# Prove for cyclic group

• Jun 5th 2011, 10:48 AM
goroner
Prove for cyclic group
I cant find answer to this:

If G is finite group of order n, G is cyclic iff there is element in G with order n, that is a^n = e (the identity in G for the operation in G).

i dont know where to start, i thought of taking it this way but im afraid that's not the answer, or is it?..

if a is any element in G, and G is of order n, and you get this element a to power n for any n E Z, and if you get this a^n = e, then you can do this..

a^n = e => a^n-1 = a^-1, so the inverse of a is a^n-1, and you can go on like this and find all the inverses for a^k, for all k =1,..,n,
since you got a^n, you can find a^1, and you can do that like this:
a^n = e => a^n+1 = a^1, and you can go on like this and get all a^k for k=1,..,n so you have all the elements and their inverses in G making a the generator of G (G = <a>).

But im afraid this is not quite true cause i cant tell if a is the generator for sure , i lack of confidence to tell that, so if someone can guide me here :) thanks!!
• Jun 5th 2011, 11:05 AM
tonio
Quote:

Originally Posted by goroner
I cant find answer to this:

If G is finite group of order n, G is cyclic iff there is element in G with order n, that is a^n = e (the identity in G for the operation in G).

This is absurd: in a group of order n we have that $x^n=e$ for ANY element in the group, and the

group may perfectly well not be cyclic. What you probably meant is that there's an element x

in the group s.t. $x^n =e$ AND $x^m\neq e\,\,\,\forall m

And this is simply the definition of cyclic group...

Tonio

i dont know where to start, i thought of taking it this way but im afraid that's not the answer, or is it?..

if a is any element in G, and G is of order n, and you get this element a to power n for any n E Z, and if you get this a^n = e, then you can do this..

a^n = e => a^n-1 = a^-1, so the inverse of a is a^n-1, and you can go on like this and find all the inverses for a^k, for all k =1,..,n,
since you got a^n, you can find a^1, and you can do that like this:
a^n = e => a^n+1 = a^1, and you can go on like this and get all a^k for k=1,..,n so you have all the elements and their inverses in G making a the generator of G (G = <a>).

But im afraid this is not quite true cause i cant tell if a is the generator for sure , i lack of confidence to tell that, so if someone can guide me here :) thanks!!

.
• Jun 5th 2011, 11:16 AM
goroner
i didnt mean what you wrote, i just try to solve task in the exercise part of the book im reading and this is what is written there:

Prove that:
"If G is group of order n, G is cyclic iff G has an element of order n"
• Jun 5th 2011, 03:01 PM
Deveno
we have to prove TWO things:

1.) if |G| = n and G has an element (let's call it g) of order n, then G is cyclic.

2.) If |G| = n and G is cyclic, G has an element of order n.

the second one is easier to prove, so we'll start there. we know G is cyclic. so G has an element x, with G = <x> = {x, x^2, x^3,.....,x^k,e}

as you can see, we have k powers of x, and the identity e. since these are ALL the elements of G, we have: k+1 = n.

this means that k = n-1. thus x, x^2, x^3,....,x^(n-1) are all NOT the identity. but we know that x^n IS the identity

(since this is true of EVERY element of a group of order n). so x is the element of order n we are looking for.

now, let's see about #1. here, we only know that G has an element g of order n. this means:

g^k ≠ e for 0 < k < n, and g^n = e. so let's consider THIS subgroup of G: <g>. since g is of order n,

{g,g^2,g^3,....,g^(n-1), g^n = e} are all different. why?

suppose g^k = g^m for 0 < k,m < n, with k ≠ m. we can assume k > m (or else just switch them around).

then (g^k)(g^(-m)) = (g^m)(g^(-m)), so

g^(k-m) = e. but k > m, so k-m > 0. and k < n, so k-m < k < n. and this would mean that k-m is a smaller (positive) power of g

equal to the identity than n. but n is the order of g, so this cannot be the case (n is the smallest positive power of g equal to e).

since this is a contradiction, we can't possibly have g^k = g^m for 0 < k,m < n and k ≠ m. so the powers of g:

{g, g^2, g^3,.....,g^(n-1), g^n = e} are all distinct. but this is n elements, which means that <g> is all of G. so g generates G,

thus G is cyclic.
• Jun 5th 2011, 07:56 PM
tonio
Quote:

Originally Posted by goroner
i didnt mean what you wrote, i just try to solve task in the exercise part of the book im reading and this is what is written there:

Prove that:
"If G is group of order n, G is cyclic iff G has an element of order n"

Well, then it all boils down to what's the definition of "cyclic group" in that book. Apparently it is that a group G is

cyclic if there's an element $g\in G\,\,s.t.\,\,\forall x\in G\,,\,\,x = g^m$ for some integer m.

The trick: if $g\in G$ has order n (this means what I wrote and NOT what you wrote) then count the elements

in the set $\{g^0,g,g^2,...,g^{n-1}\}$

Tonio
• Jun 6th 2011, 11:47 AM
goroner
thanks a lot guys =)