Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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

3 Answers 3

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.

share|improve this answer
    
@ 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.

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.