Profinite group, and why the product of subgroups is closed
$begingroup$
The following lemma is taken from Wilson's book Profinite Groups;
Let $( H_i : i in I)$ be a family of normal subgroups of a profinite
group $G$.
Assume $G = overline{< cup_{i in I} H_i >}$, and write $K_i =
overline{< cup_{j neq i } H_j >}$.
Moreover assume that $cap_{i in I}K_i = {e}$.
Then $G = prod H_i$.
He starts the proof by noting:
$K_i cap H_i = {e}$ which follows by the assumptions above.
$K_i triangleleft G$ which follows because $< cup_{i in I} H_i >$ is normal, and the closure of a normal subgroup is normal.
$K_iH_i$ is closed.
Why does 3. follow? I know that if $H_i$ was closed then it would follow directly because in a hausdorff, compact group $G$, $C,D$ closed implies that $CD$ is too.
abstract-algebra group-theory profinite-groups
asked Dec 10 '18 at 15:20
MariahMariah
2,0131718
$endgroup$
add a comment |
$begingroup$
The following lemma is taken from Wilson's book Profinite Groups;
Let $( H_i : i in I)$ be a family of normal subgroups of a profinite
group $G$.
Assume $G = overline{< cup_{i in I} H_i >}$, and write $K_i =
overline{< cup_{j neq i } H_j >}$.
Moreover assume that $cap_{i in I}K_i = {e}$.
Then $G = prod H_i$.
He starts the proof by noting:
$K_i cap H_i = {e}$ which follows by the assumptions above.
$K_i triangleleft G$ which follows because $< cup_{i in I} H_i >$ is normal, and the closure of a normal subgroup is normal.
$K_iH_i$ is closed.
Why does 3. follow? I know that if $H_i$ was closed then it would follow directly because in a hausdorff, compact group $G$, $C,D$ closed implies that $CD$ is too.
abstract-algebra group-theory profinite-groups
asked Dec 10 '18 at 15:20
MariahMariah
2,0131718
$endgroup$
$begingroup$
He does write in the beginning of the chapter that when he writes $H leq G$ he means that $H$ is closed as well. But the notation isn't used here..
$endgroup$
– Mariah
Dec 10 '18 at 15:47
add a comment |
$begingroup$
The following lemma is taken from Wilson's book Profinite Groups;
Let $( H_i : i in I)$ be a family of normal subgroups of a profinite
group $G$.
Assume $G = overline{< cup_{i in I} H_i >}$, and write $K_i =
overline{< cup_{j neq i } H_j >}$.
Moreover assume that $cap_{i in I}K_i = {e}$.
Then $G = prod H_i$.
He starts the proof by noting:
$K_i cap H_i = {e}$ which follows by the assumptions above.
$K_i triangleleft G$ which follows because $< cup_{i in I} H_i >$ is normal, and the closure of a normal subgroup is normal.
$K_iH_i$ is closed.
Why does 3. follow? I know that if $H_i$ was closed then it would follow directly because in a hausdorff, compact group $G$, $C,D$ closed implies that $CD$ is too.
abstract-algebra group-theory profinite-groups
asked Dec 10 '18 at 15:20
MariahMariah
2,0131718
$endgroup$
The following lemma is taken from Wilson's book Profinite Groups;
Let $( H_i : i in I)$ be a family of normal subgroups of a profinite
group $G$.
Assume $G = overline{< cup_{i in I} H_i >}$, and write $K_i =
overline{< cup_{j neq i } H_j >}$.
Moreover assume that $cap_{i in I}K_i = {e}$.
Then $G = prod H_i$.
He starts the proof by noting:
$K_i cap H_i = {e}$ which follows by the assumptions above.
$K_i triangleleft G$ which follows because $< cup_{i in I} H_i >$ is normal, and the closure of a normal subgroup is normal.
$K_iH_i$ is closed.
Why does 3. follow? I know that if $H_i$ was closed then it would follow directly because in a hausdorff, compact group $G$, $C,D$ closed implies that $CD$ is too.
abstract-algebra group-theory profinite-groups
abstract-algebra group-theory profinite-groups
asked Dec 10 '18 at 15:20
MariahMariah
2,0131718
asked Dec 10 '18 at 15:20
MariahMariah
2,0131718
edited Dec 10 '18 at 15:46
Mariah
asked Dec 10 '18 at 15:20
MariahMariah
2,0131718
asked Dec 10 '18 at 15:20
MariahMariah
2,0131718
asked Dec 10 '18 at 15:20
MariahMariah
2,0131718
2,0131718
$begingroup$
He does write in the beginning of the chapter that when he writes $H leq G$ he means that $H$ is closed as well. But the notation isn't used here..
$endgroup$
– Mariah
Dec 10 '18 at 15:47
add a comment |
$begingroup$
He does write in the beginning of the chapter that when he writes $H leq G$ he means that $H$ is closed as well. But the notation isn't used here..
$endgroup$
– Mariah
Dec 10 '18 at 15:47
$begingroup$
He does write in the beginning of the chapter that when he writes $H leq G$ he means that $H$ is closed as well. But the notation isn't used here..
$endgroup$
– Mariah
Dec 10 '18 at 15:47
$begingroup$
He does write in the beginning of the chapter that when he writes $H leq G$ he means that $H$ is closed as well. But the notation isn't used here..
$endgroup$
– Mariah
Dec 10 '18 at 15:47
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Surely it is intended to be assumed that the $H_i$ are closed. Indeed, the result is not true otherwise. For instance, you could take $I$ to be a singleton and the only $H$ to be a dense normal subgroup and then the result would claim that $G=H$, which certainly does not need to be true in general.
answered Dec 10 '18 at 16:43
Eric WofseyEric Wofsey
190k14216348
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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votes
$begingroup$
Surely it is intended to be assumed that the $H_i$ are closed. Indeed, the result is not true otherwise. For instance, you could take $I$ to be a singleton and the only $H$ to be a dense normal subgroup and then the result would claim that $G=H$, which certainly does not need to be true in general.
answered Dec 10 '18 at 16:43
Eric WofseyEric Wofsey
190k14216348
$endgroup$
add a comment |
$begingroup$
Surely it is intended to be assumed that the $H_i$ are closed. Indeed, the result is not true otherwise. For instance, you could take $I$ to be a singleton and the only $H$ to be a dense normal subgroup and then the result would claim that $G=H$, which certainly does not need to be true in general.
answered Dec 10 '18 at 16:43
Eric WofseyEric Wofsey
190k14216348
$endgroup$
add a comment |
$begingroup$
Surely it is intended to be assumed that the $H_i$ are closed. Indeed, the result is not true otherwise. For instance, you could take $I$ to be a singleton and the only $H$ to be a dense normal subgroup and then the result would claim that $G=H$, which certainly does not need to be true in general.
answered Dec 10 '18 at 16:43
Eric WofseyEric Wofsey
190k14216348
$endgroup$
Surely it is intended to be assumed that the $H_i$ are closed. Indeed, the result is not true otherwise. For instance, you could take $I$ to be a singleton and the only $H$ to be a dense normal subgroup and then the result would claim that $G=H$, which certainly does not need to be true in general.
answered Dec 10 '18 at 16:43
Eric WofseyEric Wofsey
190k14216348
answered Dec 10 '18 at 16:43
Eric WofseyEric Wofsey
190k14216348
answered Dec 10 '18 at 16:43
Eric WofseyEric Wofsey
190k14216348
answered Dec 10 '18 at 16:43
Eric WofseyEric Wofsey
190k14216348
190k14216348
add a comment |
add a comment |
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$begingroup$
He does write in the beginning of the chapter that when he writes $H leq G$ he means that $H$ is closed as well. But the notation isn't used here..
$endgroup$
– Mariah
Dec 10 '18 at 15:47