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.

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- January 26th 2013, 04:45 PMlovesmathShow 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.

- January 26th 2013, 05:03 PMlovesmathRe: 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}. - January 26th 2013, 11:47 PMDevenoRe: 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 a^{2}?

we have 3 choices: e, a or b.

if a^{2}= a, then a^{-1}(a^{2}) = a^{-1}a, that is:

a = e, a contradiction. so a^{2}must be e, or b. if a^{2}= 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

a^{2}b = a

eb = a

b = a, contradiction.

if ab = a, then:

a(ab) = a^{2}

a^{2}b = e

b = e, also a contradiction.

if ab = b, then:

(ab)b = b^{2}(since b = b^{-1}, b^{2}= bb^{-1}= e)

ab^{2}= e

ae = e

a = e, also a contradiction.

so setting a^{2}= e leaves us with no possible choice for ab. since we have ruled out a^{2}= a, as well, that leaves only one possible choice:

a^{2}= b, in which case:

G = {e,a,b} = {e,a,a^{2}} = <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). - January 27th 2013, 06:50 PMjakncokeRe: Show Cyclic
Let |G| = 3. . 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.