$\mathbb{Z}\ast\mathbb{Z}\ast\mathbb{Z}$ is an index two subgroup of $\mathbb{Z}\ast\mathbb{Z}$ I want to prove $\mathbb{Z}\ast\mathbb{Z}\ast\mathbb{Z}$ is an index two subgroup of $\mathbb{Z}\ast\mathbb{Z}$. Can I use covering map to prove this? Any ideas and suggestions are greatly appreciated.
 A: Directly:
Define $\color{red}{\Bbb Z}*\color{green}{\Bbb Z}*\color{blue}{\Bbb Z}\to \color{red}{\Bbb Z}*\color{green}{\Bbb Z} $ by $\color{red}1\mapsto \color{red}2$, $\color{green}1\mapsto \color{green}2$, $\color{blue}1\mapsto \color{red}1\cdot\color{green}1$.
If you use $a$, $b$, and  $c$ as the standard generators of $\mathbb{Z\ast Z\ast Z}$, and $x$ and $y$ for $\mathbb{Z\ast Z}$, then this map is given by
\begin{align}
a &\mapsto x^2 \\
b &\mapsto y^2 \\
c &\mapsto xy
\end{align}
A: Yes, you can use coverings. In fact, you can use coverings to prove that every subgroup of index two is isomorphic to $\mathbb{Z\ast Z\ast Z}$.
To use coverings, note that your group is the fundamental group of the graph with a single vertex and two loop edges.
A two-sheeted cover of this graph is a connected graph with two vertices, each with degree four. Up to isomorphism there are 2 such graphs (two edges between the two vertices, or four). In each case, the graph has fundamental group $\mathbb{Z\ast Z\ast Z}$, as required.
