Re: Application of the Sylow Theorem
- From: jankrihau@xxxxxxxxxxx
- Date: Tue, 13 Nov 2007 13:06:09 -0800
On 13 Nov, 21:57, "Tadeusz Jordan" <te...@xxxxxxxxxxxxxx> wrote:
Hello. In the book there is an example that I don't understand. Could anyone
answer my questions?
Example:
Show that any group of order 30 must have a nontrivial normal subgroup.
Answer: The number of Sylow 3-subgroups must be congruent to 1 modulo 3 and
a divisor of 10, so it must be either 1 or 10. The number of Sylow
5-subgroups must be congruent to 1 modulo 5 and a divisor of 6, so it must
be either 1 or 6. Any Sylow 3-subgroup must have order 3, so the
intersection of two distinct such subgroups must be trivial. Therefore ten
Sylow 3-subgroups would yield twenty elements of order 3. Similarly, six
Sylow 5-subgroups would yield twenty-four elements of order 5. Together,
this would simply give too many elements for the group, so we conclude that
there must be either one Sylow 3-subgroup or one Sylow 5-subgroup, showing
the existence of a nontrivial normal subgroup.
I don't understand why "Any Sylow 3-subgroup must have order 3, so the
intersection of two distinct such subgroups must be trivial" - the second
part of this sentence - why does the intersection have to be trivial?
Because an element of a group of order 3 other than the identity
generates the group. So suppose an element other than the identity
lies in two different groups of order 3. It generates both of them:
contradiction.
Also
"Similarly, six Sylow 5-subgroups would yield twenty-four elements of order
5." - why is it not 30 elements of order 5?
Because a group of order 5 contains the identity and only four
elements of order 5.
Sincerely,
Tadeusz Jordan
---
J K Haugland
http://home.no.net/zamunda
.
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