Math 4124
Monday, March 21
March 21, Ungraded Homework
Prove that a group of order 175 is abelian.
Let
G
be the group of order 175
=
7
·
5
2
. The number of Sylow 7subgroups is congruent
to 1 mod 7 and divides 25. The only possibility is 1, so
G
has a normal Sylow 7subgroup,
call it
A
. Also the number of Sylow 5subgroups is congruent to 1 mod 5 and divides 7.
Therefore
G
has a normal Sylow 5subgroup, call it
B
. Since
A
and
B
are of prime or prime
squared order, they are both abelian, in particular
A
≤
C
G
(
A
)
. Furthermore by Lagrange’s
theorem
A
∩
B
=
1. Since
A
,
B
G
, we deduce that every element of
A
commutes with every
element of
B
(recall that one proves this by considering
aba

1
b

1
= (
aba

1
)
b

1
∈
B
and
similarly
∈
A
, so
aba

1
b

1
∈
A
∩
B
=
1). Thus
B
≤
C
G
(
A
)
. By Lagrange’s theorem

A

,

B

divide

C
G
(
A
)

, so

C
G
(
A
)

=
175, consequently C
G
(
A
) =
G
. This means that
A
≤
Z
(
G
)
.
Similarly
B
≤
Z
(
G
)
and we deduce that

A

,

B

divide

Z
(
G
)

. It follows that Z
(
G
)
has order
a multiple of 7
·
25, which is only possible if Z
(
G
) =
G
. This means that
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 Spring '08
 Staff
 Math, Algebra, Group Theory, Subgroup, Sylow 3subgroups, Sylow, Sylow 7subgroups

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