Question

What do we mean by abelian,cyclic and infinite groups?

Answer 1

an ablian group is a group for which the elements commute (i.e., AB=BA for all elements A and B)...
whereas a cyclic group is a group that can be generated by a single element X (the group generator)...
and an infinite group is a group, of which the underlying set contains an infinite number of elements...

Answer 2

Can you explain cyclic and infinite group more?

Answer 3

Maybe an example?

Answer 4

hmm... a cyclic group is a group in which every element is a power of some fixed element....

Answer 5

ill give u example

Answer 6

So how do I check if it is cyclic,abelian or infinite for this question?

Answer 7

Here is the qn I am working on^

Answer 8

group G is cyclic if G = (a)... for some a belongs to G and a = {a^k / k belongs to Z}.. Such an element is called a generator
of G...

Answer 9

Can you explain using the qn I posted?

Answer 10

How do I check for cyclic and infinite in it?

Answer 11

hey is this the only info abt ur group ?

Answer 12

i mean what kind of element should G have ?

Answer 13

Yep. That's an obective qn and options are like a)abelian b)cyclic c)infinite

Answer 14

I am seeing such a question for first time. These terms are like from another planet for me.

Answer 15

well i wont guess lol
but to show if its cyclic u need info abt the elements , G is consederd to be cyclic
when all elements are generated from one element so i cant juge if its cyclic or not :O
ovs its abelian , and its infinte

Answer 16

Only 1 answer is right.

Answer 17

its not the abilaian... because the abelian group holds commutative property.. and here not that case... if a and b are tha members of abelian group, then ab must be equal to ba...

Answer 18

for exampl
f(x)=e^x
we wud have G
g={1,e,e^2,e^3,...}
i think this group apply the condition
(a^2*b^2)=(a*b)^2
(e^2a * b^2b)=(e^2(a+b))=(e^a+b)^2=(e^a *e^b)^2
so as u can see here in this exampl
its cyclic , abelian,and infinte :3
so somthing missing :O

Answer 19

I dont want to check abelian. I want cyclic and infinite!

Answer 20

for cyclic we need the members of G...?

Answer 21

so in general i dnt think that only one of them :O
whats ur course name ?

Answer 22

Just a competitive exam.

Answer 23

ok ok i said what i have , i gave u an example for such a G that have the same condition and its cyclic , abelian and infinite at the same time :O

Answer 24

hmm...

Answer 25

consider the group of n-th roots of unity under the operation of multiplication.. i.e. \( \omega ^n = 1 \) , this is an example of cyclic group generated by \( \omega \). it's elements are
\[G = \{ 1, \omega, \dots , \omega^{n-1}\}\]

consider the group of integers under the operation of addition, it is an example of infinite group.

and, for abelian group, both of above are Abelian group. Rather let me give you an example of non-abelain group.

consider a set of 2x2 matrices, under the operation of multiplication such that \(ad-bc \neq 0 \). where matrix is
\[
\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}
\]
http://groupprops.subwiki.org/wiki/Symmetric_group:S3
Also consider this group.

Answer 26

if \(a^2*b^2 = (a*b)^2 \) for two elements in a group, then it's abelian group.

Answer 27

It can be seen in following way,
\[ a^2*b^2 = (a*b)^2 \\
a^{-1}*(a^2*b^2)*b^{-1} = a^{-1}((a*b)*(a*b))*b^{-1} \]
Since both, 'a' and 'b' are elements of group, they both have inverse. using associativity property, you get \( a*b = b*a \). There is not enough information about weather it's cyclic group or infinite group.

Answer 28

Thank you for such a wonderful explanation @experimentX . I have 2 doubts after going through it:

1. How do we know that inverse exists for both a and b if they are present in the group?
2.You applied associative property while proving commutativity.How do we know that "*"obeys associative property?

Thanks again :)

Answer 29

that is the definition of group.
http://en.wikipedia.org/wiki/Group_(mathematics)#Definition

Answer 30

For a set equipped with some operation to be group ... it must satisfy those properties.

Answer 31

Oh I see it now. One more question. What do we mean by the order of a group and why we take order of e as 1?

Answer 32

e does not necessarily have to be 1. (if the operation is (algebraic) multiplicative then e=1
Just that it has to satisfy the property
e*a = a*e = a
the order of group if the number of elements present on it.

Answer 33

If you consider the group of 2x2 matrix under multiplication, then e = unit matrix.
under addition, your e=0.

Answer 34

under addition for set of real numbers, your e=0.

Answer 35

"why we take order of e as 1"


\[e^1=1\]

therefore \(|e|=1\)

Answer 36

e^1=e

Answer 37

O.o ... you meant order of e!!

Answer 38

Take a look at this:

So If it cannot be true for addition then why this qn is general?

Answer 39

@Zarkon How does that relate with number of members?

Answer 40

first of all you need to define the order of element.
for \( a \in G \) , order of a = n =>
a^n = e

Answer 41

where n is the smallest natural number that works

Answer 42

0 + 0 = 0

Answer 43

Oh wait..That doesnt look right. Why we get a^n=e?

Answer 44

hmm ... pigeonhole principle.

Answer 45

closure property ... for finite group

Answer 46

By closure property we mean the values lie in a particular range?

Answer 47

no ... for two elements \(a, b \in G \), \(a *b \in G\) ...
if suppose \( a^n \) is different from \( a^{n-1}\) for each n, what do you get?
and if Group is finite?

Answer 48

No idea about that.

Answer 49

for some 'n', all elements of G, a^n = e. and you can guess this n is less or equal to the order of group.

Answer 50

you get
\[\{ a , a^2, a^3, \dots , a^n , a^{n+1}, ... \}\]
they all are different elements and unique elements.

But in group of order k, there are only k elements, say
\[ G= \{ e, a, b, c, \dots , k\}\]

Answer 51

*k+1 elements

Answer 52

what do you see?

Answer 53

a,b,c....k product gives 1?

Answer 54

Suppose that \(|G|=n\)

suppose \(a\ne e\)

then look at the list \[\{a,a^2,a^3,\ldots,a^n\}\]

if they are all different (and not equal to e) then you have n+1 elements

this is not possible

therefore for some i,j; \(i\ne j\)

\[a^i=a^j\]

then \[a^{i-j}=e\]

Answer 55

No ... i meant to say that there are only k+1 elements on the group.

Answer 56

Why doesn't the product work out to 1 as the group contains inverse of each elements?

Answer 57

the group might not be abelian

Answer 58

Oh I see.

Answer 59

So then i-j=n?

Answer 60

maybe...maybe not

Answer 61

if a generates G then the order of a is n,

but is could be any natural number that divides n

Answer 62

depends on the group and the element

Answer 63

So order is least natural number n such that a^n=e?

Answer 64

yep!!

Answer 65

and \( o(a) \le o(G) \)

Answer 66

Okay it makes a lot of things clear!! Thank you both for clarifying my doubts!! :)

Answer 67

yw