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| thirdmojo... |
 Posted: Mon Nov 02, 2009 6:38 am |
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Let G be a finite p-group. Elements of order p in Z(G) together with 1
form a subgroup. The only reason given by the book is that Z(G) is
abelian... but still don't see why they form exactly a group.
Please help. |
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| Arturo Magidin... |
| Posted: Mon Nov 02, 2009 6:47 am |
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On Nov 2, 10:38 am, thirdmojo <thirdm... at (no spam) gmail.com> wrote:
[quote]Let G be a finite p-group. Elements of order p in Z(G) together with 1
form a subgroup. The only reason given by the book is that Z(G) is
abelian... but still don't see why they form exactly a group.
Please help.
[/quote]
If A is *any* abelian group, and n is *any* integer, then the set
H = {x in A | x^n = 1}
forms a subgroup.
To see this, note that 1 in H always, so H is nonempty. If x is in H,
then x^{-1} is in H (for (x^{-1})^n = (x^n)^{-1}); and if x and y are
in H, then xy is in H (because, since A is abelian, then (xy)^m x^m*y^m for every integer m).
Now apply that to the issue at hand.
--
Arturo Magidin |
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| thirdmojo... |
| Posted: Thu Nov 05, 2009 3:00 am |
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On 2 Nov, 17:47, Arturo Magidin <magi... at (no spam) member.ams.org> wrote:
[quote]On Nov 2, 10:38 am, thirdmojo <thirdm... at (no spam) gmail.com> wrote:
Let G be a finite p-group. Elements of order p in Z(G) together with 1
form a subgroup. The only reason given by the book is that Z(G) is
abelian... but still don't see why they form exactly a group.
Please help.
If A is *any* abelian group, and n is *any* integer, then the set
H = {x in A | x^n = 1}
forms a subgroup.
To see this, note that 1 in H always, so H is nonempty. If x is in H,
then x^{-1} is in H (for (x^{-1})^n = (x^n)^{-1}); and if x and y are
in H, then xy is in H (because, since A is abelian, then (xy)^m > x^m*y^m for every integer m).
Now apply that to the issue at hand.
[/quote]
Damn! I even knew that but did not think about it. Thank you very
much. |
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| Arturo Magidin... |
| Posted: Thu Nov 05, 2009 6:20 am |
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On Nov 5, 7:00 am, thirdmojo <thirdm... at (no spam) gmail.com> wrote:
[quote]On 2 Nov, 17:47, Arturo Magidin <magi... at (no spam) member.ams.org> wrote:
On Nov 2, 10:38 am, thirdmojo <thirdm... at (no spam) gmail.com> wrote:
Let G be a finite p-group. Elements of order p in Z(G) together with 1
form a subgroup. The only reason given by the book is that Z(G) is
abelian... but still don't see why they form exactly a group.
Please help.
If A is *any* abelian group, and n is *any* integer, then the set
H = {x in A | x^n = 1}
forms a subgroup.
To see this, note that 1 in H always, so H is nonempty. If x is in H,
then x^{-1} is in H (for (x^{-1})^n = (x^n)^{-1}); and if x and y are
in H, then xy is in H (because, since A is abelian, then (xy)^m > > x^m*y^m for every integer m).
Now apply that to the issue at hand.
Damn! I even knew that but did not think about it.
[/quote]
I've always had the conjecture that not thinking is unlikely to lead
one to an answer to a question.
But I haven't really thought about it, so I haven't proven it.
--
Arturo Magidin |
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