If $\langle a \rangle \unlhd \langle a,b \rangle$ and $a,b$ have finite order, then $\langle a, b \rangle$ is finite. Let $G$ be a group and $a,b \in G$. If $\langle a \rangle \unlhd \langle a,b \rangle$ and $a,b$ have finite order, then $\langle a, b \rangle$ is finite.
Is this true?
My attempt. Since $\langle a \rangle \unlhd \langle a,b \rangle$, we have
$$\langle a, b \rangle = \langle \langle a \rangle \cup \langle b\rangle\rangle = \langle a \rangle \langle b \rangle$$
where I used that "$H,K \leq G$ and $K \unlhd G$ implies $\langle H\cup K\rangle = KH$" with $G= \langle a,b \rangle$.
Since $a$ and $b$ have finite order, the subgroups $\langle a \rangle, \langle b\rangle$ have finite order as well. Thus
$$|\langle a,b\rangle| = |\langle a \rangle \langle b \rangle | \leq |\langle a \rangle | | \langle b \rangle | < \infty$$
and this completes the proof.
Is this correct?
 A: Yes, your proof is correct. The cornerstone of this proof is that you used $$H, K \leq G \text{, } K \lhd G \text{ implies } \langle H \cup K \rangle = KH$$ so I think that it is interesting to include a proof of that.
Indeed, we can write $hk=hkh^{-1}h = k^h h = k' h$, because $K$ is normal. Then in particular given any word in $\langle H \cup K \rangle$ we can write it as $kh$, by "moving each $k$ to the left" as we did above. Trivially, it follows that $\langle H \cup K \rangle = \langle K \rangle \langle H \rangle$. 
Note that since every word contains finitely many letters this proof does not require $H$ or $K$ to be finite.
A: Since the group generated by $a$ is normal $b a b^{-1} = a^k$ for some $k$, every element can be written in the form $b^na^m$, i.e. there are at most $ord(a)\cdot ord(b)$ elements. 
proof: in any word in $a,b$, every time you see $ba$, you can replace it by $a^kb$, since the orders are finite, this process stops.
Edit: Your proof seems alrigth, just wanted to give a more elementary proof.
