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Let $G$ be an abelian group. If all proper subgroups of $G$ are finite, what can we say about $G$. Are the any properties, $G$ is guaranteed to have.

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  • $\begingroup$ Maybe check this out math.stackexchange.com/questions/155521/… $\endgroup$
    – H_B
    Commented Sep 13, 2015 at 1:44
  • $\begingroup$ If $G$ is finitely generated it must be finite under your assumption. $\endgroup$
    – Sempliner
    Commented Sep 14, 2015 at 2:02
  • $\begingroup$ Not exactly the same thing, but a group whose finitely generated subgroups are all finite is called locally finite. It's a wider class than the one you're considering (which only contains finite groups and the Prüfer groups). $\endgroup$ Commented Sep 15, 2015 at 12:16

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$\newcommand{\bZ}{\mathbb Z} \newcommand{\soc}{\operatorname{soc}}$Obviously, finite abelian groups as well as the $p$-Prüfer groups $\bZ(p^\infty)$ for primes $p$ have this property. These are in fact all the possibilities.

Proposition. Let $G$ be an abelian group such that all proper subgroups of $G$ are finite. Then either $G$ is finite or there exists a prime $p$ such that $G \cong \bZ(p^\infty)$.

Proof: Suppose that $G$ is not finite. First note that $G$ is indecomposable: If $G \cong H \oplus K$ with nonzero $H$ and $K$, then at least one of $H$ and $K$ must be infinite. Hence $G$ has a proper infinite subgroup.

Since all cyclic subgroups of $G$ are finite, $G$ is a torsion group. Hence it decomposes as $G \cong \bigoplus_{p \in \mathbb P} G_p$, where the $G_p$s are (uniquely determined) $p$-groups. But since $G$ is indecomposable, $G \cong G_p$ for some $p$. Therefore $G$ is an indecomposable infinite abelian $p$-group.

Recall that the height, $H(x)$, of an element $x \in G$ is defined as the maximal $n \in \mathbb N_0 \cup \{\infty\}$ such that there exists $x_n \in G$ with $x=p^nx_n$. Let $G^\infty \subset G$ denote the subset of all elements of infinite height. It is easily checked that $G^\infty$ is a subgroup of $G$.

Claim: If $G$ does not have finite exponent, then $G^\infty$ is infinite.

For $k \ge 0$, let $\soc^k(G)$ denote the $k$-th socle of $G$, that is, the subgroup consisting of all elements of order at most $p^k$. By assumption, $\soc^k(G)$ is finite for all $k$. Since we also assume that $G$ does not have finite exponent, $\soc^{k-1}(G) \subsetneq \soc^{k}(G)$ for all $k \ge 1$. We will show: For all $k$, there exists an $x \in \soc^k(G) \setminus \soc^{k-1}(G)$ (i.e., $\operatorname{ord}(x) = p^k$) with $H(x) = \infty$. Note that $$ \bigcup_{l \ge k} p^{l-k}(\soc^l(G) \setminus \soc^{l-1}(G)) \subset \soc^k(G) \setminus \soc^{k-1}(G). $$ Since the set on the right hand side is finite, while each of the sets in the infinite union on the left is non-empty, it follows that there must exist at least one $x \in \soc^k(G) \setminus \soc^{k-1}(G)$ which is divisible by arbitrary high powers of $p$. But then $H(x) = \infty$. Thus $G^\infty$ is infinite and the claim is shown.

Now we have two cases: If $G$ has finite exponent, then $G=\soc^n(G)$ where $p^n = \operatorname{exp}(G)$. However, then $\soc^{n-1}(G)$ is finite as a proper subgroup, and $\soc^n(G)/\soc^{n-1}(G) \cong (\bZ/p\bZ)^{(I)}$ for some index set $I$. If $I$ is infinite, one easily produces an infinite proper subgroup of $G$. Therefore $I$, and hence also $G$, is finite.

If $G$ does not have finite exponent, then we must have $G = G^\infty$. But then every element is divisible by arbitrary high powers of $p$. Since $G$ is a $p$-group, this implies that $G$ is divisible. It follows that $G$ is a direct sum of $p$-Prüfer groups. Since $G$ is also indecomposable, in fact $G \cong \bZ(p^\infty)$.

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  • $\begingroup$ Maybe I should also add that the nonabelian picture is of course completely different: For instance, there are the Tarski monsters. $\endgroup$
    – moonlight
    Commented Sep 15, 2015 at 12:42

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