 10 at 14:50
M. VanM. Van
2,645311
answered Jan 10 at 14:50
M. VanM. Van
2,645311
answered Jan 10 at 14:50
M. VanM. Van
2,645311
2,645311
add a comment |
add a comment |
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1
$begingroup$
(c) is odd, not even.
$endgroup$
– Theo Bendit
Jan 10 at 1:55
$begingroup$
@TheoBendit Oh, yes. You're right. I'm not sure how I made that mistake. I'll edit to reflect this.
$endgroup$
– thisisourconcerndude
Jan 10 at 1:57
3
$begingroup$
For (b), notice that the permutation is essentially adding $1$ modulo $5$. Maybe look at adding $1/2$ (by which I mean, the multiplicative inverse of $2$) modulo $5$, and see if that becomes a valid square root?
$endgroup$
– Theo Bendit
Jan 10 at 1:59
1
$begingroup$
Any permutation of odd order is a square.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 4:14
1
$begingroup$
Hint: if $tau^5=id$ then $tau=tau^6=(tau^3)^2$.
$endgroup$
– bof
Jan 10 at 13:59