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We know that if $G$ is a finite group, then all subgroups of $G$ are finite and the number subgroups of $G$ is finite. Now

1) If all proper subgroups of $G$ are finite, then is $G$ finite?

2) If the number subgroups of $G$ is finite, then is $G$ finite?

Thank you

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To add some variety to the party, another counter-example to (1) would be a Tarskii monster group. A group $G$ is a Tarskii monster group if (by definition) every proper, non-trivial subgroup is cyclic of order $p$ for $p$ a fixed prime. These are clearly counter-examples of (1), and such groups (for $p>>1$) were shown to exist in the early 80s. The proof is not trivial though! – user1729 Jan 25 '13 at 12:37
up vote 3 down vote accepted

The answer to the first question is no. An abelian example is given by the Prüfer group of type $p$. It can be shown that it has precisely one subgroup of every order $p^k$ and no others (other than the trivial one). In fact, for the abelian case this covers all cases as it can be further shown that if $G$ is infinite abelian all of whose proper subgroups are finite then $G$ must be Prüfer of type $p$.

As for the second question, the answer is yes. Let us show that if $G$ is infinite then it must have infinitely many subgroups. If $G$ has an element $g$ of infinite order then $<g>$ is isomorphic to $\mathbb Z$ which has infinitely many subgroups and so $G$ has infinitely many subgroups and we are done. We may thus continue under the assumption that $G$ has no element of infinite order. Let $g_1$ be some non-trivial element in $G$ and let $G_1=<g_1>$. It is a finite subgroup (by our assumption) and thus there is some $g_2 \in G$ which is not in $G_1$. Let $G_2=<g_2>$. It is finite and different than $G_1$. So there is some $g_3\in G$ not in $G_1\cup G_2$. Let $G_3=<g_3>$ and so on. A bit more formally, suppose that we have found $n$ different subgroups $G_1,\cdots ,G_n$ of $G$. Each must be finite and so there is some $h\notin G_1\cup \cdots \cup G_n$. Let $G_{n+1}=<h>$, which is thus another subgroup. So there are infinitely many subgroups.

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@ Ittay weiss: I forgot to write " proper subgroups" – maryam Jan 24 '13 at 10:25
@maryam so I adjusted my answer. – Ittay Weiss Jan 24 '13 at 10:53

I think, the first one is wrong. Enough to consider $G=\mathbb Z(p^{\infty})$. This group is infinite, each of whose proper subgroups is finite and also cyclic.

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Yay! Nice and quick counterexample!+1 – amWhy Feb 10 '13 at 0:18

I think the Prüfer group serves as a counterexample for 1, let me attempt 2:

Assume $G$ is infinite, it therefore contains at least a countable number of elements. Let $g_i$ be these elements. Look then at $(g_i)$, the subgroups generated by the $g_i$. If all of these subgroups are finite, then there are an infinite number of subgroups, regardless of whether or not there are $(g_i) = (g_j)$ for $i \neq j$. If a single one of these subgroups are infinite, then it is isomporphic to $\mathbb{Z}$, which has an infinite number of subgroups. Therefore 2 is true.

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