Mixing
In a subgroup $Hle G$ of index $2$ every element $xin G$ satisfies $x^2in H$
$begingroup$
I was thinking about this particular proof of the fact
$$[G:H]=2implies g^2in H~forall gin G$$
1) There is a bijection $f:G/Hmapsto Hbackslash G$ defined by $f(xH)=Hx^{-1}$
- It is well defined because if $xH=yH$ for some $x,yin G$ by multiplying by $y^{-1}$ on the left $$y^{-1}xin Hiff y^{-1}xH=Hy^{-1}x=H$$
On the other hand $$Hx^{-1}=Hy^{-1}iff y^{-1}xin H~text{ by multiplying by x on the right}$$
Clearly $f$ is surjective.
By a similar reasoning as before, $f$ is injective:
If $Hx^{-1}=Hy^{-1}$ we have $H=Hy^{-1}ximplies y^{-1}xin Himplies y^{-1}xH=Himplies xH=yH$
2) $H$ has index $2$ so $G$ is the disjoint union of $H$ and $xH$ but by 1) G is also the disjoint union of $H$ and $Hx^{-1}$
Let $g$ be an element of $G$.
If $gin H$ since $H$ is a group $g^2in H$.
If $gnotin H$ then $gin xH=Hx^{-1}$ so $~~exists h_1,h_2in H~:~g=xh_1=h_2x^{-1}$ . $$g^2=xh_1h_2x^{-1}in H$$ because $H$ is normal as a subgroup of index $2$.
Is this a valid proof? I know there are many proofs of this fact but I'm interested in this one.
Edit: I just realized part 1) is unnecessary because there are only two classes so $xH=Hx=Hx^{-1}$... Unless it is not true that $xnotin H$ and $x^{-1}in H$ can be true simultaneously
abstract-algebra group-theory proof-verification normal-subgroups
edited Jan 5 at 12:01
Matt Samuel
37.8k63665
asked Jan 5 at 9:38
John CataldoJohn Cataldo
1,1721316
$endgroup$
add a comment |
$begingroup$
I was thinking about this particular proof of the fact
$$[G:H]=2implies g^2in H~forall gin G$$
1) There is a bijection $f:G/Hmapsto Hbackslash G$ defined by $f(xH)=Hx^{-1}$
- It is well defined because if $xH=yH$ for some $x,yin G$ by multiplying by $y^{-1}$ on the left $$y^{-1}xin Hiff y^{-1}xH=Hy^{-1}x=H$$
On the other hand $$Hx^{-1}=Hy^{-1}iff y^{-1}xin H~text{ by multiplying by x on the right}$$
Clearly $f$ is surjective.
By a similar reasoning as before, $f$ is injective:
If $Hx^{-1}=Hy^{-1}$ we have $H=Hy^{-1}ximplies y^{-1}xin Himplies y^{-1}xH=Himplies xH=yH$
2) $H$ has index $2$ so $G$ is the disjoint union of $H$ and $xH$ but by 1) G is also the disjoint union of $H$ and $Hx^{-1}$
Let $g$ be an element of $G$.
If $gin H$ since $H$ is a group $g^2in H$.
If $gnotin H$ then $gin xH=Hx^{-1}$ so $~~exists h_1,h_2in H~:~g=xh_1=h_2x^{-1}$ . $$g^2=xh_1h_2x^{-1}in H$$ because $H$ is normal as a subgroup of index $2$.
Is this a valid proof? I know there are many proofs of this fact but I'm interested in this one.
Edit: I just realized part 1) is unnecessary because there are only two classes so $xH=Hx=Hx^{-1}$... Unless it is not true that $xnotin H$ and $x^{-1}in H$ can be true simultaneously
abstract-algebra group-theory proof-verification normal-subgroups
edited Jan 5 at 12:01
Matt Samuel
37.8k63665
asked Jan 5 at 9:38
John CataldoJohn Cataldo
1,1721316
$endgroup$
1
$begingroup$
Yes, some arguments are unnessecary, compare with this duplicate for a much shorter version for your proof.
$endgroup$
– Dietrich Burde
Jan 5 at 11:24
add a comment |
$begingroup$
I was thinking about this particular proof of the fact
$$[G:H]=2implies g^2in H~forall gin G$$
1) There is a bijection $f:G/Hmapsto Hbackslash G$ defined by $f(xH)=Hx^{-1}$
- It is well defined because if $xH=yH$ for some $x,yin G$ by multiplying by $y^{-1}$ on the left $$y^{-1}xin Hiff y^{-1}xH=Hy^{-1}x=H$$
On the other hand $$Hx^{-1}=Hy^{-1}iff y^{-1}xin H~text{ by multiplying by x on the right}$$
Clearly $f$ is surjective.
By a similar reasoning as before, $f$ is injective:
If $Hx^{-1}=Hy^{-1}$ we have $H=Hy^{-1}ximplies y^{-1}xin Himplies y^{-1}xH=Himplies xH=yH$
2) $H$ has index $2$ so $G$ is the disjoint union of $H$ and $xH$ but by 1) G is also the disjoint union of $H$ and $Hx^{-1}$
Let $g$ be an element of $G$.
If $gin H$ since $H$ is a group $g^2in H$.
If $gnotin H$ then $gin xH=Hx^{-1}$ so $~~exists h_1,h_2in H~:~g=xh_1=h_2x^{-1}$ . $$g^2=xh_1h_2x^{-1}in H$$ because $H$ is normal as a subgroup of index $2$.
Is this a valid proof? I know there are many proofs of this fact but I'm interested in this one.
Edit: I just realized part 1) is unnecessary because there are only two classes so $xH=Hx=Hx^{-1}$... Unless it is not true that $xnotin H$ and $x^{-1}in H$ can be true simultaneously
abstract-algebra group-theory proof-verification normal-subgroups
edited Jan 5 at 12:01
Matt Samuel
37.8k63665
asked Jan 5 at 9:38
John CataldoJohn Cataldo
1,1721316
$endgroup$
I was thinking about this particular proof of the fact
$$[G:H]=2implies g^2in H~forall gin G$$
1) There is a bijection $f:G/Hmapsto Hbackslash G$ defined by $f(xH)=Hx^{-1}$
- It is well defined because if $xH=yH$ for some $x,yin G$ by multiplying by $y^{-1}$ on the left $$y^{-1}xin Hiff y^{-1}xH=Hy^{-1}x=H$$
On the other hand $$Hx^{-1}=Hy^{-1}iff y^{-1}xin H~text{ by multiplying by x on the right}$$
Clearly $f$ is surjective.
By a similar reasoning as before, $f$ is injective:
If $Hx^{-1}=Hy^{-1}$ we have $H=Hy^{-1}ximplies y^{-1}xin Himplies y^{-1}xH=Himplies xH=yH$
2) $H$ has index $2$ so $G$ is the disjoint union of $H$ and $xH$ but by 1) G is also the disjoint union of $H$ and $Hx^{-1}$
Let $g$ be an element of $G$.
If $gin H$ since $H$ is a group $g^2in H$.
If $gnotin H$ then $gin xH=Hx^{-1}$ so $~~exists h_1,h_2in H~:~g=xh_1=h_2x^{-1}$ . $$g^2=xh_1h_2x^{-1}in H$$ because $H$ is normal as a subgroup of index $2$.
Is this a valid proof? I know there are many proofs of this fact but I'm interested in this one.
Edit: I just realized part 1) is unnecessary because there are only two classes so $xH=Hx=Hx^{-1}$... Unless it is not true that $xnotin H$ and $x^{-1}in H$ can be true simultaneously
abstract-algebra group-theory proof-verification normal-subgroups
abstract-algebra group-theory proof-verification normal-subgroups
edited Jan 5 at 12:01
Matt Samuel
37.8k63665
asked Jan 5 at 9:38
John CataldoJohn Cataldo
1,1721316
edited Jan 5 at 12:01
Matt Samuel
37.8k63665
asked Jan 5 at 9:38
John CataldoJohn Cataldo
1,1721316
edited Jan 5 at 12:01
Matt Samuel
37.8k63665
edited Jan 5 at 12:01
Matt Samuel
37.8k63665
edited Jan 5 at 12:01
Matt Samuel
37.8k63665
37.8k63665
asked Jan 5 at 9:38
John CataldoJohn Cataldo
1,1721316
asked Jan 5 at 9:38
John CataldoJohn Cataldo
1,1721316
asked Jan 5 at 9:38
John CataldoJohn Cataldo
1,1721316
1,1721316
1
$begingroup$
Yes, some arguments are unnessecary, compare with this duplicate for a much shorter version for your proof.
$endgroup$
– Dietrich Burde
Jan 5 at 11:24
add a comment |
1
$begingroup$
Yes, some arguments are unnessecary, compare with this duplicate for a much shorter version for your proof.
$endgroup$
– Dietrich Burde
Jan 5 at 11:24
1
1
$begingroup$
Yes, some arguments are unnessecary, compare with this duplicate for a much shorter version for your proof.
$endgroup$
– Dietrich Burde
Jan 5 at 11:24
$begingroup$
Yes, some arguments are unnessecary, compare with this duplicate for a much shorter version for your proof.
$endgroup$
– Dietrich Burde
Jan 5 at 11:24
add a comment |
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$begingroup$
Yes, some arguments are unnessecary, compare with this duplicate for a much shorter version for your proof.
$endgroup$
– Dietrich Burde
Jan 5 at 11:24