Showing Isomorphisms in Subgroups

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Homework Help Overview

The problem involves the subgroup G of the direct product Z x Z, defined by the condition that the sum of its elements is even. The original poster seeks to establish that G is a proper subgroup and to explore whether it is isomorphic to Z x Z.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between cardinality and isomorphism, with some questioning the necessity of equal cardinalities for establishing an isomorphism. The identification of generators for both G and Z x Z is also a point of inquiry.

Discussion Status

Some participants have provided hints regarding the identification of generators for G and Z x Z, suggesting that a visual representation may aid in understanding. The discussion reflects a mix of interpretations and attempts to clarify the nature of the generators required for G.

Contextual Notes

There is a mention of the need for generators to lie within G, as well as the consideration of the parity of sums in the context of subgroup generation.

Obraz35
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Homework Statement


Let G be a subset of Z x Z (direct product) where G = {(a,b)|a+b = 2k for some integer k}.
I'd like to show that G is a proper subgroup of Z x Z and determine whether G is isomorphic to Z x Z.

I am pretty sure I have shown that it is a proper subgroup but the isomorphism part is what is giving me trouble. Here the cardinality of the two sets is equal and it seems reasonable to be able to find a bijection between them, but I am not sure on the details of showing this.
 
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It may seem reasonable that if the cardinalities are equal there should be an isomorphism, but that's not necessarily true. In this case it is. Z x Z is an abelian group which can be generated by two elements. G is also an abelian group with two generators. Can you find them? Hint: draw Z x Z in the plane and circle the elements belonging to G. It's a lattice. You can get an isomorphism by mapping generators to generators.
 
I see the visual representation, but I guess I am just not seeing how this gives you the generators. The only way I can think to generate G is with (2,0), (0,2) and (1,1). I'm also not sure how to map the rest of the elements that are not generators.
 
(0,2)=(-1)*(2,0)+2*(1,1). You don't need three generators. Clearly, (a,b)=a*(1,0)+b*(0,1). Only two generators. Your G is not that much different.
 
But don't the generators of G have to lie in G? Because 1+0 is odd.
 
No, I meant (1,0) and (0,1) generate Z x Z. The point is that you can also find two elements that generate G.
 
Okay, I see it now. Thanks.
 

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