This exercise is taken from the book "Office Hours with a Geometric Group Theorist" (Office Hour 15, exercise 8):
Exercise: Show that the group $\mathbb{Z}$ has dead end elements with respect to the presentation $\mathbb{Z} = \langle a, b \mid a^{12} = b,\ ab = ba \rangle$. What are the possible depths of dead end elements with respect to this generating set?
Definition [from Bogopolski, Infinite commensurable hyperbolic groups are bi-Lipschitz equivalent]: An element $g$ of a group $G$ is called dead in $G$ with respect to the finite generating set $X$ if $|gx| \leq |g|$ $\forall x \in X^{\pm 1}$. Here $|g|$ denotes the length of $g$ in the word metric with respect to $X$.
Intuitively, think of a dead end element as a dead end in a maze: no matter which direction you go, you'll be closer (or at least not farther away) from the exit (exit being the analogy for the neutral element here).
I have two questions:
Why does this presentation generate $\mathbb{Z}$? If you take $a = 2$ and $b = 24$, I think it does not generate $\mathbb{Z}$, since if you start from an odd number, you will only get odd numbers using $a$ and $b$.
Assuming $a = 1$ and $b = 12$, I think there is no dead end element: assume $g \in \mathbb{Z}$. Without loss of generality, we can assume $2 \leq g \leq 11$. Suppose that $|g| = n$. Then $|gb| = n + 1$ because the shortest path to go to $gb$ is to go from $0$ to $b$, and then use the same path from $0$ to $g$, but shifted by $b$. More formally, $g$ is of the form $g = a^i$, $2 \leq i \leq 6$ or $g = ba^{-j}$, $1 \leq j \leq 5$. Then $gb = ba^i$ or $g = b^2a^{-j}$, and both can't be reduced. So my question: is is possible that the exercise is actually impossible?