Thread: Isomorphism

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

Ok whatever I think I figured it out

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,a²} = {e,b²,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,a²}, and H = {e',x,x²}. what could be more natural than to define:

h(e) = e'
h(a) = x
h(a²) = x².

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*a²) = h(a)*h(a²) (this is easy if you observe that a² = a^-1, and x² = x^-1, or equivalently that:

a³ = e, and x³ = e').

h(a²*a) = h(a²)*h(a) (you might want to convince yourself that this boils down to the same thing as the product above:

a³ = a*(a*a) = (a*a)*a, and x³ = x*(x*x) = (x*x)*x, by associativity)

and finally:

h(a²*a²) = h(a²)*h(a²).

to prove this last product: note that a⁴ = (a²)*(a²) = (a*a)*(a*a) = ((a*a)*a)*a = a³*a = e*a = a, and similarly for x.

Wow that was exactly what i needed to be explained. Thanks )))