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The question is to calculate the homology groups of the chain complex:

$0 \to A \stackrel{n}{\to} A \to 0$, where $A$ is an Abelian group and $n \in \mathbb{N}$.

I don't see a nice way to get anything out of this other than $H_1 = \mathrm{ker} \ \partial_1$ and $H_0 = A / \mathrm{Im} \ \partial_1$

Some quick calculations on small groups like $\mathbb{Z}/ 4\mathbb{Z}$ seem to confirm no 'nice' pattern, but I am probably missing something

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could you elaborate more? for the case you mentioned it seems $H_{1}=a:na=0$, $H_{0}=A/nA$. –  Kerry Mar 11 '11 at 6:03
    
@user7887: I am sure you are correct. But for $n=3$ do we not have $\mathrm{Ker} \ \partial_0 = \mathbb{Z}/4\mathbb{Z}$ and $\mathrm{Im } \ \partial_1 = \mathbb{Z}/4\mathbb{Z}$? (As $\{ 0,1,2,3 \} \mapsto \{ 0,3,2,1 \}$) –  Juan S Mar 11 '11 at 6:07
    
Hi, for $n=3$ you should have $Ker \partial_{0}$ to be $0$ and $Im \partial_{1}=\mathbb{Z}/4\mathbb{Z}$. Notice multiplying by 3 should be an isomorphism in $Z_{4}$, hence injective. So $H_{0}=0$ and $H_{1}=0$ as well. –  Kerry Mar 11 '11 at 6:32
    
@user7887 Did you mean $\mathrm{Ker} \ \partial_0 = \mathbb{Z}/4\mathbb{Z}$? Clearly the map $A \to 0$ has kernel $A$. But I agree that $H_0=H_1=0$. Ok then - you are saying that if $A=\mathbb{Z}/4\mathbb{Z}$ that $A/nA=0$ because multiplication by 3 is an isomorphism? (refresh my memory is this because 3 and 4 are co-prime? –  Juan S Mar 11 '11 at 6:36
    
Hi, if I understand correctly $\partial_{0}$ is the trivial map $0\rightarrow A$ for whatever group $A$ you use. And $\partial_{1}$ is the multiplication map. Then $H_{0}$ is defined to be $\ker \partial_{1}/im \partial_{0}=0/0=0$, and $H_{1}=\ker \partial_{2}/im \partial_{1}=A/A=0$. Yes, I think it is because 3 and 4 are coprime. –  Kerry Mar 11 '11 at 6:45

2 Answers 2

up vote 4 down vote accepted

We are given a chain complex: $ \ldots \to C_n \stackrel{\partial_n}{\to} \ldots \to C_1 \stackrel{\partial_1}{\to} C_0 \stackrel{\partial_0}{\to} \ldots$ where $C_0 = C_1 = A$, $\partial_0$ is the zero map and $\partial_1$ is multiplication by $n$. Also $C_n = 0$ for $n \neq 0,1$.

Since $C_n = 0$ for $n \neq 0,1$, we must have $\mathrm{Ker}\partial_n = 0$ for $n \neq 0,1$ (because $\mathrm{Ker}\partial_n\leq C_n = 0$). So $H_n = 0$ for $n \neq 0,1$.

Next observe that, because $C_2 = 0$, we must have $\mathrm{Im}\partial_2 = 0$. So $H_1 = \mathrm{Ker}\partial_1/\mathrm{Im}\partial_2 \cong \mathrm{Ker}\partial_1 = \{a \in A | na = 0\}$.

Finally, $H_0 = \mathrm{Ker}\partial_0/\mathrm{Im}\partial_1$. Since $\partial_0: A \to A$ is the zero map, $\mathrm{Ker}\partial_0 = A$. Also since $\partial_1: A \to A$ is multiplication by $n$, $\mathrm{Im}\partial_1 = nA$. So $H_0 = A/nA$.

We conclude that for the given Chain complex:

$H_0 = A/nA$, $H_1 = \{a \in A | na =0\}$ and $H_n = 0$ otherwise.

The issue you are describing (with $\mathbb{Z}/4\mathbb{Z}$, and $n=3$) is not really a problem. The above calculation is for an arbitrary abelian group $A$ and integer $n$. You should think of the final answer as a formula of sorts: once you input a choice of abelian group $A$ and integer $n$, you are able to quickly determine the homology without having to go through the lengthy argument every time. However, just as is the case with functions, if you change your input (in this case the pair $(A,n)$) then the value of the function (in this case the Homology of the Chain complex) will also change.

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Thanks - this makes sense –  Juan S Mar 12 '11 at 2:50

You actually picked out the right class of counterexamples, since for your average abelian group (Z or R), you won't find anything interesting. $H_0$ always equals $A/nA$, but $H_1$ 'defines' something that is called torsion, i.e. the vanishing of finite-order elements in a group. In some literature (for example, Hatcher's AT), you'll see the definition $$n-{\rm Torsion}(G) = {\rm Ker}(g \mapsto n g | g \in G),$$ which coincides with your $H_1$. I hope that'll give you something to look up in your favorite algebra reference.

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Thank you, I'll have a look at this –  Juan S Mar 12 '11 at 2:50

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