Subgroups of finitely generated groups are not necessarily finitely generated I was wondering this today, and my algebra professor didn't know the answer. 

Are subgroups of finitely generated groups also finitely generated? 

I suppose it is necessarily true for finitely generated abelian groups, but is it true in general?
And if not, is there a simple example of a finitely generated group with a non-finitely generated subgroup?
NOTE:  This question has been merged with another question, asked by an undergraduate.  For an example not involving free groups, please see Andreas Caranti's answer, which was the accepted answer on the merged question.
 A: No.  The example given on Wikipedia is that the free group $F_2$ contains a subgroup generated by $y^n x y^{-n}, n \ge 1$, which is free on countably many generators.
A: It is well-known that the free group $F_2$ on two generators
has as a subgroup a group isomorphic to a free group on a countably
infinite set of generators. See Qiaochu's example.
However a finite index subgroup of a finitely generated
group is finitely generated.
A: A theorem of Higman, Neumann, and Neumann says that every countable group (no matter what horrible properties it might have) can be embedded as a subgroup of a group generated by $2$ elements.  Thus subgroups of finitely generated groups can be pretty much anything.
A: One of the easiest (counter)example is in Hungerford's Algebra.
Let $G$ be the multiplicative group generated by the real matrices 
$$a = \left(\begin{array}{l l} 
1 & 1\\
0 & 1 
\end{array}\right),
b = \left(\begin{array}{l l} 
2 & 0\\
0 & 1 
\end{array}\right)
$$
Let $H$ be the subgroup of $G$ consisting of matrices that have $1$s on the main diagonal. Then $H$ is not finitely generated. 
A: Perhaps an elementary example can be provided by the wreath product of two copies of (the additive group of) $\mathbf{Z}$.
Take copies $G_{i}$ of $\mathbf{Z}$, for $i \in \mathbf{Z}$, and let 
$$
B = \coprod_{i \in \mathbf{Z}} G_{i}
$$
be the direct sum (coproduct in the category of abelian groups).
Now let another copy $H = \langle h \rangle$ of $\mathbf{Z}$ act on $B$ by
$$
G_{i}^{h} = G_{i+1}.
$$
More precisely, conjugation by $h$ takes a generator $g_{i}$ in the copy $G_{i}$ of $\mathbf{Z}$ to a generator $g_{i+1}$ of the $(i+1)$-th copy.
Then the semidirect product $G = B \rtimes H$ is generated by $g_{0}$ and $h$, but its subgroup $B$ requires an infinite number of generators.
It is easy to see what is going on. $B$ requires an infinite number of generators $g_{i}$. Now $h$ takes one of these generators by conjugation to all others.
