1. ## Show Cyclic

Let G be a group of order three. Show that G is cyclic. I feel like this should be a very simple problem, but I am not sure how to proceed.

2. ## Re: Show Cyclic

This is what I have so far:
Let G be a group, g is an element of G. The order of g is the smallest positive integer n such that g^n=1. We know that n exists because G is finite. Furthermore, we are told the order is 3, so n=3. I have written g^n=1, and in this case, g^3=1, which implies that g=<1>. So, <1>={1,2,0}.

3. ## Re: Show Cyclic

this is what we're given:

G = {e,a,b}.

as yet, we know nothing about the order of a or b. but we can ask: which element of G is a2?

we have 3 choices: e, a or b.

if a2 = a, then a-1(a2) = a-1a, that is:

a = e, a contradiction. so a2 must be e, or b. if a2 = e, then a is its own inverse. that means b must also be its own inverse (as there are no more elements left to be its inverse: b is not the identity, so b-1 ≠ e,

and b-1 ≠ a, since a-1 is a).

so far, so good. now we ask: if a and b are their own inverses, which element of G is ab?

if ab = e, then:

a(ab) = a
a2b = a
eb = a

if ab = a, then:

a(ab) = a2
a2b = e
b = e, also a contradiction.

if ab = b, then:

(ab)b = b2 (since b = b-1, b2 = bb-1 = e)
ab2 = e
ae = e
a = e, also a contradiction.

so setting a2 = e leaves us with no possible choice for ab. since we have ruled out a2 = a, as well, that leaves only one possible choice:

a2 = b, in which case:

G = {e,a,b} = {e,a,a2} = <a>.

(i want to point out that all we have done is shown that IF |G| = 3, then G is cyclic. we haven't actually shown that any groups of order 3 actually exist. however, one can easily verify that the set {e,a,b} with the operation * defined by:

e*e = e
e*a = a
e*b = b

a*e = a
a*a = b
a*b = e

b*e = b
b*a = e
b*b = a

actually defines a group with identity e. the hard part is verifying associativity, which is a chore (there are 9 possible products, so there are 27 possible triple products, which gives 27 associations to verify).

4. ## Re: Show Cyclic

Let |G| = 3. $\displaystyle G = \{e,a,b\}$. the order of an element divides the order of the group, since only 1 element of order 1 exists (e). |a| and |b| = 3. which means a,b are generators for G. thus G is cyclic.