Abstract Nonsense

Crushing one theorem at a time

Homomorphisms Between Finitely Generated Abelian Groups (Pt. I)

Point of Post: In this post we derive the necessary information to describe \text{Hom}_\mathbb{Z}(A,B) for any finitely generated abelian groups A,B according to their cyclic group decomposition.

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In this post, using our ability to split Hom across finite products we shall show how to effectively calculate the Hom group between any two finitely generated abelian groups. The only caveat to this is that one must first decompose the finitely generated abelian groups into their cyclic decomposition as is guaranteed by the structure theorem. Probably the most useful aspect of this post is that it will, if you keep your bookkeeping orderly, the number of homomorphisms between two finite abelian groups (presented as a product of cyclic groups) and the generators that will enable one to, in theory, produce every single homomorphism.

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We shall prove this in a series of theorems, the first being the most difficult. Namely, we want to calculate \text{Hom}_\mathbb{Z}(\mathbb{Z}_{p^\ell},\mathbb{Z}_{p^k}) for a prime p. While this is the most work, the intuitive idea is clear enough. We show what homomorphisms are in \text{Hom}_\mathbb{Z}(\mathbb{Z}_{p^\ell},\mathbb{Z}_{p^k}), count them, and then just produce an element of that order. Regardless, here we go:

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Theorem: Let p be a prime. Then, \text{Hom}_\mathbb{Z}(\mathbb{Z}_{p^\ell},\mathbb{Z}_{p^k})\cong\mathbb{Z}_{p^{\min\{\ell,k\}}}.

Proof: For sake of notational convenience allow us to denote elements of \mathbb{Z}_{p^\ell} and \mathbb{Z}_{p^k} by [\cdot]_k and [\cdot]_\ell respectively. We then define, for when it makes sense, the function f_{[x]_k} to be the function \mathbb{Z}_{p^\ell}\to\mathbb{Z}_{p^k} given by f_{[x]_k}([a]_\ell)=a[x]_k. Let then X=\left\{f_{[x]_k}:|[x]_k|\mid p^\ell\right\}. We claim that, as sets, X=\text{Hom}_\mathbb{Z}(\mathbb{Z}_{p^\ell},\mathbb{Z}_{p^k}). Indeed, suppose first that f\in\text{Hom}_\mathbb{Z}(\mathbb{Z}_{p^\ell},\mathbb{Z}_{p^k}). From the identity f([a]_\ell)=f(a[1]_\ell)=af([1]_\ell) it’s clear that f=f_{f([1]_\ell)}. Moreover, from the general theorem that given a group homomorphism v:G\to H (for finite groups G,H)  one must have |v(g)|\mid |g| for all g\in G we may easily conclude that |f([1]_\ell)|\mid |[1]_\ell|=p^\ell from where it immediately follows that f\in X. Conversely, we claim that every element of X is a well-defined homomorphism. Indeed, we first must show that for [x]_k with |[x]_k|\mid p^\ell the mapping f_{[x]_k} is well-defined. To see this, suppose that [a]_\ell=[b]_\ell, then p^\ell\mid a-b and so f_{[x]_k}([a]_\ell)-f_{[x]_k}([b]_\ell)=(a-b)[x]_\ell=0 since |[x]_k|\;\mid p^\ell\mid a-b. To see that f_{[x]_k} really the is a homomorphism we merely note that

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\begin{aligned}f_{[x]_k}([a]_\ell+[b]_\ell) &=f_{[x]_k}([a+b]_\ell)\\ &=(a+b)[x]_k=a[x]_k+b[x]_k\\ &=f_{[x]_k}([a]_\ell)+f_{[x]_k}([b]_\ell)\end{aligned}

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and so f_{[x]_k}\in\text{Hom}_\mathbb{Z}(\mathbb{Z}_{p^\ell},\mathbb{Z}_{p^k})–from here the set equality we claimed follows.

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What we now claim is that \#(X)=p^{\min\{\ell,k\}}. Of course, this is equivalent to counting how many elements of \mathbb{Z}_{p^k} have order dividing p^\ell. Suppose first that k\leqslant \ell, then clearly every element of \mathbb{Z}_{p^k} has order dividing p^\ell since (by Lagrange’s theorem, of course) since any such elements order would divide p^k which divides p^\ell. Thus, if k\leqslant \ell then \#(X)=p^k. Suppose now that \ell\leqslant k. We begin then by recalling that the order of [x]_k=x[1]_k is equal to \displaystyle \frac{p^k}{(p^k,x)} so that the division of p^\ell by p^k is (p^k,x)p^{\ell-k}. For this to be an integer it is a necessary and sufficient condition that p^{k-\ell}\mid x. But, the number of such x\in\{1,\cdots,p^k\} is evidently p^\ell and so it follows that \#(X)=p^\ell. Taking these two cases into account the assertion that \#(X)=p^{\min\{\ell,k\}} follows.

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Lastly, we produce an element of X of order p^{\min\{\ell,k\}}. To do this we begin by noting that |f_{[x]_k}|=|[x]_k|. Indeed, since 0=|f_{[x]_k}|f([a]_\ell)=|f_{[x]_k}|a[x]_k for all [a]_\ell\in\mathbb{Z}_{p^\ell} we see taking a=1 that |f_{[x]_k}| annihilates [x]_k so that |[x]_k|\leqslant |f_{[x]_k}|. That said, evidently |[x]_k|f([a]_\ell)=a|[x]_k|[x]_k=0 for all [a]_\ell\in\mathbb{Z}_{p^\ell} so that |f_{[x]_k}|\leqslant |[x]_k| and so equality follows. Thus, to find an element of X of order p^{\min\{\ell,k\}} it suffices to find an element of \mathbb{Z}_{p^k} of order p^{\min\{\ell,k\}}. But, clearly \left[p^{k-\min\{\ell,k\}}\right]_k is such an element.

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From this we may conclude that \text{Hom}_{\mathbb{Z}}(\mathbb{Z}_{p^\ell},\mathbb{Z}_{p^k}) is a group of order p^{\min\{\ell,k\}} and contains an element, f_{\left[p^{k-\min\{\ell,k\}}\right]}, of order p^{\min\{\ell,k\}}. Thus, it clearly follows that \text{Hom}_\mathbb{Z}(\mathbb{Z}_{p^\ell},\mathbb{Z}_{p^k})\cong\mathbb{Z}_{p^{\min\{\ell,k\}}}. \blacksquare

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With this and our result about splitting Hom over products we can go a step further and classify the Hom between any two finite cyclic groups

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Theorem: Let m,\in\mathbb{N}, then \text{Hom}_\mathbb{Z}(\mathbb{Z}_{m},\mathbb{Z}_n)\cong\mathbb{Z}_{(m,n)}

Proof: We factorize n and m as p_1^{\alpha_1}\cdots p_k^{\alpha_k} and p_1^{\beta_1}\cdots p_k^{\beta_k}.  We next note  that \text{Hom}_\mathbb{Z}\left(\mathbb{Z}_{p_i^{\alpha_i}},\mathbb{Z}_{p_j^{\beta_j}}\right)=\{0\} for i\ne j–this follows immediately from the first isomorphism theorem which tells us that the order of the image of any homomorphism \mathbb{Z}_{p_i^{\alpha_i}}\to\mathbb{Z}_{p_j^{\beta_j}} would have to be a common divisor of p_i^{\alpha_i} and p_j^{\beta_j}, and thus have order one. From this, the Chinese remainder theorem (only the group part), the previous theorem, and the way Hom splits we can easily deduce that

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\begin{aligned}\text{Hom}_{\mathbb{Z}}(\mathbb{Z}_m,\mathbb{Z}_n) &\cong \text{Hom}_{\mathbb{Z}}\left(\prod_{i=1}^{n}\mathbb{Z}_{p_i^{\alpha_i}},\prod_{j=1}^{n}\mathbb{Z}_{p_j^{\beta_j}}\right)\\ & \cong \prod_{i,j=1}^{n}\text{Hom}_{\mathbb{Z}}\left(\mathbb{Z}_{p_i^{\alpha_i}},\mathbb{Z}_{p_j^{\beta_j}}\right)\\ & \cong \prod_{j=1}^{n}\text{Hom}_{\mathbb{Z}}\left(\mathbb{Z}_{p_j^{\alpha_j}},\mathbb{Z}_{p_j^{\beta_j}}\right)\\ & \cong \prod_{j=1}^{n}\mathbb{Z}_{p_j^{\min\{\alpha_j,\beta_j\}}}\\ &\cong \mathbb{Z}_{p_1^{\min\{\alpha_1,\beta_1\}}\cdots p_n^{\min\{\alpha_n,\beta_n\}}}\\ &=\mathbb{Z}_{(m,n)}\end{aligned}


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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 | , , , , , , , , , , ,


  1. […] Homomorphisms Between Finitely Generated Abelian Groups (Pt. II) Point of Post: This is a continuation of this post. […]

    Pingback by Homomorphisms Between Finitely Generated Abelian Groups (Pt. II) « Abstract Nonsense | November 14, 2011 | Reply

  2. […] promised, let’s use this to make a previously messy computation easy. Namely, let’s compute , or really (since we know the answer) prove this is […]

    Pingback by The Hom Functor is Left Exact « Abstract Nonsense | January 30, 2012 | Reply

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