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Do you have an example of an abelian group $G$ with a sequence of mutually distinct nontrivial subgroups $(A_n)$ such that $$\dots \le A_n\le\dots \le A_2\le A_1\le A_0=G$$ and $$\bigcap_{n=1}^\infty A_n=\{1\}$$ ?

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up vote 5 down vote accepted
  1. Mutually disjoint does not make sense for subgroups since all subgroups contain $1$.

  2. If the abelian group is finite, then any decreasing sequence of subgroups is stationary. ie. $\exists n_0 \in \mathbb{N}$ such that $A_n = A_{n_0}$ for all $n\geq n_0$. Hence what you ask for is that $A_{n_0} = \{1\}$ for some $n_0 \in \mathbb{N}$.

  3. If the group is finitely generated and infinite, then it must contain $\mathbb{Z}$, in which case you can take the subgroups $A_n = (2^n)$

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I think 1 should be editted out after I corrected a mistake in my question. – user795571 May 9 '14 at 12:22

This question becomes more interesting when you stipulate that the subgroups $A_i$ are of finite index in $G$. Your question then translates as,

Does there exist an infinite, residually finite abelian group?

The most obvious example is $\mathbb{Z}$, taking the subgroups to be $A_i=\langle n^i\rangle$, where $n$ is some fixed integer of absolute value greater than $1$.

I should point out that in this stronger setting you cannot simply say that "if your group contains a copy of $\mathbb{Z}$ then it works" (although you can in the weaker setting, which is what the other answers do). For example, dispensing with the abelian condition for a second, the Baumslag-Solitar group $BS(2, 3)=\langle a, t; t^{-1}a^2t=a^3\rangle$ contains elements of infinite order (for example, the element $t$ by Britton's lemma) but it is not Hopfian so it is not residually finite.

Now, it turns out that every finitely generated, abelian group works. That is, that every infinite, finitely-generated abelian group is residually finite. The following two exercises prove this.

Exercise 1: Prove that the following are equivalent for $G$ infinite:

  • For every element $1\neq g\in G$ there exists a homomorphism $\phi:G\rightarrow F$ where $F$ is finite and $\phi(g)\neq 1_F$.
  • $G$ is residually finite. That is, there exists a sequence of mutually distinct nontrivial subgroups $(A_n)$ of finite index in $G$ such that the following hold. $$\dots \le A_n\le\dots \le A_2\le A_1\le A_0=G$$ $$\bigcap_{n=1}^\infty A_n=\{1\}$$

Use this to complete the following exercise.

Exercise 2: Prove that every finitely-generated abelian group is residually finite.


  1. Prove that if $H$ is residually finite then $G=H\times H$ is residually finite.

  2. Prove that if $H$ is residually finite and $H$ has finite index in $G$ then $G$ is residually finite.

  3. Apply the fundamental theorem for finitely generated abelian groups.

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thanks. It can help me have more info about subgroups lattice of abelian groups. – user795571 May 9 '14 at 15:49

Let $$G= \mathbb{Z} \oplus \mathbb{Z} \oplus \cdots$$ (or replace $\mathbb{Z}$ with any abelian group) and $$G_n= \underset{n \ \text{times}}{\underbrace{\{0\} \oplus \cdots \oplus \{0\}}} \oplus \mathbb{Z} \oplus \cdots \leq G.$$

Then $ \cdots < G_2<G_1 < G_0=G$ and $\bigcap\limits_{n \geq 1} G_n =\{0\}$.

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Inspiring‌‌‌‌‌! – user795571 May 9 '14 at 16:00

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