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Math 4124
Monday, March 28
March 28, Ungraded Homework
Prove that a group of order 765 is abelian.
Let
G
be a group of order 765
=
9
*
5
*
17. We need to prove that
G
is abelian. It is
usually best to consider the largest prime dividing the order of the group ﬁrst, so we will
ﬁrst consider the number of Sylow 17subgroups. This is congruent to 1 mod 17 and divides
45, so the only possibility is 1. Thus
G
has a normal 17subgroup, which we shall call
A
. Now we cannot immediately assert that the number of Sylow 5subgroups is 1, because
51 appears possible. Instead we consider
G
/
A
, a group of order 765
/
17
=
45. Since the
number of Sylow 5subgroups is congruent to 1 mod 5 and divides 9, we see that
G
/
A
has a
normal Sylow 5subgroup
P
/
A
, where
P
±
G
and

P

=
5
*
17
=
85. The number of Sylow
5subgroups of
P
is congruent to 1 mod 5 and divides 17, so must be 1 and we see that
P
has
a normal Sylow 5subgroup
B
. Now
B
±
P
±
G
does not imply
B
±
G
in general, but it does
here because
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This note was uploaded on 01/02/2012 for the course MATH 4124 taught by Professor Staff during the Spring '08 term at Virginia Tech.
 Spring '08
 Staff
 Math, Algebra

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