Show that $sgn (f) = 1$ if and only if there is $h ∈ S_n$ such that $f = h ◦ h$.
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I could not show it, I could not define the h. I need help, please.I'm not very good at math.
Show that $sgn (f) = 1$ if and only if there is h ∈ $S_n$ such that $f = h ◦ h$.
Help me.
My proof: -> Suppose that sign(f) = 1 then f is even. Consider $h$ = $h$ o $h^-1$
$h$ o $h$ = (($h$ o $h^-1$) o ($h$ o $h^-1$))(f(x))
= (($h$ o ($h^-1$ o $h$) o $h^-1$))(f(x))
= (($h$ o $1_h$) o $h^-1$))(f(x))
= (($h$ o $h^-1$))(f(x))
= (($h$ ($h^-1(f(x)$))= f(x) =f .
Therefore there is h in $S_n$ such that f = h o h.
<- there is h in $S_n$ such that $f = $h o $h$, if h is in $S_n$, then h is bijective, but $f = $h o $h$,
therefore f is bijective, then ... $sgn(f)=sgn(ho h)= 1$ :(.
linear-algebra abstract-algebra permutations
asked Aug 9 at 23:16
Jean Paul
535
add a comment |Â
up vote
1
down vote
favorite
I could not show it, I could not define the h. I need help, please.I'm not very good at math.
Show that $sgn (f) = 1$ if and only if there is h ∈ $S_n$ such that $f = h ◦ h$.
Help me.
My proof: -> Suppose that sign(f) = 1 then f is even. Consider $h$ = $h$ o $h^-1$
$h$ o $h$ = (($h$ o $h^-1$) o ($h$ o $h^-1$))(f(x))
= (($h$ o ($h^-1$ o $h$) o $h^-1$))(f(x))
= (($h$ o $1_h$) o $h^-1$))(f(x))
= (($h$ o $h^-1$))(f(x))
= (($h$ ($h^-1(f(x)$))= f(x) =f .
Therefore there is h in $S_n$ such that f = h o h.
<- there is h in $S_n$ such that $f = $h o $h$, if h is in $S_n$, then h is bijective, but $f = $h o $h$,
therefore f is bijective, then ... $sgn(f)=sgn(ho h)= 1$ :(.
linear-algebra abstract-algebra permutations
asked Aug 9 at 23:16
Jean Paul
535
Also, I believe the result is not true? See math.stackexchange.com/questions/266569/…
– Mike Earnest
Aug 9 at 23:29
1
Let $f = (1,2)(3,4,5,6)$. Even though $textsign f=+1$, you can show there is no $h$ for which $f=hcirc h$ by considering all the possible cycle structures of a permutation in $hin S_6$, and that for each of them, $hcirc h$ does not have the cycle structure of $f$.
– Mike Earnest
Aug 9 at 23:54
Thank you Mike :)
– Jean Paul
Aug 10 at 0:00
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I could not show it, I could not define the h. I need help, please.I'm not very good at math.
Show that $sgn (f) = 1$ if and only if there is h ∈ $S_n$ such that $f = h ◦ h$.
Help me.
My proof: -> Suppose that sign(f) = 1 then f is even. Consider $h$ = $h$ o $h^-1$
$h$ o $h$ = (($h$ o $h^-1$) o ($h$ o $h^-1$))(f(x))
= (($h$ o ($h^-1$ o $h$) o $h^-1$))(f(x))
= (($h$ o $1_h$) o $h^-1$))(f(x))
= (($h$ o $h^-1$))(f(x))
= (($h$ ($h^-1(f(x)$))= f(x) =f .
Therefore there is h in $S_n$ such that f = h o h.
<- there is h in $S_n$ such that $f = $h o $h$, if h is in $S_n$, then h is bijective, but $f = $h o $h$,
therefore f is bijective, then ... $sgn(f)=sgn(ho h)= 1$ :(.
linear-algebra abstract-algebra permutations
asked Aug 9 at 23:16
Jean Paul
535
I could not show it, I could not define the h. I need help, please.I'm not very good at math.
Show that $sgn (f) = 1$ if and only if there is h ∈ $S_n$ such that $f = h ◦ h$.
Help me.
My proof: -> Suppose that sign(f) = 1 then f is even. Consider $h$ = $h$ o $h^-1$
$h$ o $h$ = (($h$ o $h^-1$) o ($h$ o $h^-1$))(f(x))
= (($h$ o ($h^-1$ o $h$) o $h^-1$))(f(x))
= (($h$ o $1_h$) o $h^-1$))(f(x))
= (($h$ o $h^-1$))(f(x))
= (($h$ ($h^-1(f(x)$))= f(x) =f .
Therefore there is h in $S_n$ such that f = h o h.
<- there is h in $S_n$ such that $f = $h o $h$, if h is in $S_n$, then h is bijective, but $f = $h o $h$,
therefore f is bijective, then ... $sgn(f)=sgn(ho h)= 1$ :(.
linear-algebra abstract-algebra permutations
asked Aug 9 at 23:16
Jean Paul
535
edited Aug 9 at 23:45
asked Aug 9 at 23:16
Jean Paul
535
asked Aug 9 at 23:16
Jean Paul
535
asked Aug 9 at 23:16
Jean Paul
535
535
Also, I believe the result is not true? See math.stackexchange.com/questions/266569/…
– Mike Earnest
Aug 9 at 23:29
1
Let $f = (1,2)(3,4,5,6)$. Even though $textsign f=+1$, you can show there is no $h$ for which $f=hcirc h$ by considering all the possible cycle structures of a permutation in $hin S_6$, and that for each of them, $hcirc h$ does not have the cycle structure of $f$.
– Mike Earnest
Aug 9 at 23:54
Thank you Mike :)
– Jean Paul
Aug 10 at 0:00
add a comment |Â
Also, I believe the result is not true? See math.stackexchange.com/questions/266569/…
– Mike Earnest
Aug 9 at 23:29
1
Let $f = (1,2)(3,4,5,6)$. Even though $textsign f=+1$, you can show there is no $h$ for which $f=hcirc h$ by considering all the possible cycle structures of a permutation in $hin S_6$, and that for each of them, $hcirc h$ does not have the cycle structure of $f$.
– Mike Earnest
Aug 9 at 23:54
Thank you Mike :)
– Jean Paul
Aug 10 at 0:00
Also, I believe the result is not true? See math.stackexchange.com/questions/266569/…
– Mike Earnest
Aug 9 at 23:29
Also, I believe the result is not true? See math.stackexchange.com/questions/266569/…
– Mike Earnest
Aug 9 at 23:29
1
1
Let $f = (1,2)(3,4,5,6)$. Even though $textsign f=+1$, you can show there is no $h$ for which $f=hcirc h$ by considering all the possible cycle structures of a permutation in $hin S_6$, and that for each of them, $hcirc h$ does not have the cycle structure of $f$.
– Mike Earnest
Aug 9 at 23:54
Let $f = (1,2)(3,4,5,6)$. Even though $textsign f=+1$, you can show there is no $h$ for which $f=hcirc h$ by considering all the possible cycle structures of a permutation in $hin S_6$, and that for each of them, $hcirc h$ does not have the cycle structure of $f$.
– Mike Earnest
Aug 9 at 23:54
Thank you Mike :)
– Jean Paul
Aug 10 at 0:00
Thank you Mike :)
– Jean Paul
Aug 10 at 0:00
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Hint: Try to show that there is no $h in S_6$ such that
$$h^2= (1,2) (3,4,5,6)=beginbmatrix 1 & 2 &3 &4 &5 &6 \
2 & 1 &4 &5 &6 &3
endbmatrix$$
answered Aug 9 at 23:53
N. S.
98k5105197
N.S I have a proof, thank you for your time.
– Jean Paul
Aug 10 at 4:35
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint: Try to show that there is no $h in S_6$ such that
$$h^2= (1,2) (3,4,5,6)=beginbmatrix 1 & 2 &3 &4 &5 &6 \
2 & 1 &4 &5 &6 &3
endbmatrix$$
answered Aug 9 at 23:53
N. S.
98k5105197
N.S I have a proof, thank you for your time.
– Jean Paul
Aug 10 at 4:35
add a comment |Â
up vote
2
down vote
accepted
Hint: Try to show that there is no $h in S_6$ such that
$$h^2= (1,2) (3,4,5,6)=beginbmatrix 1 & 2 &3 &4 &5 &6 \
2 & 1 &4 &5 &6 &3
endbmatrix$$
answered Aug 9 at 23:53
N. S.
98k5105197
N.S I have a proof, thank you for your time.
– Jean Paul
Aug 10 at 4:35
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint: Try to show that there is no $h in S_6$ such that
$$h^2= (1,2) (3,4,5,6)=beginbmatrix 1 & 2 &3 &4 &5 &6 \
2 & 1 &4 &5 &6 &3
endbmatrix$$
answered Aug 9 at 23:53
N. S.
98k5105197
Hint: Try to show that there is no $h in S_6$ such that
$$h^2= (1,2) (3,4,5,6)=beginbmatrix 1 & 2 &3 &4 &5 &6 \
2 & 1 &4 &5 &6 &3
endbmatrix$$
answered Aug 9 at 23:53
N. S.
98k5105197
answered Aug 9 at 23:53
N. S.
98k5105197
answered Aug 9 at 23:53
N. S.
98k5105197
answered Aug 9 at 23:53
N. S.
98k5105197
98k5105197
N.S I have a proof, thank you for your time.
– Jean Paul
Aug 10 at 4:35
add a comment |Â
N.S I have a proof, thank you for your time.
– Jean Paul
Aug 10 at 4:35
N.S I have a proof, thank you for your time.
– Jean Paul
Aug 10 at 4:35
N.S I have a proof, thank you for your time.
– Jean Paul
Aug 10 at 4:35
add a comment |Â
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Also, I believe the result is not true? See math.stackexchange.com/questions/266569/…
– Mike Earnest
Aug 9 at 23:29
1
Let $f = (1,2)(3,4,5,6)$. Even though $textsign f=+1$, you can show there is no $h$ for which $f=hcirc h$ by considering all the possible cycle structures of a permutation in $hin S_6$, and that for each of them, $hcirc h$ does not have the cycle structure of $f$.
– Mike Earnest
Aug 9 at 23:54
Thank you Mike :)
– Jean Paul
Aug 10 at 0:00