# Isomorphism Test

• Oct 31st 2012, 02:26 PM
jzellt
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
• Oct 31st 2012, 04:46 PM
TheEmptySet
Re: Isomorphism Test
Quote:

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 $b \in B$ 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 $a \in A$ such that $f(a)=b$.

Now use the fact that $a$ can be generated by two elements and the isomorphism preserve the group operations to show that $b$ can be generated by two elements.
• Oct 31st 2012, 04:49 PM
jzellt
Re: Isomorphism Test
Thanks. Can you fix your latex.. Its not showing correctly in the last sentence.
• Oct 31st 2012, 07:23 PM
jzellt
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
• Oct 31st 2012, 07:33 PM
TheEmptySet
Re: Isomorphism Test
Quote:

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

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

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

but this shows that b is generated by two elements. Contradiction!
• Nov 1st 2012, 06:19 AM
Deveno
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):

$a = \prod_{i \in S} x^{k_i}y^{m_i}$ 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).