Covering map lifting correspondence Let $p : (E,e_0) -> (B,b_0)$ be a 3-fold covering map. Suppose that $[f*f]=[e_{b_0}]$ for each $[f]$ in $\pi_1(B,b_0)$. Prove that E is not path connected.
My attempt it to show that the lifting correspondence is not surjective.
Suppose $\phi $ is the lifting correspondence between $\pi_1(B,b_0)$ and $p^{-1}(b_0)$, we know that there are only three elements in $p^{-1}(b_0)$, we can show that $\phi(\pi_1(B,b_0))$ has less than three elements. 
But I have some difficulties to show that. Can someone help me? Thanks.
 A: Assume for a contradiction that $E$ is path connected.  Then $p(E) = B$ is path connected as well.
From Hatcher's Algebraic Topology book, we have Proposition 3.2:

If both $E$ and $B$ are path connected, the number of sheets of a covering $p:E\rightarrow B$ is equal to the index $[\pi_1(B):p_\ast(\pi_1(E))]$.

Since the number of sheets is $3$, this contradicts the following group theoretic result:

If $G$ is a group in which every element squares to the identity, and $H$ is a subgroup of $G$ of finite index, then $[G:H]$ is a power of $2$.

Proof (of the group theoretic result):  First, since the square of every element is the identity, we first claim $G$ is abelian.  To see this, note $e = (ab)^2 = abab$, so $e = abab$.  Now, multiply on the left by $ba$ to get $ba = ab$.  We will now use additive notation.
Now, suppose for a contradiction that there is a subgroup $H$ of $G$ with finite index, but $[G:H]$ is not a power of $2$.  Since $G$ is abelian, $H$ is normal, so we have a surjective homomorphism $G\rightarrow G/H$ with $G/H$ a finite group.  By assumption, the order of $G/H$ is not a power of $2$ so there is prime $p>2$ which divides the order of $G/H$.  By Cauchy's theorem, there is an element $[h]\in G/H$ of order $p$.
But $h+h = 0$ in $G$, so $[h]+[h]=[0]\in G/H$.  That is, the order of $[h]$ is 2, not $p$, giving a contradiction.  $\square$
