I want to show that for any language $L \subseteq \{ a \}^* $, $L$ satisfies the pumping lemma for context free languages if and only if it satisfies the pumping lemma for regular languages.I know that every regular language is also a context free language so I tried to show that direction of the proof first but ran into some difficulties. Is there a more logical approach to this? Would I have to show that the conditions for both the pumping lemma for regular languages and the pumping lemma for context free grammars are equivalent for this language? 
Show that the pumping lemmas for context-free and regular languages are equivalent for unary languages
formal languages;regular languages;context free;pumping lemma
let $ z \in L$ such that $ |z| \geq n $ (where $n$ is the lemma's constant).By the pumping lemma for CFL, we know that we can write $z=uvwxy$ such that:$|vwx| \leq n$$|vx| \geq 1$for all $i \geq 0$: $uv^iwx^iy \in L$Because $L$ is over an unary alphabet, we can change the order of the sub-words and the word ($z$) will not change, meaning we can also write $z=wvxuy$.So. for all $i \geq 0$, $uv^iwx^iy = wv^ix^iuy = w(vx)^iuy$.Let's write a little different:$u' = w, v' = vx, w' = uy$. and we have that $u'(v')^iw' \in L$.It's easy to see that $|u'v'| \leq n$ and $|v'| \geq 1$, So we can conclude that for the same $n$, the conditions of pumping lemma for regular languages holds. It might be worth mentioning that every CFL of unary alphabet is also regular (We know that even though a language satisfies the pumping lemma for regular/CF languages, it does't mean that the language is regular/CF).It can be shown using Parikh's theorem, and showing that for every semi-linear set $S \subseteq \mathbb{N}  $ there is a regular language $L$ such that $\Psi (L)=S$ (or $p(L)$ using wikipedia's notations) 