The Wikipedia page for "Relatively hyperbolic group" lists this as a property of relatively hyperbolic groups:
"If a group $G$ is relatively hyperbolic with respect to a hyperbolic group $H$, then $G$ itself is hyperbolic."
Wikipedia does not provide a proof or reference for this property. Can anybody help me prove it?
I found another equivalent definition for relatively hyperbolic groups in https://arxiv.org/pdf/math/0404040v1.pdf (found in the Wikipedia page's references section). By this equivalent definition $G$ is relatively hyperbolic with respect to $H$ means that
(1) There exist finite sets $X,R$ such that $X\cup H$ generates $G$ and the kernel of the natural map $F(X)*H\to G$ is the normal closure of $R$ in $F:= F(X)*H$.
(2) For a word $W \in (X \cup H)^∗$ such that $W$ represents $1$ in G, there exists an expression $W =\prod_{i=1}^{k}f_{i}^{−1}R_{i}f_{i}$ with the equality in the group $F$, where $R_{i} ∈ R$ and $f_i ∈ F$ for any $i$. A function $f : \mathbb{N} → \mathbb{N}$ is a relative isoperimetric function with respect to subgroup $H$ if for any $n ∈ \mathbb{N}$ and any word $W ∈ (X ∪H)^∗$ of length $||W|| ≤ n$ representing the identity in the group $G$, one can write $W$ in the form of that expression with $k ≤ f(n)$. The second condition is that there exists a linear relative isoperimetric function with this definition.
It seems like it would be easier to prove the property by showing that $G$ being relatively hyperbolic with respect to a hyperbolic group $H$ implies that $G$ has a linear ordinary/non-relative isoperimetric function for $G$. $H$ being hyperbolic tells us that it is generated by some finite subset $Y$. Then $G$ being hyperbolic relative to $H$ implies that $X\cup Y$ is a finite generating set for $G$. I wanted to show that there is a finite group presentation for $G$ with generating set $X\cup Y$ that we can use to get a linear ordinary/non-relative isoperimetric function for $G$. I got stuck. Can anyone help me?