Show that $phi(g)$ is an even permutation
$begingroup$
Let $G$ be a group of order $n$ then $G$ is isomorphic to a subgroup of $S_n$,Denote it by $phi$, $phi:Gto S_n$ be an monomorphism of Cayleys Theorem
Let $gin G$ has order $k$.Show that $phi(g)$ is a product of disjoint cycles of length $k$.Moreovver Show that $phi(g)$ is an even permutation unless $g$ has even order and $langle grangle $ has odd index in $G$.
My Try:
Since $g$ has order $k$ so $phi(g)$ also has order $k$ and hence is either a $k-$ cycle or a product of $k$ cycles.
Given that $g$ has even order and so $phi(g)$ has even order and hence it is a product of odd number of transpositions.
But I need to prove that it is even permutation?
Where am I missing the point?
Please help.
abstract-algebra group-theory finite-groups symmetric-groups
$endgroup$
|
show 2 more comments
$begingroup$
Let $G$ be a group of order $n$ then $G$ is isomorphic to a subgroup of $S_n$,Denote it by $phi$, $phi:Gto S_n$ be an monomorphism of Cayleys Theorem
Let $gin G$ has order $k$.Show that $phi(g)$ is a product of disjoint cycles of length $k$.Moreovver Show that $phi(g)$ is an even permutation unless $g$ has even order and $langle grangle $ has odd index in $G$.
My Try:
Since $g$ has order $k$ so $phi(g)$ also has order $k$ and hence is either a $k-$ cycle or a product of $k$ cycles.
Given that $g$ has even order and so $phi(g)$ has even order and hence it is a product of odd number of transpositions.
But I need to prove that it is even permutation?
Where am I missing the point?
Please help.
abstract-algebra group-theory finite-groups symmetric-groups
$endgroup$
5
$begingroup$
Either you have a different meaning for "isomorphism" than the usual one (perhaps a $;1-1;$ homomorphism=a monomorphism...?), or else there is a mistake in the question, since a group of order $;n;$ cannot be isomorphic with a group of order $;n!;$ , for $;nge3;$ ...
$endgroup$
– DonAntonio
Dec 6 '18 at 15:09
$begingroup$
@DonAntonio;question edited
$endgroup$
– user596656
Dec 6 '18 at 15:14
$begingroup$
Still, where you mention Cayley's theorem it should be $;phi:Gto Kle S_n;$ or something similar, and not what you wrote: $;G;$ cannot be isomorphic with $;S_n;$ .
$endgroup$
– DonAntonio
Dec 6 '18 at 15:20
$begingroup$
@Jean-ClaudeArbaut;why is that not true ,isomorphic image of an element has the same order as the element
$endgroup$
– user596656
Dec 6 '18 at 15:22
2
$begingroup$
It seems that the question is meant to be about the specific homomorphism coming from the usual proof of Cayley's Theorem. That is the one arising from the group acting on itself by left multiplication
$endgroup$
– Tobias Kildetoft
Dec 6 '18 at 15:52
|
show 2 more comments
$begingroup$
Let $G$ be a group of order $n$ then $G$ is isomorphic to a subgroup of $S_n$,Denote it by $phi$, $phi:Gto S_n$ be an monomorphism of Cayleys Theorem
Let $gin G$ has order $k$.Show that $phi(g)$ is a product of disjoint cycles of length $k$.Moreovver Show that $phi(g)$ is an even permutation unless $g$ has even order and $langle grangle $ has odd index in $G$.
My Try:
Since $g$ has order $k$ so $phi(g)$ also has order $k$ and hence is either a $k-$ cycle or a product of $k$ cycles.
Given that $g$ has even order and so $phi(g)$ has even order and hence it is a product of odd number of transpositions.
But I need to prove that it is even permutation?
Where am I missing the point?
Please help.
abstract-algebra group-theory finite-groups symmetric-groups
$endgroup$
Let $G$ be a group of order $n$ then $G$ is isomorphic to a subgroup of $S_n$,Denote it by $phi$, $phi:Gto S_n$ be an monomorphism of Cayleys Theorem
Let $gin G$ has order $k$.Show that $phi(g)$ is a product of disjoint cycles of length $k$.Moreovver Show that $phi(g)$ is an even permutation unless $g$ has even order and $langle grangle $ has odd index in $G$.
My Try:
Since $g$ has order $k$ so $phi(g)$ also has order $k$ and hence is either a $k-$ cycle or a product of $k$ cycles.
Given that $g$ has even order and so $phi(g)$ has even order and hence it is a product of odd number of transpositions.
But I need to prove that it is even permutation?
Where am I missing the point?
Please help.
abstract-algebra group-theory finite-groups symmetric-groups
abstract-algebra group-theory finite-groups symmetric-groups
edited Dec 6 '18 at 15:29
asked Dec 6 '18 at 15:05
user596656
5
$begingroup$
Either you have a different meaning for "isomorphism" than the usual one (perhaps a $;1-1;$ homomorphism=a monomorphism...?), or else there is a mistake in the question, since a group of order $;n;$ cannot be isomorphic with a group of order $;n!;$ , for $;nge3;$ ...
$endgroup$
– DonAntonio
Dec 6 '18 at 15:09
$begingroup$
@DonAntonio;question edited
$endgroup$
– user596656
Dec 6 '18 at 15:14
$begingroup$
Still, where you mention Cayley's theorem it should be $;phi:Gto Kle S_n;$ or something similar, and not what you wrote: $;G;$ cannot be isomorphic with $;S_n;$ .
$endgroup$
– DonAntonio
Dec 6 '18 at 15:20
$begingroup$
@Jean-ClaudeArbaut;why is that not true ,isomorphic image of an element has the same order as the element
$endgroup$
– user596656
Dec 6 '18 at 15:22
2
$begingroup$
It seems that the question is meant to be about the specific homomorphism coming from the usual proof of Cayley's Theorem. That is the one arising from the group acting on itself by left multiplication
$endgroup$
– Tobias Kildetoft
Dec 6 '18 at 15:52
|
show 2 more comments
5
$begingroup$
Either you have a different meaning for "isomorphism" than the usual one (perhaps a $;1-1;$ homomorphism=a monomorphism...?), or else there is a mistake in the question, since a group of order $;n;$ cannot be isomorphic with a group of order $;n!;$ , for $;nge3;$ ...
$endgroup$
– DonAntonio
Dec 6 '18 at 15:09
$begingroup$
@DonAntonio;question edited
$endgroup$
– user596656
Dec 6 '18 at 15:14
$begingroup$
Still, where you mention Cayley's theorem it should be $;phi:Gto Kle S_n;$ or something similar, and not what you wrote: $;G;$ cannot be isomorphic with $;S_n;$ .
$endgroup$
– DonAntonio
Dec 6 '18 at 15:20
$begingroup$
@Jean-ClaudeArbaut;why is that not true ,isomorphic image of an element has the same order as the element
$endgroup$
– user596656
Dec 6 '18 at 15:22
2
$begingroup$
It seems that the question is meant to be about the specific homomorphism coming from the usual proof of Cayley's Theorem. That is the one arising from the group acting on itself by left multiplication
$endgroup$
– Tobias Kildetoft
Dec 6 '18 at 15:52
5
5
$begingroup$
Either you have a different meaning for "isomorphism" than the usual one (perhaps a $;1-1;$ homomorphism=a monomorphism...?), or else there is a mistake in the question, since a group of order $;n;$ cannot be isomorphic with a group of order $;n!;$ , for $;nge3;$ ...
$endgroup$
– DonAntonio
Dec 6 '18 at 15:09
$begingroup$
Either you have a different meaning for "isomorphism" than the usual one (perhaps a $;1-1;$ homomorphism=a monomorphism...?), or else there is a mistake in the question, since a group of order $;n;$ cannot be isomorphic with a group of order $;n!;$ , for $;nge3;$ ...
$endgroup$
– DonAntonio
Dec 6 '18 at 15:09
$begingroup$
@DonAntonio;question edited
$endgroup$
– user596656
Dec 6 '18 at 15:14
$begingroup$
@DonAntonio;question edited
$endgroup$
– user596656
Dec 6 '18 at 15:14
$begingroup$
Still, where you mention Cayley's theorem it should be $;phi:Gto Kle S_n;$ or something similar, and not what you wrote: $;G;$ cannot be isomorphic with $;S_n;$ .
$endgroup$
– DonAntonio
Dec 6 '18 at 15:20
$begingroup$
Still, where you mention Cayley's theorem it should be $;phi:Gto Kle S_n;$ or something similar, and not what you wrote: $;G;$ cannot be isomorphic with $;S_n;$ .
$endgroup$
– DonAntonio
Dec 6 '18 at 15:20
$begingroup$
@Jean-ClaudeArbaut;why is that not true ,isomorphic image of an element has the same order as the element
$endgroup$
– user596656
Dec 6 '18 at 15:22
$begingroup$
@Jean-ClaudeArbaut;why is that not true ,isomorphic image of an element has the same order as the element
$endgroup$
– user596656
Dec 6 '18 at 15:22
2
2
$begingroup$
It seems that the question is meant to be about the specific homomorphism coming from the usual proof of Cayley's Theorem. That is the one arising from the group acting on itself by left multiplication
$endgroup$
– Tobias Kildetoft
Dec 6 '18 at 15:52
$begingroup$
It seems that the question is meant to be about the specific homomorphism coming from the usual proof of Cayley's Theorem. That is the one arising from the group acting on itself by left multiplication
$endgroup$
– Tobias Kildetoft
Dec 6 '18 at 15:52
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
As pointed out in the comments, the homomorphism $phi:Gto Sym(G)$ is the one arising from the action of $G$ on itself by left multiplication. The cycle decomposition of $phi(g)$ is obtained from the orbits of the action of $g$ on $G$. Let $mathcal{X}$ be a set of representatives of the orbits of $g$. Then, as $g$ has order $k$,
$$phi(g)=prod_{xinmathcal{X}}(x,gx,g^2x,ldots,g^{k-1}x)$$
which is a product of disjoint cycles.
Now, cycles of odd length are even. Therefore, if $k$ is odd, $phi(g)in A_n$. If $k$ is even, then $phi(g)in A_n$ if, and only if $|X|$ is even (why?). The result now follows from the following lemma:
Lemma: $|X|=[G:langle grangle]$.
Proof: For $x,yin G$, $g^rx=y$ if, and only if $yx^{-1}inlangle grangle$. Therefore, the map $xmapsto langle grangle x$ defines a bijection between $mathcal{X}$ and the set of right cosets $langle granglebackslash G$.
answered Dec 6 '18 at 18:31
David HillDavid Hill
9,2261619
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add a comment |
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$begingroup$
As pointed out in the comments, the homomorphism $phi:Gto Sym(G)$ is the one arising from the action of $G$ on itself by left multiplication. The cycle decomposition of $phi(g)$ is obtained from the orbits of the action of $g$ on $G$. Let $mathcal{X}$ be a set of representatives of the orbits of $g$. Then, as $g$ has order $k$,
$$phi(g)=prod_{xinmathcal{X}}(x,gx,g^2x,ldots,g^{k-1}x)$$
which is a product of disjoint cycles.
Now, cycles of odd length are even. Therefore, if $k$ is odd, $phi(g)in A_n$. If $k$ is even, then $phi(g)in A_n$ if, and only if $|X|$ is even (why?). The result now follows from the following lemma:
Lemma: $|X|=[G:langle grangle]$.
Proof: For $x,yin G$, $g^rx=y$ if, and only if $yx^{-1}inlangle grangle$. Therefore, the map $xmapsto langle grangle x$ defines a bijection between $mathcal{X}$ and the set of right cosets $langle granglebackslash G$.
answered Dec 6 '18 at 18:31
David HillDavid Hill
9,2261619
$endgroup$
add a comment |
$begingroup$
As pointed out in the comments, the homomorphism $phi:Gto Sym(G)$ is the one arising from the action of $G$ on itself by left multiplication. The cycle decomposition of $phi(g)$ is obtained from the orbits of the action of $g$ on $G$. Let $mathcal{X}$ be a set of representatives of the orbits of $g$. Then, as $g$ has order $k$,
$$phi(g)=prod_{xinmathcal{X}}(x,gx,g^2x,ldots,g^{k-1}x)$$
which is a product of disjoint cycles.
Now, cycles of odd length are even. Therefore, if $k$ is odd, $phi(g)in A_n$. If $k$ is even, then $phi(g)in A_n$ if, and only if $|X|$ is even (why?). The result now follows from the following lemma:
Lemma: $|X|=[G:langle grangle]$.
Proof: For $x,yin G$, $g^rx=y$ if, and only if $yx^{-1}inlangle grangle$. Therefore, the map $xmapsto langle grangle x$ defines a bijection between $mathcal{X}$ and the set of right cosets $langle granglebackslash G$.
answered Dec 6 '18 at 18:31
David HillDavid Hill
9,2261619
$endgroup$
add a comment |
$begingroup$
As pointed out in the comments, the homomorphism $phi:Gto Sym(G)$ is the one arising from the action of $G$ on itself by left multiplication. The cycle decomposition of $phi(g)$ is obtained from the orbits of the action of $g$ on $G$. Let $mathcal{X}$ be a set of representatives of the orbits of $g$. Then, as $g$ has order $k$,
$$phi(g)=prod_{xinmathcal{X}}(x,gx,g^2x,ldots,g^{k-1}x)$$
which is a product of disjoint cycles.
Now, cycles of odd length are even. Therefore, if $k$ is odd, $phi(g)in A_n$. If $k$ is even, then $phi(g)in A_n$ if, and only if $|X|$ is even (why?). The result now follows from the following lemma:
Lemma: $|X|=[G:langle grangle]$.
Proof: For $x,yin G$, $g^rx=y$ if, and only if $yx^{-1}inlangle grangle$. Therefore, the map $xmapsto langle grangle x$ defines a bijection between $mathcal{X}$ and the set of right cosets $langle granglebackslash G$.
answered Dec 6 '18 at 18:31
David HillDavid Hill
9,2261619
$endgroup$
As pointed out in the comments, the homomorphism $phi:Gto Sym(G)$ is the one arising from the action of $G$ on itself by left multiplication. The cycle decomposition of $phi(g)$ is obtained from the orbits of the action of $g$ on $G$. Let $mathcal{X}$ be a set of representatives of the orbits of $g$. Then, as $g$ has order $k$,
$$phi(g)=prod_{xinmathcal{X}}(x,gx,g^2x,ldots,g^{k-1}x)$$
which is a product of disjoint cycles.
Now, cycles of odd length are even. Therefore, if $k$ is odd, $phi(g)in A_n$. If $k$ is even, then $phi(g)in A_n$ if, and only if $|X|$ is even (why?). The result now follows from the following lemma:
Lemma: $|X|=[G:langle grangle]$.
Proof: For $x,yin G$, $g^rx=y$ if, and only if $yx^{-1}inlangle grangle$. Therefore, the map $xmapsto langle grangle x$ defines a bijection between $mathcal{X}$ and the set of right cosets $langle granglebackslash G$.
answered Dec 6 '18 at 18:31
David HillDavid Hill
9,2261619
answered Dec 6 '18 at 18:31
David HillDavid Hill
9,2261619
answered Dec 6 '18 at 18:31
David HillDavid Hill
9,2261619
answered Dec 6 '18 at 18:31
David HillDavid Hill
9,2261619
9,2261619
add a comment |
add a comment |
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5
$begingroup$
Either you have a different meaning for "isomorphism" than the usual one (perhaps a $;1-1;$ homomorphism=a monomorphism...?), or else there is a mistake in the question, since a group of order $;n;$ cannot be isomorphic with a group of order $;n!;$ , for $;nge3;$ ...
$endgroup$
– DonAntonio
Dec 6 '18 at 15:09
$begingroup$
@DonAntonio;question edited
$endgroup$
– user596656
Dec 6 '18 at 15:14
$begingroup$
Still, where you mention Cayley's theorem it should be $;phi:Gto Kle S_n;$ or something similar, and not what you wrote: $;G;$ cannot be isomorphic with $;S_n;$ .
$endgroup$
– DonAntonio
Dec 6 '18 at 15:20
$begingroup$
@Jean-ClaudeArbaut;why is that not true ,isomorphic image of an element has the same order as the element
$endgroup$
– user596656
Dec 6 '18 at 15:22
2
$begingroup$
It seems that the question is meant to be about the specific homomorphism coming from the usual proof of Cayley's Theorem. That is the one arising from the group acting on itself by left multiplication
$endgroup$
– Tobias Kildetoft
Dec 6 '18 at 15:52