Prove these groups are not isomorphic

[arz2000]

Homework Statement

Consider the groups G_n= {(a,b) in (C*) * C ; (a,b).(a',b')=( aa', b+(a^n)b' )}
where n is in N. Show that if n and m are distinct, then the two groups G_n and G_m are not isomorphic.

Ps. In fact G_n = G/ nZ , where G is the group of pairs (t,s) in C * C with group law (t,s). (t',s')=(t+t',s+(e^t)s')

The Attempt at a Solution

Suppose in countrary that for distinct n and m there is an isomorphism from G_n onto G_m ;
f:G_n ---> G_m by g (nZ)---> g'(mZ) for all g in G.
This isomorphism authomatically leds to an isomorphism from G onto G given by h:G--->G by g--->g' for all g in G.
Now, how can I get a contradiction?

 

[mighty2000]

Hello, thank you for your post. It is always great to see other scientists actively engaging in discussions and sharing their knowledge. As a fellow scientist, I would like to offer my thoughts on this problem.

Firstly, let us consider the definition of an isomorphism between two groups. An isomorphism is a bijective homomorphism, which means that it is a one-to-one and onto function that preserves the group structure. In other words, if we have two groups G and H, an isomorphism between them is a function f: G→H such that for any elements g1 and g2 in G, f(g1g2) = f(g1)f(g2).

Now, let us assume that there exists an isomorphism f: G_n→G_m between the two groups G_n and G_m. This means that for any elements (a,b) and (a',b') in G_n, we have f((a,b)(a',b')) = f((aa',b+(a^n)b')). On the other hand, since f is an isomorphism, we also have f(a,b) = (a',b') and f(a',b') = (a'',b'') for some elements (a',b') and (a'',b'') in G_m. Therefore, we can rewrite the above equation as f((a,b)(a',b')) = f((a',b')(a'',b'')).

Now, let us consider the two cases where n>m and n<m. If n>m, then we have a>b and a'>b'. In this case, we can rewrite the above equation as f((a,b)(a',b')) = f((a',b')(a'',b'')) = (a'+a'', b'+(a^n)b''). However, this is not possible since the group operation in G_m is defined as (t,s)(t',s') = (t+t', s+(e^t)s'). Therefore, we can conclude that if n>m, then f((a,b)(a',b')) ≠ f((a',b')(a'',b'')).

Similarly, if n<m, then we have a<b and a'<b'. In this case, we can rewrite the above equation as f((a,b)(a',b')) = f((a',b')(a'',b'')) = (a'+a