Mixing
Let H be a subgroup of G. Show that if G/H is Abelian, then ghg^-1h^-1 is in H for all g, h in G.
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Let H be a subgroup of G. Show that if G/H is Abelian, then ghg^-1h^-1 is in H for all g, h in G.
I know if G/H is Abelian, (g1H)(g2H)=(g2H)(g1H). But I don't know how to approach from here.
abstract-algebra
asked 18 hours ago
david D
835
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^-1h^-1 is in H for all g, h in G.
I know if G/H is Abelian, (g1H)(g2H)=(g2H)(g1H). But I don't know how to approach from here.
abstract-algebra
asked 18 hours ago
david D
835
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^-1h^-1 is in H for all g, h in G.
I know if G/H is Abelian, (g1H)(g2H)=(g2H)(g1H). But I don't know how to approach from here.
abstract-algebra
asked 18 hours ago
david D
835
Let H be a subgroup of G. Show that if G/H is Abelian, then ghg^-1h^-1 is in H for all g, h in G.
I know if G/H is Abelian, (g1H)(g2H)=(g2H)(g1H). But I don't know how to approach from here.
abstract-algebra
abstract-algebra
asked 18 hours ago
david D
835
asked 18 hours ago
david D
835
asked 18 hours ago
david D
835
asked 18 hours ago
david D
835
asked 18 hours ago
david D
835
835
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 17 hours ago
Ted Shifrin
62k44489
Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
– david D
17 hours ago
1
If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
– Ted Shifrin
17 hours ago
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 17 hours ago
Ted Shifrin
62k44489
Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
– david D
17 hours ago
1
If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
– Ted Shifrin
17 hours ago
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 17 hours ago
Ted Shifrin
62k44489
Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
– david D
17 hours ago
1
If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
– Ted Shifrin
17 hours ago
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 17 hours ago
Ted Shifrin
62k44489
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 17 hours ago
Ted Shifrin
62k44489
answered 17 hours ago
Ted Shifrin
62k44489
answered 17 hours ago
Ted Shifrin
62k44489
answered 17 hours ago
Ted Shifrin
62k44489
62k44489
Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
– david D
17 hours ago
1
If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
– Ted Shifrin
17 hours ago
add a comment |
Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
– david D
17 hours ago
1
If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
– Ted Shifrin
17 hours ago
Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
– david D
17 hours ago
Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
– david D
17 hours ago
1
1
If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
– Ted Shifrin
17 hours ago
If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
– Ted Shifrin
17 hours ago
add a comment |
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