Composition of isometries of trees which is loxodromic(hyperbolic) with nonempty intersection of axis. Question: If $\gamma,\delta\in$ Isom $T$ and Min$(\gamma)\cap$ Min$(\delta)=\emptyset$, then $\gamma\delta$ is loxodromic, Min$(\gamma\delta)\cap$ Min $(\gamma)\neq\emptyset$, and Min$(\gamma\delta)\cap$Min$(\delta)\neq\emptyset$.
What I know: part $1)$ that $\gamma\delta$ is loxodromic(hyperbolic) (i.e., the translation length of $\gamma\delta$ is attained and is strictly greater then zero), I have solved it using the fact that if $\widehat{ab}$ denotes the midpoint of the segment ab, then $\widehat{x\psi(x)}\in$ Min($\psi$) for any isometry $\psi$.
For part $2)$ and part$3)$ which are to show that  Min$(\gamma\delta)\cap$ Min $(\gamma)\neq\emptyset$, and Min$(\gamma\delta)\cap$Min$(\delta)\neq\emptyset$, I do not have much clues except the fact that I need to use the axis of $\gamma\delta$ which is the set of all $x$ such that $d(\gamma\delta x,x)>0$. I need some help on how to proceed from this point. Thank you!
Here's the claim: For any isometry $\psi$ , $\widehat{x\psi(x)} \in$ Min ($\psi$), where $x$ is an element of the tree $X$. I will use the claim to show part 1) that $\gamma \delta$ is loxodromic. Suppose by way of contradiction that $\gamma \delta$ is elliptic. Then,  $\gamma \delta(x)=x$ for some $x$. So, let $\delta(x)=a$ so that $\gamma(a)=x$ where $x\neq a$, because by assumption Min ($\gamma )\cap$ Min($\delta)=\emptyset$. By the claim, we know that $\widehat{a\gamma (a)}\in $ Min($\gamma$). But $\widehat{a\gamma(a)}$ is exactly $\widehat{\delta (x) x}\in $ Min($\delta$), a contradiction. Hence, $\gamma \delta$ must be loxodromic as any isometry of a tree $X$ to itself is either elliptic or loxodromic. This solves part 1).
 A: Your proof for part 1) is short and clever, but it leaves out some additional information which is useful for deriving parts 2) and 3). Let me sketch an alternate proof of part 1) which exhibits that information.
To say that an isometry $\zeta : T \to T$ is loxodromic is equivalent to saying that $\text{Min}(\zeta)$ is a line called the axis of $\zeta$ that I will denote $\mathcal L_\zeta$, and that the restriction $\zeta \mid \mathcal L_\zeta$ is topologically conjugate to a translation action of the real line, i.e. a map $\mathbb R \mapsto \mathbb R$ of the form $x \mapsto x+L$.
So, another way that one can prove that $\zeta = \gamma\delta$ is loxodromic is to actually construct its axis $\mathcal L_\zeta$. Having done that, using the information we get from the construction we will deduce that $\mathcal L_\zeta = \text{Min}(\gamma\delta)$ has nonempty intersection with both $\text{Min}(\gamma)$ and $\text{Min}(\delta)$.
One annoying feature of the alternate proof is that it is long. It has to be done in separate cases: Case 1) where $\gamma,\delta$ are both elliptic; Case 2) where one is elliptic and the other is loxodromic; and Case 3) where both are loxodromic. Furthermore, each of the cases is long. However, it has the great advantage that the axis of $\gamma\delta$ is described in great detail, and that is the extra information that we will make use of.
I'm going to do only Case 1) in detail. After that I'll sketch the other cases. 
Let's assume that $\gamma,\delta$ are elliptic, so $\text{Min}(\gamma)$ and $\text{Min}(\delta)$ are their respective fixed point sets.
From the assumption that $\text{Min}(\gamma)$ and $\text{Min}(\delta)$ are disjoint, it follows that there exists a unique pair of points $x_0 \ne y_0$ bounding a segment $[x_0,y_0] \subset T$ such that $[x_0,y_0] \cap \text{Min}(\delta)=\{x_0\}$ and $[x_0,y_0] \cap \text{Min}(\gamma)=y_0$. Intuitively, the point $x_0$ is the unique point of $\text{Min}(\delta)$ that is closest to the set $\text{Min}(\gamma)$, and $y_0$ is the unique point of $\text{Min}(\gamma)$ that is closest to $\text{Min}(\delta)$.
Now let's make some deductions. Define $x_1 = \gamma(x_0) = \gamma\delta(x_0)$. Consider the segment $[y_0,x_1]$. It follows that $[y_0,x_1] \cap \text{Min}(\gamma)=\{y_0\}$. From this it follows that $[x_0,y_0] \cap [y_0,x_1] = \{y_0\}$, because otherwise, since $\gamma$ is an isometry, any other point in this intersection would be in $\text{Min}(\gamma)$, contradicting that $[x_0,y_0]$ intersects $\text{Min}(\gamma)$ uniquely in the point $y_0$.
We now have the segment $[x_0,x_1] = [x_0,y_0] \cup [y_0,x_1]$. The next step is to prove that this segment $[x_0,x_1]$ is a fundamental domain for an axis of $\zeta = \gamma\delta$, thereby proving that $\zeta$ is loxodromic. In other words, letting $x_i = \zeta^i(x_0)$, one must that $[x_i,x_{i+1}] \cap [x_{i+1},x_{i+2}]=\{x_{i+1}\}$ and that if $i < j+1$ then $[x_i,x_{i+1}] \cap [x_j,x_{j+1}] = \emptyset$; these are not hard to do by induction. Having done that, it follows that
$$\mathcal L_\zeta = \bigcup_{x \in \mathbb Z} [x_i,x_{i+1}]
$$
is a properly embedded line and it is an axis for $\zeta$, and therefore $\zeta = \gamma\delta$ is loxodromic. Furthermore, your desired conclusions 2) and 3) are true, because $\min(\gamma\delta) \cap \min(\gamma)$ contains the point $y_0$ and $\min(\gamma\delta) \cap \min(\delta)$ contains the point $x_0$.
Here's a brief sketch of cases 2) and 3). One still gives an explicit description of the axis $L_\zeta$, and the axis is still constructed by starting with its subsegment $[x_0,y_0]$ described above, and at the end of the proof one still obtains $x_0 \in \text{Min}(\gamma\delta) \cap \text{Min}(\delta)$ and similarly for $y_0$. Also, one starts with $x_1 = \gamma\delta(x_0)$ as before. The only trouble being that the segment $[x_0,x_1]$ is harder to analyze and the axis $\mathcal L_\zeta$ is more complicated to describe. 
Here's one idea of the added complications. If, say, $\gamma$ is loxodromic, then it does not fix $y_0$. However since $y_0 \in \mathcal L_\gamma$ then we can conclude that $[y_0,\gamma \cdot y_0]$ is a subsegment of $\mathcal L_\gamma$, in fact it is a fundamental domain of $\mathcal L_\gamma$. From this, we can conclude that $[x_0,x_1]$ is a concatenation of three segments meeting only at their endpoints:
$$[x_0,x_1] = [x_0,y_0] \, [y_0 \gamma \cdot y_0] \, [\gamma \cdot y_0,x_1]
$$
That should be enough to start you off in the cases that $\gamma$ is loxodromic.
One last word: perhaps this proof can be couched in a language which is "case free". One starts with $[x_0,y_0]$ as described above. Then one defines $z_0 = \gamma(y_0)$, and $x_1 = \gamma\delta(x_0)$. Then one proves that $[x_0,x_1]$ can be written as a concatenation $[x_0,y_0] \, [y_0,z_0] \, [z_0,x_1]$; when $\gamma$ is elliptic, the middle segment $[y_0,z_0]$ will degenerate to a point, but who cares?
