I am trying to prove that if $G := \langle a_k : k \in \mathbb{Z} \rangle$ and $H:= \langle a_{k+1} - a_k : k \in\mathbb{Z} \rangle$ with both of them being free abelian groups, then $G/H \cong \mathbb{Z}$. Here, $a_k$ is just some arbitrary element ,so that (for example) $\langle a_1, a_2 \rangle = \mathbb{Z}^2$. This questions comes in the context of finding the simplicial homology of $\mathbb{R}$. I know that the homology group $H_0(\mathbb{R})$ is $\mathbb{Z}$ since $\mathbb{R}$ is path connected. However, since I haven't covered the equivalence of singular and simplicial homology in class yet, this problem - which was given before the equivalence - should be provable by using a simplicial structure. Here, I gave the delta simplex structure to $\mathbb{R}$ by having the $1$-simplicies as the interval $[n, n+1]$ for all $n \in \mathbb Z$, so that the 1st simplicial homology group is exactly the group $G/H$ as given. How can I find the generator of $G/H$? It is tricky as we have infinitely many elements.

If you don't know anything about homology groups, then you can still attempt this problem.

Thank you

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    $\begingroup$ Umm, what is $a_k?$ $\endgroup$ Commented May 17 at 1:07
  • $\begingroup$ @ThomasAndrews Generators of the free group $G$, indexed by $\mathbb{Z}$. $\endgroup$
    – Tom
    Commented May 17 at 1:12
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    $\begingroup$ By the way, perhaps you have a typo: $H_1(\mathbb R)=0$ since $\mathbb R$ is contractible. $\endgroup$
    – Kenta S
    Commented May 17 at 1:15
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    $\begingroup$ @tom I have no doubt this is likely true, but let OP answer. The question doesn't state what OP means, and OP should clarify. Never once are the words "free" or "abelian" used. $\endgroup$ Commented May 17 at 1:21
  • $\begingroup$ How do you know $H \lhd G$? If you don't know that, it doesn't make sense to talk about $G/H$. Or is $G$ the free abelian group on generators $\{a_k \mid k \in \Bbb Z \}$? $\endgroup$ Commented May 17 at 2:56

1 Answer 1


There is a surjective homomorphism $G\to\mathbb Z$ sending $\sum_{k\in\mathbb Z}N_ka_k$ (where all but finitely many $N_k$ are zero) to $\sum_{k\in\mathbb Z}N_k$.

The kernel consists of $\sum_{k\in\mathbb Z}N_ka_k$ such that $\sum_{k\in\mathbb Z}N_k=0$. Such elements are linear combinations of $a_k-a_{k+1}$. Indeed, $$\sum_{k\in\mathbb Z}N_ka_k=\sum_{k\in\mathbb Z}\big(\sum_{i\ge k}N_i\big)(a_k-a_{k+1}),$$ where the sum $\displaystyle \sum_{i\ge k}N_i$ is well-defined since all but finitely many $N_k$ are zero. Thus the kernel is $H$.


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