If $H$ and $K$ are subgroups of $G$, and the index of $K$ in $G$ finite, then $[ H : H cap K ] leq [G:K]$












1












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















1












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














1












1








1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










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