Question

If f from g to g' is a isomorphism, and g is abelianthen prove that g' is also an abelian group

Answer 1

Solution:
Given that f in g→g' is an isomorphism.
Proof:
(A) If g' is abelian, we have xy=yx, for any x,y∈g'.
Consider any two element in g. say c and d.
Since f is an isomorphism (more specifically a surjection), we have
c=f(a) and d=f(b) for some a,b∈g'.

Therefore,
cd=f(a)f(b) =f(ab) =f(ba) =f(b)f(a) =dc
This means g is also abelian.

(B) If g is abelian, we have xy=yx, for any x,y∈g. Now consider two elements in g', say a and b. Now, since f is a mapping, we have 
c=f(a) and d=f(b) for some c,d∈g.

Therefore,
f(ab) =f(a)f(b) =cd= dc =f(b)f(a) =f(ba)

Since f is an isomorphism (more specifically injective), we have ab=ba.
This means g' is also abelian.

Proved.