On subgroups of a group 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
 A: 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. 
A: 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.
A: 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. 
