Are free groups simple? Let $F_n$ be the free group of finite rank $n>1$. Is $F_n$ simple? I know that every subgroup of $F_n$ is free, but I don't know how to tell whether it's a normal subgroup. 
 A: Free groups have lots of normal subgroups. This can be seen by combining the normal subgroup/homomorphism duality given by the first isomorphism theorem to the universal property of free groups, which says that every group is the quotient of some free group. In particular, if $G$ is an $n$-generated group then there a normal subgroup $N$ of the free group on $n$-generators $F_n$ such that $F_n/N\cong G$.
For example, the subgroup $N=\langle a^2, b^2, ba^2b^{-1}, ab^2a^{-1}, (ab)^2\rangle$ of the free group $F(a, b)$ is normal, and the quotient group $F(a, b)/N$ is the Klein $4$-group. (Proving that $N$ is normal is not difficult, and in fact this is the subgroup $F_2(X^2)$ mentioned at the end of this post.)
Lets just consider the free groups of finite rank $n\geq 2$, denoted $F_n$. Now, as I said above, we know that these groups have lots of normal subgroups...so what do they look like? Well, we know that they are free, but the following result says that their generators are hard to write down:

Theorem.
  A non-trivial normal subgroup $N$ of $F_n$ has finite index if and only if it is finitely generated.

Hence, if $F_n/N$ is infinite then $N$ has infinite rank; it is hard to write down!
There are other results which say that the subgroups are hard to write down. For example, there are finitely generated groups $F_n/N$ which are not recursively presentable (as there are uncountably many $2$-generated groups but only countably many recursively presentable $2$-generated groups); this means that we cannot algorithmically (use a computer to) describe the corresponding normal subgroup $N$.
Also, weirdly, there are normal subgroups $N$ of $F_n$ such that we cannot determine if an arbitrary element $W$ of $F_n$ is contained in $N$. This statement is just a re-phrasing of the insolubility of the word problem for groups.

Now, for some positive results. A verbal subgroup is a subgroup generated by all words of a specific form. More precisely: Let $W_{\mu}(X_{\lambda})$, where $\mu=1, 2, \ldots$, be a set of words in the symbols $X_{\lambda}$, where $\lambda=1, 2,\ldots$. Then the $\{W_{\mu}\}$-verbal subgroup $G(W_{\mu})$ of a group $G$ is the subgroup of $G$ generated by all the elements of the form $W_{\mu}(g_{\lambda})$ where $g_{\lambda}$ ranges over $G$.
Verbal subgroups are always normal (why?). The first examples are:


*

*$G(X_1X_2X_1^{-1}X_2^{-1})$ is the derived subgroup

*$G(X)$ is the whole group $G$

*$G(X_1X_2)$ is also the whole group $G$ (take $X_1=g$ and $X_2=1$ for all $g\in G$)

*$F_n(X^2)$ has finite index, and so has finite rank. This is because every non-trivial element of $F_n/F_n(X^2)$ has order two, and so (by a standard exercise) this group is abelian. It is then also finite (this uses that $F_n$ is finitely generated).

