Group operation
$begingroup$
I have learnt here that, given a group $G$ and a subgroup $H le G$, if we denote by $f$ the group operation and by $complement_GH$ the complement of $H$ in $G$, we have that $f(complement_GH times complement_GH)=G$ whenever $[G:H]>2$. Let's now suppose $G$ finite and $[G:H]>2$. How many pairs $(g,g') in complement_GH times complement_GH$ are there such that $gg'^{-1} in H$ ?
group-theory
asked Jan 11 at 15:58
LucaLuca
27319
$endgroup$
add a comment |
$begingroup$
I have learnt here that, given a group $G$ and a subgroup $H le G$, if we denote by $f$ the group operation and by $complement_GH$ the complement of $H$ in $G$, we have that $f(complement_GH times complement_GH)=G$ whenever $[G:H]>2$. Let's now suppose $G$ finite and $[G:H]>2$. How many pairs $(g,g') in complement_GH times complement_GH$ are there such that $gg'^{-1} in H$ ?
group-theory
asked Jan 11 at 15:58
LucaLuca
27319
$endgroup$
1
$begingroup$
The number is $left|Hright|^2 left(left[G:Hright] - 1right)$. Hint for the proof: Argue that $gleft(g^{prime}right)^{-1} in H$ is equivalent to $Hg = Hg^{prime}$, i.e., equivalent to $g$ and $g^{prime}$ lying in the same $Hx$-coset of $H$ in $G$. Now, how many cosets are there, and for each coset, how many pairs of elements both belong to this coset?
$endgroup$
– darij grinberg
Jan 11 at 16:04
$begingroup$
I rephrase/deploy your hint to see if I got it: $complement_GH$ is made of $[G:H]-1$ cosets; within each of them, we can built up $|H|^2$ products; so there are $|H|^2([G:H]-1)$ pairs $(g,g') in complement_GH times complement_GH$ such that $Hg=Hg'$ $(Leftrightarrow gg'^{-1} in H$). Does it fit?
$endgroup$
– Luca
Jan 11 at 17:28
$begingroup$
Yes, that's exactly it.
$endgroup$
– darij grinberg
Jan 11 at 17:53
add a comment |
$begingroup$
I have learnt here that, given a group $G$ and a subgroup $H le G$, if we denote by $f$ the group operation and by $complement_GH$ the complement of $H$ in $G$, we have that $f(complement_GH times complement_GH)=G$ whenever $[G:H]>2$. Let's now suppose $G$ finite and $[G:H]>2$. How many pairs $(g,g') in complement_GH times complement_GH$ are there such that $gg'^{-1} in H$ ?
group-theory
asked Jan 11 at 15:58
LucaLuca
27319
$endgroup$
I have learnt here that, given a group $G$ and a subgroup $H le G$, if we denote by $f$ the group operation and by $complement_GH$ the complement of $H$ in $G$, we have that $f(complement_GH times complement_GH)=G$ whenever $[G:H]>2$. Let's now suppose $G$ finite and $[G:H]>2$. How many pairs $(g,g') in complement_GH times complement_GH$ are there such that $gg'^{-1} in H$ ?
group-theory
group-theory
asked Jan 11 at 15:58
LucaLuca
27319
asked Jan 11 at 15:58
LucaLuca
27319
asked Jan 11 at 15:58
LucaLuca
27319
asked Jan 11 at 15:58
LucaLuca
27319
asked Jan 11 at 15:58
LucaLuca
27319
27319
1
$begingroup$
The number is $left|Hright|^2 left(left[G:Hright] - 1right)$. Hint for the proof: Argue that $gleft(g^{prime}right)^{-1} in H$ is equivalent to $Hg = Hg^{prime}$, i.e., equivalent to $g$ and $g^{prime}$ lying in the same $Hx$-coset of $H$ in $G$. Now, how many cosets are there, and for each coset, how many pairs of elements both belong to this coset?
$endgroup$
– darij grinberg
Jan 11 at 16:04
$begingroup$
I rephrase/deploy your hint to see if I got it: $complement_GH$ is made of $[G:H]-1$ cosets; within each of them, we can built up $|H|^2$ products; so there are $|H|^2([G:H]-1)$ pairs $(g,g') in complement_GH times complement_GH$ such that $Hg=Hg'$ $(Leftrightarrow gg'^{-1} in H$). Does it fit?
$endgroup$
– Luca
Jan 11 at 17:28
$begingroup$
Yes, that's exactly it.
$endgroup$
– darij grinberg
Jan 11 at 17:53
add a comment |
1
$begingroup$
The number is $left|Hright|^2 left(left[G:Hright] - 1right)$. Hint for the proof: Argue that $gleft(g^{prime}right)^{-1} in H$ is equivalent to $Hg = Hg^{prime}$, i.e., equivalent to $g$ and $g^{prime}$ lying in the same $Hx$-coset of $H$ in $G$. Now, how many cosets are there, and for each coset, how many pairs of elements both belong to this coset?
$endgroup$
– darij grinberg
Jan 11 at 16:04
$begingroup$
I rephrase/deploy your hint to see if I got it: $complement_GH$ is made of $[G:H]-1$ cosets; within each of them, we can built up $|H|^2$ products; so there are $|H|^2([G:H]-1)$ pairs $(g,g') in complement_GH times complement_GH$ such that $Hg=Hg'$ $(Leftrightarrow gg'^{-1} in H$). Does it fit?
$endgroup$
– Luca
Jan 11 at 17:28
$begingroup$
Yes, that's exactly it.
$endgroup$
– darij grinberg
Jan 11 at 17:53
1
1
$begingroup$
The number is $left|Hright|^2 left(left[G:Hright] - 1right)$. Hint for the proof: Argue that $gleft(g^{prime}right)^{-1} in H$ is equivalent to $Hg = Hg^{prime}$, i.e., equivalent to $g$ and $g^{prime}$ lying in the same $Hx$-coset of $H$ in $G$. Now, how many cosets are there, and for each coset, how many pairs of elements both belong to this coset?
$endgroup$
– darij grinberg
Jan 11 at 16:04
$begingroup$
The number is $left|Hright|^2 left(left[G:Hright] - 1right)$. Hint for the proof: Argue that $gleft(g^{prime}right)^{-1} in H$ is equivalent to $Hg = Hg^{prime}$, i.e., equivalent to $g$ and $g^{prime}$ lying in the same $Hx$-coset of $H$ in $G$. Now, how many cosets are there, and for each coset, how many pairs of elements both belong to this coset?
$endgroup$
– darij grinberg
Jan 11 at 16:04
$begingroup$
I rephrase/deploy your hint to see if I got it: $complement_GH$ is made of $[G:H]-1$ cosets; within each of them, we can built up $|H|^2$ products; so there are $|H|^2([G:H]-1)$ pairs $(g,g') in complement_GH times complement_GH$ such that $Hg=Hg'$ $(Leftrightarrow gg'^{-1} in H$). Does it fit?
$endgroup$
– Luca
Jan 11 at 17:28
$begingroup$
I rephrase/deploy your hint to see if I got it: $complement_GH$ is made of $[G:H]-1$ cosets; within each of them, we can built up $|H|^2$ products; so there are $|H|^2([G:H]-1)$ pairs $(g,g') in complement_GH times complement_GH$ such that $Hg=Hg'$ $(Leftrightarrow gg'^{-1} in H$). Does it fit?
$endgroup$
– Luca
Jan 11 at 17:28
$begingroup$
Yes, that's exactly it.
$endgroup$
– darij grinberg
Jan 11 at 17:53
$begingroup$
Yes, that's exactly it.
$endgroup$
– darij grinberg
Jan 11 at 17:53
add a comment |
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1
$begingroup$
The number is $left|Hright|^2 left(left[G:Hright] - 1right)$. Hint for the proof: Argue that $gleft(g^{prime}right)^{-1} in H$ is equivalent to $Hg = Hg^{prime}$, i.e., equivalent to $g$ and $g^{prime}$ lying in the same $Hx$-coset of $H$ in $G$. Now, how many cosets are there, and for each coset, how many pairs of elements both belong to this coset?
$endgroup$
– darij grinberg
Jan 11 at 16:04
$begingroup$
I rephrase/deploy your hint to see if I got it: $complement_GH$ is made of $[G:H]-1$ cosets; within each of them, we can built up $|H|^2$ products; so there are $|H|^2([G:H]-1)$ pairs $(g,g') in complement_GH times complement_GH$ such that $Hg=Hg'$ $(Leftrightarrow gg'^{-1} in H$). Does it fit?
$endgroup$
– Luca
Jan 11 at 17:28
$begingroup$
Yes, that's exactly it.
$endgroup$
– darij grinberg
Jan 11 at 17:53