If $H$ and $K$ are subgroups of $G$, and the index of $K$ in $G$ finite, then $[ H : H cap K ] leq [G:K]$
$begingroup$
I'm working on this problem and may have found a solution, but I have a step which is not rigorous enough.
My answer: Since $H$ and $K$ are subgroups of $G$, we know that $$|HK|cdot |H cap K|=|H|cdot |K|$$
$$frac{|HK|}{|K|}=frac{|H|}{|H cap K|}$$
Since $K$ is a subgroup of $HK$, and $H cap K$ a subgroup of $H$:
$$[H:H cap K]=[HK:K]stackrel{(*)}{=} [HK:G][G:K]leq[G:K]$$
My questions: Does the step $(*)$ hold? If $G neq HK$ is $[HK:G]$ well defined? Can we also conclude that there is equality iff $HK = G$?
abstract-algebra group-theory
asked Jan 6 at 7:28
NotAbelianGroupNotAbelianGroup
18211
$endgroup$
|
show 2 more comments
$begingroup$
I'm working on this problem and may have found a solution, but I have a step which is not rigorous enough.
My answer: Since $H$ and $K$ are subgroups of $G$, we know that $$|HK|cdot |H cap K|=|H|cdot |K|$$
$$frac{|HK|}{|K|}=frac{|H|}{|H cap K|}$$
Since $K$ is a subgroup of $HK$, and $H cap K$ a subgroup of $H$:
$$[H:H cap K]=[HK:K]stackrel{(*)}{=} [HK:G][G:K]leq[G:K]$$
My questions: Does the step $(*)$ hold? If $G neq HK$ is $[HK:G]$ well defined? Can we also conclude that there is equality iff $HK = G$?
abstract-algebra group-theory
asked Jan 6 at 7:28
NotAbelianGroupNotAbelianGroup
18211
$endgroup$
1
$begingroup$
Notice that it is not assumed in this question that $G$ is finite, only that $|G:K|$ is finite. Your proposed solution works only for finite groups.
$endgroup$
– Derek Holt
Jan 6 at 8:18
$begingroup$
Thank you for your answer, I think a correct way to solve it is showing that the homomorphism $H/Hcap K rightarrow G/K$ is injective.
$endgroup$
– NotAbelianGroup
Jan 6 at 8:21
2
$begingroup$
That only works if $K$ is a normal subgroup of $G$, but the approach does work. You can just prove the map is injective. It is not a group homomorphism.
$endgroup$
– Derek Holt
Jan 6 at 8:23
$begingroup$
Oh sorry you are right, thank you for your help.
$endgroup$
– NotAbelianGroup
Jan 6 at 8:27
$begingroup$
For arbitrary subgroups, $HK$ need not be a group/subgroup at all! So you cannot say “Since $K$ is a subgroup of $HK$”, as that may not make sense.
$endgroup$
– Arturo Magidin
Jan 6 at 21:21
|
show 2 more comments
$begingroup$
I'm working on this problem and may have found a solution, but I have a step which is not rigorous enough.
My answer: Since $H$ and $K$ are subgroups of $G$, we know that $$|HK|cdot |H cap K|=|H|cdot |K|$$
$$frac{|HK|}{|K|}=frac{|H|}{|H cap K|}$$
Since $K$ is a subgroup of $HK$, and $H cap K$ a subgroup of $H$:
$$[H:H cap K]=[HK:K]stackrel{(*)}{=} [HK:G][G:K]leq[G:K]$$
My questions: Does the step $(*)$ hold? If $G neq HK$ is $[HK:G]$ well defined? Can we also conclude that there is equality iff $HK = G$?
abstract-algebra group-theory
asked Jan 6 at 7:28
NotAbelianGroupNotAbelianGroup
18211
$endgroup$
I'm working on this problem and may have found a solution, but I have a step which is not rigorous enough.
My answer: Since $H$ and $K$ are subgroups of $G$, we know that $$|HK|cdot |H cap K|=|H|cdot |K|$$
$$frac{|HK|}{|K|}=frac{|H|}{|H cap K|}$$
Since $K$ is a subgroup of $HK$, and $H cap K$ a subgroup of $H$:
$$[H:H cap K]=[HK:K]stackrel{(*)}{=} [HK:G][G:K]leq[G:K]$$
My questions: Does the step $(*)$ hold? If $G neq HK$ is $[HK:G]$ well defined? Can we also conclude that there is equality iff $HK = G$?
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 6 at 7:28
NotAbelianGroupNotAbelianGroup
18211
asked Jan 6 at 7:28
NotAbelianGroupNotAbelianGroup
18211
asked Jan 6 at 7:28
NotAbelianGroupNotAbelianGroup
18211
asked Jan 6 at 7:28
NotAbelianGroupNotAbelianGroup
18211
asked Jan 6 at 7:28
NotAbelianGroupNotAbelianGroup
18211
18211
1
$begingroup$
Notice that it is not assumed in this question that $G$ is finite, only that $|G:K|$ is finite. Your proposed solution works only for finite groups.
$endgroup$
– Derek Holt
Jan 6 at 8:18
$begingroup$
Thank you for your answer, I think a correct way to solve it is showing that the homomorphism $H/Hcap K rightarrow G/K$ is injective.
$endgroup$
– NotAbelianGroup
Jan 6 at 8:21
2
$begingroup$
That only works if $K$ is a normal subgroup of $G$, but the approach does work. You can just prove the map is injective. It is not a group homomorphism.
$endgroup$
– Derek Holt
Jan 6 at 8:23
$begingroup$
Oh sorry you are right, thank you for your help.
$endgroup$
– NotAbelianGroup
Jan 6 at 8:27
$begingroup$
For arbitrary subgroups, $HK$ need not be a group/subgroup at all! So you cannot say “Since $K$ is a subgroup of $HK$”, as that may not make sense.
$endgroup$
– Arturo Magidin
Jan 6 at 21:21
|
show 2 more comments
1
$begingroup$
Notice that it is not assumed in this question that $G$ is finite, only that $|G:K|$ is finite. Your proposed solution works only for finite groups.
$endgroup$
– Derek Holt
Jan 6 at 8:18
$begingroup$
Thank you for your answer, I think a correct way to solve it is showing that the homomorphism $H/Hcap K rightarrow G/K$ is injective.
$endgroup$
– NotAbelianGroup
Jan 6 at 8:21
2
$begingroup$
That only works if $K$ is a normal subgroup of $G$, but the approach does work. You can just prove the map is injective. It is not a group homomorphism.
$endgroup$
– Derek Holt
Jan 6 at 8:23
$begingroup$
Oh sorry you are right, thank you for your help.
$endgroup$
– NotAbelianGroup
Jan 6 at 8:27
$begingroup$
For arbitrary subgroups, $HK$ need not be a group/subgroup at all! So you cannot say “Since $K$ is a subgroup of $HK$”, as that may not make sense.
$endgroup$
– Arturo Magidin
Jan 6 at 21:21
1
1
$begingroup$
Notice that it is not assumed in this question that $G$ is finite, only that $|G:K|$ is finite. Your proposed solution works only for finite groups.
$endgroup$
– Derek Holt
Jan 6 at 8:18
$begingroup$
Notice that it is not assumed in this question that $G$ is finite, only that $|G:K|$ is finite. Your proposed solution works only for finite groups.
$endgroup$
– Derek Holt
Jan 6 at 8:18
$begingroup$
Thank you for your answer, I think a correct way to solve it is showing that the homomorphism $H/Hcap K rightarrow G/K$ is injective.
$endgroup$
– NotAbelianGroup
Jan 6 at 8:21
$begingroup$
Thank you for your answer, I think a correct way to solve it is showing that the homomorphism $H/Hcap K rightarrow G/K$ is injective.
$endgroup$
– NotAbelianGroup
Jan 6 at 8:21
2
2
$begingroup$
That only works if $K$ is a normal subgroup of $G$, but the approach does work. You can just prove the map is injective. It is not a group homomorphism.
$endgroup$
– Derek Holt
Jan 6 at 8:23
$begingroup$
That only works if $K$ is a normal subgroup of $G$, but the approach does work. You can just prove the map is injective. It is not a group homomorphism.
$endgroup$
– Derek Holt
Jan 6 at 8:23
$begingroup$
Oh sorry you are right, thank you for your help.
$endgroup$
– NotAbelianGroup
Jan 6 at 8:27
$begingroup$
Oh sorry you are right, thank you for your help.
$endgroup$
– NotAbelianGroup
Jan 6 at 8:27
$begingroup$
For arbitrary subgroups, $HK$ need not be a group/subgroup at all! So you cannot say “Since $K$ is a subgroup of $HK$”, as that may not make sense.
$endgroup$
– Arturo Magidin
Jan 6 at 21:21
$begingroup$
For arbitrary subgroups, $HK$ need not be a group/subgroup at all! So you cannot say “Since $K$ is a subgroup of $HK$”, as that may not make sense.
$endgroup$
– Arturo Magidin
Jan 6 at 21:21
|
show 2 more comments
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1
$begingroup$
Notice that it is not assumed in this question that $G$ is finite, only that $|G:K|$ is finite. Your proposed solution works only for finite groups.
$endgroup$
– Derek Holt
Jan 6 at 8:18
$begingroup$
Thank you for your answer, I think a correct way to solve it is showing that the homomorphism $H/Hcap K rightarrow G/K$ is injective.
$endgroup$
– NotAbelianGroup
Jan 6 at 8:21
2
$begingroup$
That only works if $K$ is a normal subgroup of $G$, but the approach does work. You can just prove the map is injective. It is not a group homomorphism.
$endgroup$
– Derek Holt
Jan 6 at 8:23
$begingroup$
Oh sorry you are right, thank you for your help.
$endgroup$
– NotAbelianGroup
Jan 6 at 8:27
$begingroup$
For arbitrary subgroups, $HK$ need not be a group/subgroup at all! So you cannot say “Since $K$ is a subgroup of $HK$”, as that may not make sense.
$endgroup$
– Arturo Magidin
Jan 6 at 21:21