Show that a finite generated subgroup $H$ of $G$ with index 3 exists which is not normal in $G$.

> 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

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).
• @pershinaolad So the general recipe is: (i) pick your favourite point in the base space (it doesn't matter which one you pick), (ii) find the fibre above this point, (iii) decide whether the deck transformations act transitively on this fibre. In your example, I picked $p$ purely for convenience. – Kenny Wong Jan 18 at 14:57