Let $H$ be a subgroup of $G$. Show that if $G/H$ is abelian, then $ghg^{-1}h^{-1}$ is in $H$ for all $g, h$...
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Let $H$ be a subgroup of $G$. Show that if $G/H$ is abelian, then $ghg^{-1}h^{-1}$ is in $H$ for all $g, h$ in $G$..
I know if $G/H$ is abelian, $(g_1H)(g_2H)=(g_2H)(g_1H)$. But I don't know how to approach from here.
abstract-algebra
edited Nov 20 at 11:24
1ENİGMA1
960316
asked Nov 19 at 4:34
david D
875
add a comment |
up vote
1
down vote
favorite
Let $H$ be a subgroup of $G$. Show that if $G/H$ is abelian, then $ghg^{-1}h^{-1}$ is in $H$ for all $g, h$ in $G$..
I know if $G/H$ is abelian, $(g_1H)(g_2H)=(g_2H)(g_1H)$. But I don't know how to approach from here.
abstract-algebra
edited Nov 20 at 11:24
1ENİGMA1
960316
asked Nov 19 at 4:34
david D
875
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $H$ be a subgroup of $G$. Show that if $G/H$ is abelian, then $ghg^{-1}h^{-1}$ is in $H$ for all $g, h$ in $G$..
I know if $G/H$ is abelian, $(g_1H)(g_2H)=(g_2H)(g_1H)$. But I don't know how to approach from here.
abstract-algebra
edited Nov 20 at 11:24
1ENİGMA1
960316
asked Nov 19 at 4:34
david D
875
Let $H$ be a subgroup of $G$. Show that if $G/H$ is abelian, then $ghg^{-1}h^{-1}$ is in $H$ for all $g, h$ in $G$..
I know if $G/H$ is abelian, $(g_1H)(g_2H)=(g_2H)(g_1H)$. But I don't know how to approach from here.
abstract-algebra
abstract-algebra
edited Nov 20 at 11:24
1ENİGMA1
960316
asked Nov 19 at 4:34
david D
875
edited Nov 20 at 11:24
1ENİGMA1
960316
asked Nov 19 at 4:34
david D
875
edited Nov 20 at 11:24
1ENİGMA1
960316
edited Nov 20 at 11:24
1ENİGMA1
960316
edited Nov 20 at 11:24
1ENİGMA1
960316
960316
asked Nov 19 at 4:34
david D
875
asked Nov 19 at 4:34
david D
875
asked Nov 19 at 4:34
david D
875
875
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
You have it. So what is $(g_1g_2g_1^{-1}g_2^{-1})H$? (P.S. Probably it's better not to use the letter $h$ for an element of $G$; it suggests an element of $H$. So I think your switching to $g_1$ and $g_2$ was preferable.)
answered Nov 19 at 4:44
Ted Shifrin
62.5k44489
Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
– david D
Nov 19 at 4:56
1
If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
– Ted Shifrin
Nov 19 at 4:59
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You have it. So what is $(g_1g_2g_1^{-1}g_2^{-1})H$? (P.S. Probably it's better not to use the letter $h$ for an element of $G$; it suggests an element of $H$. So I think your switching to $g_1$ and $g_2$ was preferable.)
answered Nov 19 at 4:44
Ted Shifrin
62.5k44489
Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
– david D
Nov 19 at 4:56
1
If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
– Ted Shifrin
Nov 19 at 4:59
add a comment |
up vote
2
down vote
accepted
You have it. So what is $(g_1g_2g_1^{-1}g_2^{-1})H$? (P.S. Probably it's better not to use the letter $h$ for an element of $G$; it suggests an element of $H$. So I think your switching to $g_1$ and $g_2$ was preferable.)
answered Nov 19 at 4:44
Ted Shifrin
62.5k44489
Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
– david D
Nov 19 at 4:56
1
If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
– Ted Shifrin
Nov 19 at 4:59
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You have it. So what is $(g_1g_2g_1^{-1}g_2^{-1})H$? (P.S. Probably it's better not to use the letter $h$ for an element of $G$; it suggests an element of $H$. So I think your switching to $g_1$ and $g_2$ was preferable.)
answered Nov 19 at 4:44
Ted Shifrin
62.5k44489
You have it. So what is $(g_1g_2g_1^{-1}g_2^{-1})H$? (P.S. Probably it's better not to use the letter $h$ for an element of $G$; it suggests an element of $H$. So I think your switching to $g_1$ and $g_2$ was preferable.)
answered Nov 19 at 4:44
Ted Shifrin
62.5k44489
answered Nov 19 at 4:44
Ted Shifrin
62.5k44489
answered Nov 19 at 4:44
Ted Shifrin
62.5k44489
answered Nov 19 at 4:44
Ted Shifrin
62.5k44489
62.5k44489
Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
– david D
Nov 19 at 4:56
1
If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
– Ted Shifrin
Nov 19 at 4:59
add a comment |
Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
– david D
Nov 19 at 4:56
1
If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
– Ted Shifrin
Nov 19 at 4:59
Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
– david D
Nov 19 at 4:56
Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
– david D
Nov 19 at 4:56
1
1
If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
– Ted Shifrin
Nov 19 at 4:59
If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
– Ted Shifrin
Nov 19 at 4:59
add a comment |
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