Let $H$ be a subgroup of a group $G$ and suppose that $g_1,g_2 ∈ G$. Prove that the following conditions are equivalent: The Next CEO of Stack OverflowHow to show a map between sets is well-defined?$G$ is a finite group and $G_1$,$G_2$ are subgroup that $G_1 cap G_2=1$, then $|G_1|.|G_2| big| |G|$?Can one prove $(g_1H)(g_2H) = (g_1g_2)H$, $g_1, g_2 in G$ if and only if $gH = Hg$ for all $g in G$?About the statement $g_1 sim g_2 iff Hg_1=Hg_2$.Prove the equality of left and right coset of a group rigorouslyEquivalence of basic lemmas about cosetsBijection between left and right cosetsProve that if $Hg_1^-1=Hg_2^-1$ then $g_1 H subseteq g_2 H$Let $G=G_1times G_2$, prove $G'=G_1'times G_2'$Quotient group implies normality
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Let $H$ be a subgroup of a group $G$ and suppose that $g_1,g_2 ∈ G$. Prove that the following conditions are equivalent:
The Next CEO of Stack OverflowHow to show a map between sets is well-defined?$G$ is a finite group and $G_1$,$G_2$ are subgroup that $G_1 cap G_2=1$, then $|G_1|.|G_2| big| |G|$?Can one prove $(g_1H)(g_2H) = (g_1g_2)H$, $g_1, g_2 in G$ if and only if $gH = Hg$ for all $g in G$?About the statement $g_1 sim g_2 iff Hg_1=Hg_2$.Prove the equality of left and right coset of a group rigorouslyEquivalence of basic lemmas about cosetsBijection between left and right cosetsProve that if $Hg_1^-1=Hg_2^-1$ then $g_1 H subseteq g_2 H$Let $G=G_1times G_2$, prove $G'=G_1'times G_2'$Quotient group implies normality
$begingroup$
Let $H$ be a subgroup of a group $G$ and suppose that $g_1,g_2 ∈ G$. Prove that the following conditions are equivalent:
(a) $g_1H = g_2H$,
(b) $Hg_1^-1=Hg_2^-1$,
(c) $g_1H subset g_2H$,
(d) $g_2 in g_1H$,
(e) $g_1^-1g_2 in H$.
I'm beyond confused with this problem, all I know is that to prove the conditions are equivalent, I need to show that (a) implies (b), (b) implies (c), (c) implies (d), (d) implies (e), and (e) implies (a).
I have now gotten answers for (a) implying (e) and (e) implying (d). I'm overthinking all of this and am still confused about (d) implying (c) and (c) implying (b). When it comes to (b) implying (a), I thought I was getting somewhere but it doesn't seem to be working.
abstract-algebra group-theory
asked Mar 28 at 0:50
ClaireClaire
896
$endgroup$
add a comment |
$begingroup$
Let $H$ be a subgroup of a group $G$ and suppose that $g_1,g_2 ∈ G$. Prove that the following conditions are equivalent:
(a) $g_1H = g_2H$,
(b) $Hg_1^-1=Hg_2^-1$,
(c) $g_1H subset g_2H$,
(d) $g_2 in g_1H$,
(e) $g_1^-1g_2 in H$.
I'm beyond confused with this problem, all I know is that to prove the conditions are equivalent, I need to show that (a) implies (b), (b) implies (c), (c) implies (d), (d) implies (e), and (e) implies (a).
I have now gotten answers for (a) implying (e) and (e) implying (d). I'm overthinking all of this and am still confused about (d) implying (c) and (c) implying (b). When it comes to (b) implying (a), I thought I was getting somewhere but it doesn't seem to be working.
abstract-algebra group-theory
asked Mar 28 at 0:50
ClaireClaire
896
$endgroup$
$begingroup$
Have you started with anything? What have you tried to get from (a) to (b)?
$endgroup$
– amsmath
Mar 28 at 1:01
add a comment |
$begingroup$
Let $H$ be a subgroup of a group $G$ and suppose that $g_1,g_2 ∈ G$. Prove that the following conditions are equivalent:
(a) $g_1H = g_2H$,
(b) $Hg_1^-1=Hg_2^-1$,
(c) $g_1H subset g_2H$,
(d) $g_2 in g_1H$,
(e) $g_1^-1g_2 in H$.
I'm beyond confused with this problem, all I know is that to prove the conditions are equivalent, I need to show that (a) implies (b), (b) implies (c), (c) implies (d), (d) implies (e), and (e) implies (a).
I have now gotten answers for (a) implying (e) and (e) implying (d). I'm overthinking all of this and am still confused about (d) implying (c) and (c) implying (b). When it comes to (b) implying (a), I thought I was getting somewhere but it doesn't seem to be working.
abstract-algebra group-theory
asked Mar 28 at 0:50
ClaireClaire
896
$endgroup$
Let $H$ be a subgroup of a group $G$ and suppose that $g_1,g_2 ∈ G$. Prove that the following conditions are equivalent:
(a) $g_1H = g_2H$,
(b) $Hg_1^-1=Hg_2^-1$,
(c) $g_1H subset g_2H$,
(d) $g_2 in g_1H$,
(e) $g_1^-1g_2 in H$.
I'm beyond confused with this problem, all I know is that to prove the conditions are equivalent, I need to show that (a) implies (b), (b) implies (c), (c) implies (d), (d) implies (e), and (e) implies (a).
I have now gotten answers for (a) implying (e) and (e) implying (d). I'm overthinking all of this and am still confused about (d) implying (c) and (c) implying (b). When it comes to (b) implying (a), I thought I was getting somewhere but it doesn't seem to be working.
abstract-algebra group-theory
abstract-algebra group-theory
asked Mar 28 at 0:50
ClaireClaire
896
asked Mar 28 at 0:50
ClaireClaire
896
edited Mar 28 at 3:04
Claire
asked Mar 28 at 0:50
ClaireClaire
896
asked Mar 28 at 0:50
ClaireClaire
896
asked Mar 28 at 0:50
ClaireClaire
896
896
$begingroup$
Have you started with anything? What have you tried to get from (a) to (b)?
$endgroup$
– amsmath
Mar 28 at 1:01
add a comment |
$begingroup$
Have you started with anything? What have you tried to get from (a) to (b)?
$endgroup$
– amsmath
Mar 28 at 1:01
$begingroup$
Have you started with anything? What have you tried to get from (a) to (b)?
$endgroup$
– amsmath
Mar 28 at 1:01
$begingroup$
Have you started with anything? What have you tried to get from (a) to (b)?
$endgroup$
– amsmath
Mar 28 at 1:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'll get you started, maybe you're having trouble with the order of the clauses, but you should take the effort to do the rest yourself, as this is a basic question:
Assume $g_1H = g_2H$. Then $g_1^-1g_2H = g_1^-1g_1H = H$ so $g_1^-1g_2 in H$. So (a) implies (e).
Assume (e). Then there is $h in H$ such that $g_1^-1g_2 = h$. Thus $g_2 = g_1hin g_1H$. So (e) implies (d).
answered Mar 28 at 1:05
MariahMariah
2,1431718
$endgroup$
$begingroup$
it’s definitely the order of the clauses, because this seems so much simpler to me when you go in this order
$endgroup$
– Claire
Mar 28 at 1:10
$begingroup$
@Claire I suspected so, as going from a to b is weird without e :)
$endgroup$
– Mariah
Mar 28 at 1:12
$begingroup$
ok im still lost when it comes to finding (d) implies (c) and (c) implies (b) so far. i know how to find (b) implies (a) though @Mariah
$endgroup$
– Claire
Mar 28 at 2:00
$begingroup$
@Claire you should make an equivalence between (e) and (a). Can you see that if you assume that $g_1^-1g_2 in H$ then also $g_2^-1g_1 in H$ as well? Many of the conditions above are symmetric, and will help you proving (d) implies (c) and (c) implies (b) for instance.
$endgroup$
– Mariah
Mar 28 at 2:06
$begingroup$
okay yes i do see how we can assume that
$endgroup$
– Claire
Mar 28 at 2:11
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'll get you started, maybe you're having trouble with the order of the clauses, but you should take the effort to do the rest yourself, as this is a basic question:
Assume $g_1H = g_2H$. Then $g_1^-1g_2H = g_1^-1g_1H = H$ so $g_1^-1g_2 in H$. So (a) implies (e).
Assume (e). Then there is $h in H$ such that $g_1^-1g_2 = h$. Thus $g_2 = g_1hin g_1H$. So (e) implies (d).
answered Mar 28 at 1:05
MariahMariah
2,1431718
$endgroup$
$begingroup$
it’s definitely the order of the clauses, because this seems so much simpler to me when you go in this order
$endgroup$
– Claire
Mar 28 at 1:10
$begingroup$
@Claire I suspected so, as going from a to b is weird without e :)
$endgroup$
– Mariah
Mar 28 at 1:12
$begingroup$
ok im still lost when it comes to finding (d) implies (c) and (c) implies (b) so far. i know how to find (b) implies (a) though @Mariah
$endgroup$
– Claire
Mar 28 at 2:00
$begingroup$
@Claire you should make an equivalence between (e) and (a). Can you see that if you assume that $g_1^-1g_2 in H$ then also $g_2^-1g_1 in H$ as well? Many of the conditions above are symmetric, and will help you proving (d) implies (c) and (c) implies (b) for instance.
$endgroup$
– Mariah
Mar 28 at 2:06
$begingroup$
okay yes i do see how we can assume that
$endgroup$
– Claire
Mar 28 at 2:11
add a comment |
$begingroup$
I'll get you started, maybe you're having trouble with the order of the clauses, but you should take the effort to do the rest yourself, as this is a basic question:
Assume $g_1H = g_2H$. Then $g_1^-1g_2H = g_1^-1g_1H = H$ so $g_1^-1g_2 in H$. So (a) implies (e).
Assume (e). Then there is $h in H$ such that $g_1^-1g_2 = h$. Thus $g_2 = g_1hin g_1H$. So (e) implies (d).
answered Mar 28 at 1:05
MariahMariah
2,1431718
$endgroup$
$begingroup$
it’s definitely the order of the clauses, because this seems so much simpler to me when you go in this order
$endgroup$
– Claire
Mar 28 at 1:10
$begingroup$
@Claire I suspected so, as going from a to b is weird without e :)
$endgroup$
– Mariah
Mar 28 at 1:12
$begingroup$
ok im still lost when it comes to finding (d) implies (c) and (c) implies (b) so far. i know how to find (b) implies (a) though @Mariah
$endgroup$
– Claire
Mar 28 at 2:00
$begingroup$
@Claire you should make an equivalence between (e) and (a). Can you see that if you assume that $g_1^-1g_2 in H$ then also $g_2^-1g_1 in H$ as well? Many of the conditions above are symmetric, and will help you proving (d) implies (c) and (c) implies (b) for instance.
$endgroup$
– Mariah
Mar 28 at 2:06
$begingroup$
okay yes i do see how we can assume that
$endgroup$
– Claire
Mar 28 at 2:11
add a comment |
$begingroup$
I'll get you started, maybe you're having trouble with the order of the clauses, but you should take the effort to do the rest yourself, as this is a basic question:
Assume $g_1H = g_2H$. Then $g_1^-1g_2H = g_1^-1g_1H = H$ so $g_1^-1g_2 in H$. So (a) implies (e).
Assume (e). Then there is $h in H$ such that $g_1^-1g_2 = h$. Thus $g_2 = g_1hin g_1H$. So (e) implies (d).
answered Mar 28 at 1:05
MariahMariah
2,1431718
$endgroup$
I'll get you started, maybe you're having trouble with the order of the clauses, but you should take the effort to do the rest yourself, as this is a basic question:
Assume $g_1H = g_2H$. Then $g_1^-1g_2H = g_1^-1g_1H = H$ so $g_1^-1g_2 in H$. So (a) implies (e).
Assume (e). Then there is $h in H$ such that $g_1^-1g_2 = h$. Thus $g_2 = g_1hin g_1H$. So (e) implies (d).
answered Mar 28 at 1:05
MariahMariah
2,1431718
answered Mar 28 at 1:05
MariahMariah
2,1431718
answered Mar 28 at 1:05
MariahMariah
2,1431718
answered Mar 28 at 1:05
MariahMariah
2,1431718
2,1431718
$begingroup$
it’s definitely the order of the clauses, because this seems so much simpler to me when you go in this order
$endgroup$
– Claire
Mar 28 at 1:10
$begingroup$
@Claire I suspected so, as going from a to b is weird without e :)
$endgroup$
– Mariah
Mar 28 at 1:12
$begingroup$
ok im still lost when it comes to finding (d) implies (c) and (c) implies (b) so far. i know how to find (b) implies (a) though @Mariah
$endgroup$
– Claire
Mar 28 at 2:00
$begingroup$
@Claire you should make an equivalence between (e) and (a). Can you see that if you assume that $g_1^-1g_2 in H$ then also $g_2^-1g_1 in H$ as well? Many of the conditions above are symmetric, and will help you proving (d) implies (c) and (c) implies (b) for instance.
$endgroup$
– Mariah
Mar 28 at 2:06
$begingroup$
okay yes i do see how we can assume that
$endgroup$
– Claire
Mar 28 at 2:11
add a comment |
$begingroup$
it’s definitely the order of the clauses, because this seems so much simpler to me when you go in this order
$endgroup$
– Claire
Mar 28 at 1:10
$begingroup$
@Claire I suspected so, as going from a to b is weird without e :)
$endgroup$
– Mariah
Mar 28 at 1:12
$begingroup$
ok im still lost when it comes to finding (d) implies (c) and (c) implies (b) so far. i know how to find (b) implies (a) though @Mariah
$endgroup$
– Claire
Mar 28 at 2:00
$begingroup$
@Claire you should make an equivalence between (e) and (a). Can you see that if you assume that $g_1^-1g_2 in H$ then also $g_2^-1g_1 in H$ as well? Many of the conditions above are symmetric, and will help you proving (d) implies (c) and (c) implies (b) for instance.
$endgroup$
– Mariah
Mar 28 at 2:06
$begingroup$
okay yes i do see how we can assume that
$endgroup$
– Claire
Mar 28 at 2:11
$begingroup$
it’s definitely the order of the clauses, because this seems so much simpler to me when you go in this order
$endgroup$
– Claire
Mar 28 at 1:10
$begingroup$
it’s definitely the order of the clauses, because this seems so much simpler to me when you go in this order
$endgroup$
– Claire
Mar 28 at 1:10
$begingroup$
@Claire I suspected so, as going from a to b is weird without e :)
$endgroup$
– Mariah
Mar 28 at 1:12
$begingroup$
@Claire I suspected so, as going from a to b is weird without e :)
$endgroup$
– Mariah
Mar 28 at 1:12
$begingroup$
ok im still lost when it comes to finding (d) implies (c) and (c) implies (b) so far. i know how to find (b) implies (a) though @Mariah
$endgroup$
– Claire
Mar 28 at 2:00
$begingroup$
ok im still lost when it comes to finding (d) implies (c) and (c) implies (b) so far. i know how to find (b) implies (a) though @Mariah
$endgroup$
– Claire
Mar 28 at 2:00
$begingroup$
@Claire you should make an equivalence between (e) and (a). Can you see that if you assume that $g_1^-1g_2 in H$ then also $g_2^-1g_1 in H$ as well? Many of the conditions above are symmetric, and will help you proving (d) implies (c) and (c) implies (b) for instance.
$endgroup$
– Mariah
Mar 28 at 2:06
$begingroup$
@Claire you should make an equivalence between (e) and (a). Can you see that if you assume that $g_1^-1g_2 in H$ then also $g_2^-1g_1 in H$ as well? Many of the conditions above are symmetric, and will help you proving (d) implies (c) and (c) implies (b) for instance.
$endgroup$
– Mariah
Mar 28 at 2:06
$begingroup$
okay yes i do see how we can assume that
$endgroup$
– Claire
Mar 28 at 2:11
$begingroup$
okay yes i do see how we can assume that
$endgroup$
– Claire
Mar 28 at 2:11
add a comment |
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$begingroup$
Have you started with anything? What have you tried to get from (a) to (b)?
$endgroup$
– amsmath
Mar 28 at 1:01