Group Theory: Proving Non-Isomorphism of G_n and G_m

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To prove that the groups G_n and G_m are not isomorphic for distinct n and m, one must consider the structure of G_n as G/nZ, where G consists of pairs (t, s) in C * C with a specific group law. An isomorphism between G_n and G_m would imply a corresponding isomorphism from G to itself, which is not possible due to the differing structures imposed by n and m. The hint suggests using short exact sequences to analyze the semidirect product nature of the groups. By demonstrating that any lift from G_n to G_m would lead to contradictions in the group structures, the non-isomorphism can be established. This analysis confirms that G_n and G_m maintain fundamentally different algebraic properties.
arz2000
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Hi all,

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')

Hint. An isomorphism G_n ~ G_m would lift to a map from G to G, show that this map whould have to be an isomorphism.
(re. exercise 10.2 of Representation Theory by Fulton and Harris)


Many Thanks
 
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To use the hint, write down the corresponding two short exact sequences (the groups are semidirect products!) and then formulate what "lift" means and why you get two different structures on ##G##.
 
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