let g be a group of order 48 and h be a subgroup of order 24.then
Answers
Answer:
I don't understand what you tell
Answer:
The subgroups are all cyclic, with orders dividing
48
Explanation:
All subgroups of a cyclic group are themselves cyclic, with orders which are divisors of the order of the group.
To see why, suppose
G
=
<
a
>
is cyclic with order
N
and
H
⊆
G
is a subgroup.
If
a
m
∈
H
and
a
n
∈
H
, then so is
a
±
+
q
n
for any integers
p
,
q
.
So
a
k
∈
H
where
k
=
G
C
F
(
m
,
n
)
and both
a
m
and
a
n
are in
<
a
k
>
.
In particular, if
a
k
∈
H
with
G
C
F
(
k
,
N
)
=
1
then
H
=
<
a
>
=
G
.
Also not that if
m
n
=
N
then
<
a
m
>
is a subgroup of
G
with order
n
.
We can deduce:
H
has no more than
1
generator.
The order of
H
is a factor of
N
.
In our example
N
=
48
and the subgroups are isomorphic to:
C
1
,
C
2
,
C
3
,
C
4
,
C
6
,
C
8
,
C
12
,
C
16
,
C
24
,
C
48
being:
<
>
,
<
a
24
>
,
<
a
16
>
,
<
a
12
>
,
<
a
8
>
,
<
a
6
>
,
<
a
4
>
,
<
a
3
>
,
<
a
2
>
,
<
a
>