Mixing
How many solutions does $x$ have in $x^3 = sigma^6$ in $S_{7}$
$begingroup$
Let $sigma=begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \ 2 & 3 & 5 & 7 & 1 & 4 & 6end{pmatrix} in S_{7}.$
How many solutions does $x^3 = sigma^6$ have?
Here is what I have done so far:
If we decompose $sigma$ into disjoint cycles we get $sigma = (1 2 3 5)(4 7 6)$. From this remark we deduce that $ord(sigma) = lcm(3, 4) = 12$.
Since $ord(sigma) = 12$, this implies that $ord(sigma^6) = dfrac{12}{gcd(12, 6)} = 2$, which results in $ord(x^3) = 2.$
Suppose $m = ord(x)$. We can calculate $m$ from this equation: $ord(x^3) = dfrac{m}{gcd(m, 3)} = 2.$
That equation implies $m = {2, 6}$ and elements of order $2$ and $6$ indeed exist in $S_{7}.$
So basically what I understand from this is that we have an unknown permutation, $x$, which has the order of either $2$ or $6$ which is equal to a permutation of order $2.$
Does this mean that $x$ has $2$ solutions? Sorry if it's obvious but I barely grasp the concepts of modern algebra.
permutations cyclic-groups permutation-cycles
asked Jan 27 at 15:03
bitadeptbitadept
475
$endgroup$
add a comment |
$begingroup$
Let $sigma=begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \ 2 & 3 & 5 & 7 & 1 & 4 & 6end{pmatrix} in S_{7}.$
How many solutions does $x^3 = sigma^6$ have?
Here is what I have done so far:
If we decompose $sigma$ into disjoint cycles we get $sigma = (1 2 3 5)(4 7 6)$. From this remark we deduce that $ord(sigma) = lcm(3, 4) = 12$.
Since $ord(sigma) = 12$, this implies that $ord(sigma^6) = dfrac{12}{gcd(12, 6)} = 2$, which results in $ord(x^3) = 2.$
Suppose $m = ord(x)$. We can calculate $m$ from this equation: $ord(x^3) = dfrac{m}{gcd(m, 3)} = 2.$
That equation implies $m = {2, 6}$ and elements of order $2$ and $6$ indeed exist in $S_{7}.$
So basically what I understand from this is that we have an unknown permutation, $x$, which has the order of either $2$ or $6$ which is equal to a permutation of order $2.$
Does this mean that $x$ has $2$ solutions? Sorry if it's obvious but I barely grasp the concepts of modern algebra.
permutations cyclic-groups permutation-cycles
asked Jan 27 at 15:03
bitadeptbitadept
475
$endgroup$
1
$begingroup$
It's probably better to start by computing $ sigma ^6 = (13)(25)$ and then seeing how many cube roots that product of transpositions has.
$endgroup$
– Ethan Bolker
Jan 27 at 15:20
$begingroup$
@EthanBolker With the cube root of a permutation do you mean something like $sigma^{-3}$?
$endgroup$
– bitadept
Jan 27 at 15:23
1
$begingroup$
I mean experiment to see what permutations satisfy $x^3 = (13)(25)$.
$endgroup$
– Ethan Bolker
Jan 27 at 15:52
add a comment |
$begingroup$
Let $sigma=begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \ 2 & 3 & 5 & 7 & 1 & 4 & 6end{pmatrix} in S_{7}.$
How many solutions does $x^3 = sigma^6$ have?
Here is what I have done so far:
If we decompose $sigma$ into disjoint cycles we get $sigma = (1 2 3 5)(4 7 6)$. From this remark we deduce that $ord(sigma) = lcm(3, 4) = 12$.
Since $ord(sigma) = 12$, this implies that $ord(sigma^6) = dfrac{12}{gcd(12, 6)} = 2$, which results in $ord(x^3) = 2.$
Suppose $m = ord(x)$. We can calculate $m$ from this equation: $ord(x^3) = dfrac{m}{gcd(m, 3)} = 2.$
That equation implies $m = {2, 6}$ and elements of order $2$ and $6$ indeed exist in $S_{7}.$
So basically what I understand from this is that we have an unknown permutation, $x$, which has the order of either $2$ or $6$ which is equal to a permutation of order $2.$
Does this mean that $x$ has $2$ solutions? Sorry if it's obvious but I barely grasp the concepts of modern algebra.
permutations cyclic-groups permutation-cycles
asked Jan 27 at 15:03
bitadeptbitadept
475
$endgroup$
Let $sigma=begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \ 2 & 3 & 5 & 7 & 1 & 4 & 6end{pmatrix} in S_{7}.$
How many solutions does $x^3 = sigma^6$ have?
Here is what I have done so far:
If we decompose $sigma$ into disjoint cycles we get $sigma = (1 2 3 5)(4 7 6)$. From this remark we deduce that $ord(sigma) = lcm(3, 4) = 12$.
Since $ord(sigma) = 12$, this implies that $ord(sigma^6) = dfrac{12}{gcd(12, 6)} = 2$, which results in $ord(x^3) = 2.$
Suppose $m = ord(x)$. We can calculate $m$ from this equation: $ord(x^3) = dfrac{m}{gcd(m, 3)} = 2.$
That equation implies $m = {2, 6}$ and elements of order $2$ and $6$ indeed exist in $S_{7}.$
So basically what I understand from this is that we have an unknown permutation, $x$, which has the order of either $2$ or $6$ which is equal to a permutation of order $2.$
Does this mean that $x$ has $2$ solutions? Sorry if it's obvious but I barely grasp the concepts of modern algebra.
permutations cyclic-groups permutation-cycles
permutations cyclic-groups permutation-cycles
asked Jan 27 at 15:03
bitadeptbitadept
475
asked Jan 27 at 15:03
bitadeptbitadept
475
edited Jan 27 at 15:11
bitadept
asked Jan 27 at 15:03
bitadeptbitadept
475
asked Jan 27 at 15:03
bitadeptbitadept
475
asked Jan 27 at 15:03
bitadeptbitadept
475
475
1
$begingroup$
It's probably better to start by computing $ sigma ^6 = (13)(25)$ and then seeing how many cube roots that product of transpositions has.
$endgroup$
– Ethan Bolker
Jan 27 at 15:20
$begingroup$
@EthanBolker With the cube root of a permutation do you mean something like $sigma^{-3}$?
$endgroup$
– bitadept
Jan 27 at 15:23
1
$begingroup$
I mean experiment to see what permutations satisfy $x^3 = (13)(25)$.
$endgroup$
– Ethan Bolker
Jan 27 at 15:52
add a comment |
1
$begingroup$
It's probably better to start by computing $ sigma ^6 = (13)(25)$ and then seeing how many cube roots that product of transpositions has.
$endgroup$
– Ethan Bolker
Jan 27 at 15:20
$begingroup$
@EthanBolker With the cube root of a permutation do you mean something like $sigma^{-3}$?
$endgroup$
– bitadept
Jan 27 at 15:23
1
$begingroup$
I mean experiment to see what permutations satisfy $x^3 = (13)(25)$.
$endgroup$
– Ethan Bolker
Jan 27 at 15:52
1
1
$begingroup$
It's probably better to start by computing $ sigma ^6 = (13)(25)$ and then seeing how many cube roots that product of transpositions has.
$endgroup$
– Ethan Bolker
Jan 27 at 15:20
$begingroup$
It's probably better to start by computing $ sigma ^6 = (13)(25)$ and then seeing how many cube roots that product of transpositions has.
$endgroup$
– Ethan Bolker
Jan 27 at 15:20
$begingroup$
@EthanBolker With the cube root of a permutation do you mean something like $sigma^{-3}$?
$endgroup$
– bitadept
Jan 27 at 15:23
$begingroup$
@EthanBolker With the cube root of a permutation do you mean something like $sigma^{-3}$?
$endgroup$
– bitadept
Jan 27 at 15:23
1
1
$begingroup$
I mean experiment to see what permutations satisfy $x^3 = (13)(25)$.
$endgroup$
– Ethan Bolker
Jan 27 at 15:52
$begingroup$
I mean experiment to see what permutations satisfy $x^3 = (13)(25)$.
$endgroup$
– Ethan Bolker
Jan 27 at 15:52
add a comment |
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1
$begingroup$
It's probably better to start by computing $ sigma ^6 = (13)(25)$ and then seeing how many cube roots that product of transpositions has.
$endgroup$
– Ethan Bolker
Jan 27 at 15:20
$begingroup$
@EthanBolker With the cube root of a permutation do you mean something like $sigma^{-3}$?
$endgroup$
– bitadept
Jan 27 at 15:23
1
$begingroup$
I mean experiment to see what permutations satisfy $x^3 = (13)(25)$.
$endgroup$
– Ethan Bolker
Jan 27 at 15:52