> Suppose that $a$ & $b$ are the generators of a free group $G$.Show that a finite generated subgroup $H$ of $G$ with index 3 exists which is not normal in $G$.
The way i try to solve the problem was that trying to find a covering space for $S^1\vee S^1$ which is not regular and reach the graph i attached but i do'nt
know it works or not and still i didn't learn how to write the covering space from the following graph and the other similar graphs.Thanks in advance
Your example does work.
Let $\widetilde X$ be the graph you drew, let $X = S^1 \vee S^1$ be the original space, and let $\pi : \widetilde X \to X$ be the covering map. Also define $p \in X$ to be the point $p=\pi(p_1) = \pi(p_2) = \pi(p_3)$.
We want to show that the covering $\pi : \widetilde X \to X$ is not regular. Consider the fibre above the $p \in X$, which is $\pi^{-1}(p) = \{ p_1, p_2, p_3 \}$. To show that the covering is not regular, it suffices to show that there does not exist a deck transformation $\tau : \widetilde X \to \widetilde X$ such that $\tau(p_1) = p_2$. [A deck transformation is a homeomorphism $\tau : \widetilde X \to \widetilde X$ such that $\pi \circ \tau = \pi$.]
Since $\tau$ obeys $\pi \circ \tau = \pi$, it must permute the points in $\pi^{-1}(p)$. So if $\tau(p_1) = p_2$, then there are two options: either $\tau$ sends $p_1 \mapsto p_2$, $p_2 \mapsto p_3$ and $p_3 \mapsto p_1$, or $\tau$ sends $p_1 \mapsto p_2$, $p_2 \mapsto p_1$ and $p_3 \mapsto p_3$.
Both options are impossible. In $\widetilde X$, every path from $p_1$ to $p_3$ passes through $p_2$. But there is a path from $p_2$ to $p_1$ that doesn't pass through $p_3$, and there is a path from $p_2$ to $p_3$ hat doesn't pass through $p_1$. Therefore, there cannot exist a homeomorphism $\tau : \widetilde X \to \widetilde X$ sending $p_1 \mapsto p_2$, $p_2 \mapsto p_3$ and $p_3 \mapsto p_1$, or sending $p_1 \mapsto p_2$, $p_2 \mapsto p_1$ and $p_3 \mapsto p_3$.
The intuition is that the points $p_1, p_2, p_3$ are in different "environments". In my proof, I exploited the fact that $p_1$ is an "end-point", whereas $p_2$ is a "middle-point" (excuse the poor terminology).