Ok whatever I think I figured it out
Hey There
I need to show that 2 groups with order 3 is isomorphic
I basically have the answer I just wanted to double check with you guys
I have proven that ab=identity and that simularly cd=identity. I am getting stuck just at the last piece at the end. Just showing how its isomorphic
suppose we have a group of order 3. suppose that's all we know. well, the first thing is: our group is of the form:
G = {e,a,b}.
this gives 9 possible products:
e*e
e*a
e*b
a*e
a*a
a*b
b*e
b*a
b*b
what must these be?
from the definition of the identity e, we get some of these at once:
e*e = e
e*a = a
e*b = b
a*e = a
b*e = b, which leaves just 4 more to figure out.
so what is a*a ? well, we might guess that a*a = e (after all, that's what it is in a 2-element group). at first, there doesn't seem anything wrong with this.
so then we look at a*b. we have 3 choices: a*b = a, a*b = b, and a*b = e.
if a*b = a, then:
a^{-1}*(a*b) = a^{-1}*a
(a^{-1}*a)*b = e
e*b = e
b = e. so that's no good.
if a*b = b, then:
(a*b)*b^{-1} = b*b^{-1}
a*(b*b^{-1}) = e
a*e = e
a = e. so that can't be true, either.
so we have to have a*b = e. now if a*a = e as well:
a*b = e = a*a
a^{-1}*(a*b) = a^{-1}*(a*a)
(a^{-1}*a)*b = (a^{-1}*a)*a
e*b = e*a
b = a. huh. apparently if a*b = e (which is the only possibility that makes sense) we can't have a*a = e. since (as we saw in the 2-element case) a*a = a implies a = e, the only viable option is a*a = b
so now we know that:
a*a = b
a*b = e
so that just leaves 2 more products to figure out. but..we actually leaned something new: a*b = e means b must be a^{-1} (in a group, there is only ONE inverse for each element). this means b*a must equal e as well (since a*a^{-1} = a^{-1}*a, and both are equal to e).
which brings us to the last product: b*b.
b*b = e means b*b = b*a, which forces b = a (why?)
b*b = b means b = e (why?)
since neither of these makes any sense, we must have:
b*b = a.
this means we can write G as: {e,a,a*a} = {e,a,a^{2}} = {e,b^{2},b} = {e,b*b,b}.
note that all of this follows just from the fact that G is a group with 3 elements.
so, suppose G and H are two groups with 3 elements. we can write G = {e,a,a^{2}}, and H = {e',x,x^{2}}. what could be more natural than to define:
h(e) = e'
h(a) = x
h(a^{2}) = x^{2}.
it should be obvious that h preserves all products involving e. again there are really only 4 products to check:
h(a*a) = h(a)*h(a) (obvious, because...?)
h(a*a^{2}) = h(a)*h(a^{2}) (this is easy if you observe that a^{2} = a^{-1}, and x^{2} = x^{-1}, or equivalently that:
a^{3} = e, and x^{3} = e').
h(a^{2}*a) = h(a^{2})*h(a) (you might want to convince yourself that this boils down to the same thing as the product above:
a^{3} = a*(a*a) = (a*a)*a, and x^{3} = x*(x*x) = (x*x)*x, by associativity)
and finally:
h(a^{2}*a^{2}) = h(a^{2})*h(a^{2}).
to prove this last product: note that a^{4} = (a^{2})*(a^{2}) = (a*a)*(a*a) = ((a*a)*a)*a = a^{3}*a = e*a = a, and similarly for x.