When I was proving the following exercise with a technical method (although it could be solved by a very elementary technique), I ran accross a problem. The exercise is:
Let $F$ be a free group and let $N$ be the subgroup generated by the set {$x^n$| $x\in F$, $n$ is a fixed integer}. Show that $N\lhd F$.
My solution: For each $x\in X$ on which $F$ is free, <$x$> is a free group on {$x$} and the map $f$ taking $x$ to 1 in $\mathbb Z_n$ induces a homomorphism $\bar f$ from <$x$> to $\mathbb Z_n$ with kernal <$x^n$>. Hence, $\prod_{x\in X}^* <x^n> \lhd \prod_{x\in X}^*<x>=F$ (where $\prod^*$ is the notation for free product). But $N$ is $\prod_{x\in X}^*<x^n>$ so that $N\lhd F$. I think this solution is true provided that the following assumption is true:
If for each $i\in I$, $N_i\lhd G_i$ then $\prod_{i\in I}^*N_i \lhd \prod_{i\in I}^* G_i$. I think this can be easily checked that it is correct.