# Thread: Elements which correspond under an isomorphism.

1. ## Elements which correspond under an isomorphism.

Let G1 and G2 be groups and let f:G1 -> G2, be an isomorphism. If G1 is a cyclic group with generator a, prove G2 is also a cyclic group with generator f(a).

Ok so i did this, teacher said it was wrong and that i need to show G1 and G2 have the same order. And that G1=<a> so i need to prove G2=(f(a))

So where do i start and how do i end?

2. ## Re: Elements which correspond under an isomorphism.

It is sufficient to prove that for every y ∈ G2 there exists a natural number n such that f(a)^n = y. The fact that G1 and G2 have the same order (i.e., cardinality) holds immediately by assumption that there is an isomorphism (i.e., a bijection) between G1 and G2.

3. ## Re: Elements which correspond under an isomorphism.

I said this: since fG1-->G2) is an isomorphism, there is bijection from G1 to G2 and the order is preserved from G1 to G2, therefore they have the same order. So for the generator <a> in G1, the function <a>=<f(a)>. So G2=<a>=<f(a)>

4. ## Re: Elements which correspond under an isomorphism.

I said this: since f: (G1-->G2) is an isomorphism, there is bijection from G1 to G2 and the order is preserved from G1 to G2, therefore they have the same order.
Agree, though I would remove "the order is preserved from G1 to G2."

So for the generator <a> in G1, the function <a>=<f(a)>.
Did not understand this. First, <a> is not a generator, it is the whole group G1. Second, where is a verb in this sentence? What does "the function <a>=<f(a)>" mean?

So G2=<a>=<f(a)>
G2 ≠ <a>; in fact, a may not belong to G2. I don't see the proof that every element of G2 is a power of f(a). This is primarily what you need to show. When you are constructing a proof, make sure you identify where you use the facts that: (1) f is a surjection; (2) a is a generator of G1 and (3) f is a homomorphism.

5. ## Re: Elements which correspond under an isomorphism.

Ok, i will work on it, this post what i have.

6. ## Re: Elements which correspond under an isomorphism.

you know that <f(a)> is a subgroup of G2. suppose that it was not all of G2 (our eventual goal is to show this can never happen). pick some element y in G2 that is not in <f(a)>.

now, f is a bijection, so y = f(x) for some x in G1. but G1 is cyclic, so...(some fact about "a" goes here).

now f is also a homomorphism, so f(ak) = [f(a)]k, for any integer k. do you see the contradiction?

note that no mention of the orders of G1 and G2 are made here, because that is not what your original question says to prove.