1. ## Isomorphism Test

Let A be a group that can be generated by two elements in A. Suppose B is a group that can't be generated by any two elements in B. Show that A is NOT isomorphic to B.

Proof:

Assume, on the contrary, that f: A--> B is an isomorphism. Let A be generated by x,y e A.

What do I do from here? Thanks

2. ## Re: Isomorphism Test

Originally Posted by jzellt
Let A be a group that can be generated by two elements in A. Suppose B is a group that can't be generated by any two elements in B. Show that A is NOT isomorphic to B.

Proof:

Assume, on the contrary, that f: A--> B is an isomorphism. Let A be generated by x,y e A.

What do I do from here? Thanks
Since you are using a proof by contradiction, let $\displaystyle b \in B$ be an element that cannot be generated by two elements. Since an isomorphism is a bijection b must have a pre image. So there must be a uniquie $\displaystyle a \in A$ such that $\displaystyle f(a)=b$.

Now use the fact that $\displaystyle a$ can be generated by two elements and the isomorphism preserve the group operations to show that $\displaystyle b$ can be generated by two elements.

3. ## Re: Isomorphism Test

Thanks. Can you fix your latex.. Its not showing correctly in the last sentence.

4. ## Re: Isomorphism Test

So, here it would I've done from your advice. Please check it out and let me know how it can improve.

Assume, on the contrary, that f: A--> B is an isomorphism. Let b e B be an element that cannot be generated by two elements in B. Since f is onto, there exists a e A such that f(a) = b. Since a e A, it can be generated by two elements, say x,y e A. So, a = x^n*y^m, n,m e Z.
Thus, f(x^n*y^m) = f(a) = b. This is a contradiction. QED

5. ## Re: Isomorphism Test

Originally Posted by jzellt
So, here it would I've done from your advice. Please check it out and let me know how it can improve.

Assume, on the contrary, that f: A--> B is an isomorphism. Let b e B be an element that cannot be generated by two elements in B. Since f is onto, there exists a e A such that f(a) = b. Since a e A, it can be generated by two elements, say x,y e A. So, a = x^n*y^m, n,m e Z.
Thus, f(x^n*y^m) = f(a) = b. This is a contradiction. QED
Not quite. Since isomorphism preserve the group operation we get that

$\displaystyle f(a)=f( x^n\cdot y^m)=f(x^n)\cdot f(y^m)=[f(x)]^n[f(y)]^m=b$

but $\displaystyle f(x) \in B$ and $\displaystyle f(y) \in B$

but this shows that b is generated by two elements. Contradiction!

6. ## Re: Isomorphism Test

actually, a could be ANY "word" in x and y, for example:

yx3y-2

xyx2yx-1y4

yxyxyx

because there is no reason to suppose that x and y commute.

there is no stipulation that A is finite, so no guarantee that yx can be written any other way (A could be the free group on 2 generators).

what you CAN say, is that for some finite index set S (which we may take to be the first n integers):

$\displaystyle a = \prod_{i \in S} x^{k_i}y^{m_i}$ where ki and mi are integers that aren't BOTH 0, for each i (this excludes the identity element, but the identity element is trivially generated finitely by the empty set- this word isn't "reduced" because we might have, for example, mi = 0, ki+1 ≠ 0, in which case we could "combine" some "x terms", but it will serve for your proof).