algebraic topology, groups and covering

Hi everyone!

I would like to solve some questions:

Classify up to isomorphism the four-sheeted normal coverings of a wedge of circles. describe them.

i tried to to this and it is my understanding that such four sheeted normal coverings have four vertices and there are loops at each of the vertices. For this graph to be a normal covering the deck transformation group needs to be $\displaystyle Z/2*Z/2$.

second problem: show that a normal subgroup H of a free group L which has infinite index can't be generated by a finite subset. (use the deck transformation of an appropriate covering).

I have no ideas for this one.

Third problem: what is the fundamental group of $\displaystyle RP^2$v$\displaystyle RP^2$? can you find a normal cover of $\displaystyle RP^2$ v $\displaystyle RP^2$ which deck transformation group is $\displaystyle Z/4Z$?

First of all the fundamental group of $\displaystyle RP^2$v$\displaystyle RP^2$ is Z/2*Z/2.

Second it is my understanding that this problem reduces to knowing whether we can have $\displaystyle Z/4Z$ isomorphic to (Z/2*Z/2)/H with H a normal subgroup of $\displaystyle Z/2*Z/2$. Does such a normal subgroup exist?

Fourth problem finally: we have p:E-->B a fibration. Analyse the exact sequence :

$\displaystyle pi_1(E,e)->pi_1(B,b)->pi_0(F,e)->pi_0(E,e)->pi_0(B,b)$

what does exact sequence mean at $\displaystyle pi_0(F,e)$?

For this one I tried to say that there is an action of $\displaystyle pi_0(B,b)$ on the fiber, an action which is transitive.In addtion pi_0(F,e) represent the connected components of the fiber. But I need to show that two points in the fiber F have same image by the induced inclusion i* of pi_0(F) in pi_0(E) if and only if they are in the same orbit under the action of the fundamental group of the base B

Sorry for this long post and thx for any help or explanation!!!