Is
**P**
a group under the operation composition of classes? First, we must
define composition of classes. Taking an element from each class,
and joining them by unattaching a strand from each element and joining
the ends composes the elements. Clearly composing two classes [p1] and
[p2] in **P**
yields a class of paths which is in **S**
and travels from b to b. (see figure 3)

Also,
let [i] be the class of paths which never leaves b. Observe that
composing [i] with any class [p] in **P**
gives the path [p]. (see figure 4)

In addition, note that composition of classes of paths is associative.

Also, for any [p] in
**P**,
an element [p]' of **P **exists for which [p]*[p]' = [i]. In particular,
[p]` = [p`], where p` is the inverse of p in P. That is, p*p`=i.
We know p` exists, because p` is simply the path p traced in the opposite
direction. Notice that for any q in [p], q can be rewritten as p,
and hence we can find an element of [p]` that is the inverse of q, namely
p`. Therefore, [p]`=[p`].

So we have shown that **P**
is indeed a group under the operation composition of classes of paths.

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