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This is problem 2.25(ii) from Rotman's Homological Algebra Text.

1. Problem

Let $p$ be a prime, $B_n$ be a cycle group of order $p^n$, and $A = \bigoplus_{n=1}^\infty B_n$. Show that $$ \mathcal S := Hom_\mathbb Z(A, A) \ncong \bigoplus Hom_\mathbb Z(A, B_n) =: \mathcal T $$

Hint: $\mathcal S$ has an element of infinite order while $\mathcal T$ has elements of only finite order.

2. Partial Proof

For $\mathcal S$, note that $1_A$ has the desired property because for each $n \in \mathbb N$ there exists an $a \in A$ such that $p^n 1_A(a) = p^n a \ne 0$.

Update

See my attempted answer.

IsaacR24
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    Re question 1, "finite order" is meant with respect to pointwise addition, that is, the group operation $(f,g)\mapsto f+g$ given by $(f+g)(x)=f(x)+g(x)$. – David Moews Sep 24 '23 at 15:50
  • @DavidMoews: Under your definition, wouldn't both sets have elements order $\le p^n$? Both images are contained in $A=\bigoplus B$, whose elements have order that divides $p^n$. – IsaacR24 Sep 24 '23 at 17:28
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    Keep in mind that you are taking a sum over $n$. – David Moews Sep 24 '23 at 17:29
  • I still don't see it. Let $(k_i b) \in A$. Then $p^n (k_i b) = (p^n k_i b) = 0$. What am I missing? – IsaacR24 Sep 24 '23 at 17:50
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    You are not using one single $n$! You are taking a sum over $n$. Outside a specific summand of $A$ or a specific $B_n$, there is no "$p^n$". – David Moews Sep 24 '23 at 17:52
  • Ah hah! You're perhaps looking at the text itself, because I didn't even use $B_n$ here. Thanks! Will think on this a bit. – IsaacR24 Sep 24 '23 at 17:55
  • $A$ is a direct sum, for each $(b_i) \ \in A$ there exists a $N$ such that $b_n = 0$ for each $n \ge N$. Then $p^N(b_i) = 0$. – IsaacR24 Sep 24 '23 at 18:03
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    Yes, that's true. So, every element of $A$ has finite order. – David Moews Sep 24 '23 at 18:05
  • That's bad! I need to show there's an element in $\mathcal S$ that has infinite order, but the image of every element is contained in $A$ which has only finite order elements.. – IsaacR24 Sep 24 '23 at 18:06
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    Fortunately, $\cal S$ and $A$ are not the same, so it's possible for $\cal S$ to have an element of infinite order although $A$ does not. – David Moews Sep 24 '23 at 18:07
  • Under your definition of order, it seems like you're saying $f^2(x) = f(x) + f(x)$. Then $f^{p^N}(x) = 0$ for each $x \in A$. I must be misunderstanding what order means. – IsaacR24 Sep 24 '23 at 18:10
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    For $p^n f\in{\cal S}$ to vanish, there must be some fixed $n$ such that $p^nf(x)=0$ for all $x$. Otherwise, if $n$ is not fixed, even writing down the expression "$p^n f$" makes no sense. – David Moews Sep 24 '23 at 18:12

1 Answers1

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For $\mathcal S$, note that $1_A$ has the desired property because for each $n \in \mathbb N$ there exists an $a \in A$ such that $(1_A)(a)$ has order greater than $p^n$. For example, $a = (0, \cdots, b_{n+1}, 0, \cdots )$. Hence, $\mathcal S$ has an element of infinite order.

Let $(f_i) \in \mathcal T$ Because $(f_i)$ is a direct sum, there exists a $N$ such that $f_n = 0$ for all $n \ge N$. But then $(f_i)^{p^N} = 0$. So $\mathcal T$ has all elements of finite order.

IsaacR24
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