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.
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}.
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).