Non-normal coating of the Klein bottle

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(a) Provide an example of a non-normal connected covering map $p:(\tilde X, \tilde x_0) \to (K, x_0)$ where $K$ is the Klein bottle.

(b) Choose $x_0 \in K$ and $\tilde x_0 \in p^{-1}(x_0)$. State what $H = p_{\ast}(\pi_1((\tilde X, \tilde x_0)))$ corresponds to in $\pi_1(K, x_0)$ for the selected covering.

(c) Is it true that $H$ does not depend on the chosen base point $\tilde x_0$?

My attempt is as follows:

(a) Consider the glide reflection $a: (x, y) \mapsto (-x, y + 1)$ and the translation $b: (x, y) \mapsto (x + 1, y)$. It can be shown that $\pi_1(K)$ is isomorphic to the group of isometries of $\mathbb{R}^2$ generated by $a$ and $b$. The covering with $\tilde X = \mathbb{R}^2_{/ \langle a \rangle}$ is non-normal because $\langle a \rangle$ is not normal in $\pi_1(K)$, and $\mathbb{R}^2_{/ \langle a \rangle}$ is homeomorphic to the Möbius strip.

(b) Consider the following depiction of the chosen covering:

https://imgur.com/J0FELn9

In this case, $H ~= \langle a \rangle = \mathbb{Z}$ in $\pi_1(K) ~= \langle a \rangle \rtimes \langle b \rangle ~= \mathbb{Z} \rtimes \mathbb{Z}$.

(c) No, $H$ depends on the chosen base point. Indeed, for $\tilde x_0, \tilde x_1 \in p^{-1}(x_0)$, it holds that $p_{\ast}(\pi_1(\tilde X, \tilde x_1)) = g \cdot p_{\ast}(\pi_1(\tilde X, \tilde x_0)) \cdot g^{-1}$, where $g = [p \circ \gamma]$ and $\gamma$ is a path from $\tilde x_0$ to $\tilde x_1$. If $H$ were independent of the chosen base point, then $p_{\ast}(\pi_1(\tilde X, \tilde x_0)) = p_{\ast}(\pi_1(\tilde X, \tilde x_1))$, implying that $p_{\ast}(\pi_1(\tilde X, \tilde x_0))$ is a normal subgroup of $\pi_1(K, x_0)$, which contradicts the observations made in the previous points.