Abstract Nonsense

Crushing one theorem at a time

Homomorphisms Between Finitely Generated Abelian Groups (Pt. II)

Point of Post: This is a continuation of this post.

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The next step is to figure out how Hom works between finite and infinite cyclic groups. In particular we have the following two little theorems:

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Theorem: Let A be a finite abelian group then \text{Hom}_\mathbb{Z}(A,\mathbb{Z})=\{0\}, in particular \text{Hom}_\mathbb{Z}(\mathbb{Z}_m,\mathbb{Z})\cong0 for all m\in\mathbb{N}.

Proof: We merely note that if f:A\to \mathbb{Z} is a \mathbb{Z}-map then f(x) has finite order for each x\in A, but the only elements of \mathbb{Z} of finite order are 0 and so f(x)=0 for all x\in A. The conclusion follows. \blacksquare

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Now, for the other direction we just recall the more general result that for commutative untial left R-modules M one has that \text{Hom}_R(R,M)\cong R. From this we get the following:

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Theorem: For any m\in\mathbb{N} one has that \text{Hom}_\mathbb{Z}(\mathbb{Z},\mathbb{Z}_m)\cong\mathbb{Z}_m.

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Thus, with all these results we are finally able to state the following theorem quite unequivocally:

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Theorem: Let \ell_1,\cdots,\ell_n,k_1,\cdots,k_m\in\mathbb{N} and r,s\in\mathbb{N}\cup\{0\}. Then if 

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A\cong \mathbb{Z}^r\times\mathbb{Z}_{\ell_1}\times\cdots\times\mathbb{Z}_{\ell_ n}

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\displaystyle B\cong \mathbb{Z}^s\times\mathbb{Z}_{k_1}\times\cdots\times\mathbb{Z}_{k_m}

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\displaystyle \text{Hom}_\mathbb{Z}\left(A,B\right)\cong \mathbb{Z}^{rs}\times\prod_{j=1}^{m}\mathbb{Z}_{k_j}^r\times\prod_{i=1}^{n}\prod_{j=1}^{m}\mathbb{Z}_{(\ell_i,k_j)}

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Proof: By the functorality of Hom it suffices to prove this for when A,B equal the groups they’re isomorphic to. So, from there it’s just pure computation using the three theorems we’ve developed to take into account all the permutations we get. So, expanding we get

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\displaystyle \begin{aligned}\text{Hom}(A,B) &= \text{Hom}_\mathbb{Z}(\mathbb{Z}^r,\mathbb{Z}^s)\times\prod_{i=1}^{n}\text{Hom}_\mathbb{Z}(\mathbb{Z}_{\ell_i})^s\times\prod_{j=1}^{m}\text{Hom}_\mathbb{Z}\left(\mathbb{Z},\mathbb{Z}_{k_j}\right)^r\times\prod_{i=1}^{n}\prod_{j=1}^{m}\text{Hom}{_\mathbb{Z}}\left(\mathbb{Z}_{\ell_i},\mathbb{Z}_{k_j}\right)\\ &\cong \mathbb{Z}^{rs}\times\prod_{i=1}^{n}\{0\}^s\times\prod_{j=1}^{m}\mathbb{Z}_{k_j}^r\prod_{i=1}^{n}\prod_{j=1}^{m}\mathbb{Z}_{(\ell_i,k_j)}\\ &\cong \mathbb{Z}^{rs}\times\prod_{j=1}^{r}\mathbb{Z}_{k_j}^r\times\prod_{i=1}^{n}\prod_{j=1}^{m}\mathbb{Z}_{(\ell_i,k_j)}\end{aligned}

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from where the conclusion follows. \blacksquare

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So, what allows us to make the jump from this theorem to finding the homomorphism group between any two finitely generated abelian groups? Well, it all is a matter of perspective. Realistically, this doesn’t quite give us a algorithmic way to find the homomorphism group–it isn’t a formula that we can just take two finitely generated abelian groups and spit this out. No. But, theoretically this does answer the pertinent question. Indeed, the structure theorem tells us that every finitely generated abelian group fits into the hypothesis criteria of this past theorem–being isomorphic to a finite product of cyclic groups.

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[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. Module Theory. Clarendon, 1990. Print


November 14, 2011 - Posted by | Algebra, Group Theory, Module Theory, Ring Theory | , , , , , , , , , , , ,

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