2.4: Cyclic groups
( \newcommand{\kernel}{\mathrm{null}\,}\)
Order of group element
Let
Let
Consider the group
Consider
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The order of a group is the number of elements in the group.
The order of an element
Let
Let
Let
Let
Note:
Let
Let
- Proof
-
Let
be a group. Let . If then is the smallest subgroup of that contains .Suppose
Clearly Let Then for some and for some .Consider
Since ,Thus,
Suppose is a subgroup of G that contains . Clearly,
Let
Let
If
Remember the definition of a unitary group:
Let
The following is an infinite cyclic group. Any infinite cyclic group is isomorphic to this group.
Is
Yes,
Proof of being a cyclic group:
Since the group generated by
Possible Generators:
Note
We shall show
Consider that
It should be clear that
Thus
Similarly, we will show that
Consider that
It should be clear that
Thus the generators of
Properties:
Cyclic groups are abelian.
- Proof:
-
Let
be a group.If
is cyclic, then is abelian.Assume that
is cyclic.Then
for some .Let
.We will show
.Then
and , .Consider
Hence,
is abelian.◻
The converse is not true. Specifically, if a group is abelian, it is not necessarily cyclic. A counterexample is
Every subgroup of a cyclic group will be cyclic.
Let
- Proof
-
Let
.We will show
for some .There are two cases to consider.
Case 1:
Let
, then we are done since thus .Case 2:
Let
.Thus
s.t. .Since
, .Hence
for some .Without loss of generality, we may assume
.Define
.Since
by the well ordering principle has a smallest element .We shall show that
.Since
, .Let
.We shall show that
.Since
, .Since
, by the division algorithm, there exists s.t. .Then
and .Since
, then .Since
is the smallest, . (I think this is because )Therefore
.Thus
.◻
The Converse is not true. Every sub-group of the Klein 4-group is cyclic, but not the group.
Subgroups of
Let
Then
- Proof
-
Let
with . .Let
.Since by the division algorithm, unique integers q and r exist such that
.Consider
Note: is our assumption & is the order of group. .Hence
.Thus
.Therefore
.Conversely, assume that
.Then
. . .Hence the result.◻
Consider the group
1. Is
3. Find all subgroups of
- Answer
-
1. Consider
with addition modulo Note that is the identity and its order isElement
Order
Calculations
, , , , and . , , , and . , , and . , , , , , , , , , , , and . , and . , , , , , , , , , , , and . , and . , , and . , , , , , and . , , , , and . , , , , , , , , , , , and .From the table,
is a cyclic group, the set of all possible generators are . Hence, the number of generators is the same as the number of integers relatively prime to . That is, the number of generators is the Euler totient number2. From the above table, we see that the orders of the subgroups are
and 12, which are all divisors of the order of . Specifically, the order of the subgroup generated by isNote that
. Then the cyclic subgroups are with orders and
Let
Let
If
- Answer
-
Let
with and .Let
.We will show that
.Let
.Since
for some , and .Thus,
.Consider an arbitrary power of
, . .Let
.Using the division algorithm,
.Thus
.Now we want to show
.Consider
.Thus
.Now consider
.Now consider
.Thus
, where is positive since it is the greatest common denominator.Thus
for .Then we can show
.Consider
.Hence
.
Let
- Proof
-
Let
be a cyclic group with . Then
and .Assume that
. Then for some .Thus,
By Theorem 2.4.5, . Therefore,
for some
Which implies,
hence,
Conversly, we shall show that if
, then .
Suppose that . Then
, for some
By using the properties of powers in a group and , we get,
Hence
.Therefore, we have
. Since ,
Let
a) If
b) If
- Proof
-
a) Let
with Assume that . By Theorem 2.4.3, is cyclic. Hence for some integer Let . We shall show that . Since , , for some Consider, Thus . Therefore
Since for some Thus, Hence . Thus andb) Let
with Assume that . Suppose be subgroup of of order . We shall show that . Since is a subgroup of , , for some, By part a, Then . We shall show that is the smallest positive integer with this property.Now,
. Thus, . Thus which is unique.
A group of prime order is cyclic.
- Proof
-
Let
be a group s.t. , where is a prime number.
Let s.t. is a subgroup of .Since
, .Since
is cyclic, , for some .By the division algorithm,
where .Hence,
.Since the smallest positive integer
such that is , . Thus .Conversely, if
, then , .Consequently,
.Thus,
divides .Consider that the only two divisors of a prime number are the prime number itself and 1.
Since
, .Since
, is cyclic.◻
List all the cyclic subgroups of
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17 |
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19 |
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23 |
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29 |
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2 |
Since
However the following subgroups of
Let