Can a group have more subgroups than it has elements? I'm looking for a group for which the number of subgroups is more than the number of elements in the group! I tried a few possibilities - it can't be cyclic, and I think we'll have to consider group of infinite order.
 A: $C_2 \times C_2 \times C_2$ has 8 elements. Each of the 7 non-identity elements generates a subgroup of order 2. Any pair of non identity elements also generates a subgroup isomorphic to $C_2 \times C_2$. There are 7 of these subgroups, for a total of 14 nontrivial proper subgroups.
A: Consider the product of $n \gt 2$ copies of $\mathbb{Z}_2$, a group of order $2^n$.  Each nonzero element in this (additive) group has order two, so in addition to the trivial subgroup, there are $2^n - 1$ subgroups of order two.
Of course there are also proper subgroups of order greater than two, so more subgroups than elements.
A: Number of subgroups of $D_{2n}=\sigma(n)+\tau(n)$, where $\sigma$ and $\tau $ are sum of divisors and number of divisors of $n$, so determine all $n$ for which $\sigma(n)+\tau(n)>2n$ and you will have alot many examples (just do not try primes bigger than $3$) 
For $n=4$, $\sigma(n)+\tau(n)=10>4$, so $D_4 $ is an example.
For $n=5$, $\sigma(n)+\tau(n)$, not true.
For $n=6$, $\sigma(n)+\tau(n)=16>12$. so $D_6 $ is an example..
and so on....
A: As alluded to in the comment by Mikko Korhonen, 
the smallest example is the Klein Vierergruppe $V=\{1,a,b,c\}$, where $a^2=b^2=c^2=1$.
It has four elements and five subgroups:  $\{1\}, V, \{1,a\}, \{1,b\}, $ and $\{1,c\}$.
A: Every uncountable abelian group has more subgroups than elements.  First, suppose $G$ is an uncountable abelian group which is a direct sum of cyclic groups.  The number of summands must be equal to $|G|$, and every subset of the summands generates a different subgroup, so $G$ has $2^{|G|}$ different subgroups.
Now suppose $G$ is an arbitrary uncountable abelian group.  By the previous argument, it suffices to show $G$ has a subgroup of the same cardinality that is a direct sum of cyclic groups.  Suppose $H\subseteq G$ is any subgroup such that $|H|<|G|$.  We can enlarge $H$ to a subgroup $H'$ with $|H'|\leq |H|+\aleph_0$ such that if $h\in H'$, $n\in\mathbb{Z}$, and $h$ is divisible by $n$ in $G$, then $h$ is divisible by $n$ in $H'$ (just iterate adding witnesses to all such divisibilities to the group $\omega$ times).  In particular, since $G$ is uncountable, we still have $|H'|<|G|$, so we can pick some $g\in G\setminus H'$.  Let $n\in\mathbb{Z}_+$ be minimal such that $ng\in H'$ (or if no such $n$ exists, let $C$ be the subgroup generated by $g$ and jump to the next paragraph).  Since $ng$ is divisible by $n$ in $G$, it is also divisible by $n$ in $H'$, so there is some $h\in H'$ such that $nh=ng$.  By minimality of $n$, $g-h$ has order $n$, and the cyclic group $C$ that it generates has trivial intersection with $H'$.
We have thus shown that if $H\subset G$ is a subgroup with $|H|<|G|$, then there is a nontrivial cyclic subgroup $C\subseteq G$ such that $C\cap H=0$.  If $H$ is a direct sum of cyclic groups, then $C+H$ will be as well, with $H$ as a direct summand.  Iterating this transfinitely, we can keep finding larger and larger direct sums of cyclic groups which are subgroups of $G$, until we reach one of the same cardinality as $G$.
(Incidentally, this is not true for nonabelian groups, or at least it cannot be proved in ZFC: according to this answer, for instance, it follows from CH that there is a group of cardinality $\aleph_1$ with only $\aleph_1$ subgroups.  I don't know whether a nonabelian counterexample can be proved to exist in ZFC.)
A: Intuitively, I would think there would be a pretty good chance.  Considering all the ($2^n\gt n$) subsets.
Granted this is kind of naive, not all subsets are subgroups.  Nevertheless, we have seen examples, both finite and infinite, where alot are.
And as the answers above show, including elementary abelian groups (the Klein four group included) and dihedral groups, the answer is yes.
