non-normal covering of wedge of three circles How might I systematically approach the task of finding a three-fold, non-normal covering of a wedge of three circles?  My instinct is to find a non-normal subgroup of the free group on 3 generators and try to sketch a space whose loops realize that subgroup.  As this is in preparation for an exam in algebraic topology, I think this is not the best way of approaching the problem and think there is perhaps some topological insight I am not exploiting.  Tips for how to approach such problems in general would also be greatly appreciated.  Prior coursework in abstract algebra is not presupposed for the course, so although I am not ignorant of abstract algebra, I am not as fluent as I once was with its more advanced techniques.
 A: I think that if you can do it with a bouquet of two circles, you can
do it with a bouquet of three. Just hang an extra circle at the preimages
of the base-point.
Let $B$ be a bouquet of two circles, and $x$ the point where
they meet. The fundamental group $\pi_1$ is free on two generators, $g$ and $h$
corresponding to the two circles $C_1$ and $C_2$ say. Then $\pi_1$ has a transitive
action on the three-point set $\{1,2,3\}$ via $g\mapsto(1,2)$
and $h\mapsto(2,3)$. Let us construct a covering space of $B$
embodying this action. Take three copies of $B$; call them $B_1$, $B_2$ and $B_3$. Do some cut-and-paste on these. In $B_1$ and $B_2$ snip the circles $C_1$ and join each to the other one. In $B_2$ and $B_3$ snip the circles $C_2$ and join each to the other one. You get a covering space $B'$
of $B$ which is connected, triple, and one can check, non-abelian.
For a concrete model, take the four Euclidean circles, radii all one,
and centres $(1,0)$, $(3,0)$, $(5,0)$ and $(7,0)$. This covers $B$
with $\{(2,0),(4,0),(6,0)\}$ forming the fibre of $x$.
For the three-circle bouquet, you could add circles radius $1/2$
with centres at $(5/2,0)$, $(9/2,0)$ and $(13/2,0)$.
