I would do this computation by thinking about covering spaces of graphs and applying Stallings's folding algorithm.
We realise $F_2$ as the fundamental group of a graph $X$ with one vertex $v$ and two edges---this is sometimes called the rose with two petals. We orient the edges, and label one by $a$ and the other by $b$. This fixes an identification $\langle a,b\rangle\equiv\pi_1(X,v)$.
Your subgroup $H$ can be thought of similarly. It is the fundamental group of a graph $Y$, which we construct as follows:
- Fix a base vertex $*$.
- Attach both ends of an oriented edge labelled $a$ to $*$ .
- Attach both ends of an oriented interval, consisting of $2k$ edges labelled $a$ and $b$ alternately, to $*$.
The orientations and labels define a natural map $Y\to X$, and the image of $Y$ is your subgroup $H$.
(For more on this sort of construction see, for instance, this blog post.)
Stallings' folding algorithm is a way of turning this map into an immersion---that is, a local embedding. The algorithm is easy:
- If two edges with the same label are both oriented into the same vertex, identify them.
- If two edges with the same label are both oriented away from the same vertex, identify them.
- Repeat.
At the end of this procedure, we have a new oriented, labelled graph $Y'$, and the map $Y'\to X$ is an immersion. There are essentially two possibilites:
Every vertex of $Y'$ has valence four. If so, then $Y'\to X$ is a covering map and $H$ is a finite-index subgroup of $F_2$. The index of $H$ is equal to the degree of the covering map, which is equal to the number of vertices of $Y'$.
Some vertex of $Y'$ has valence less than four. If so, then $H$ is of infinite index. (To see this, note that you can complete it to an infinite-sheeted covering space.)
In your case, you can quickly see that you need to perform exactly one fold to turn $Y$ into $Y'$. If $k=1$ then $Y'$ is isomorphic to $X$ and your subgroup $H$ is equal to $F_2$. Otherwise, $Y'$ has a vertex of valence two and $H$ is of infinite index.
Yes, the free product of the free group of rank $n$ and the free group of rank $m$ is isomorphic to the free group of rank $n+m$. Note also that if $G$ is free of rank $2$, and two elements $a$ and $b$ generate $G$, then $a$ and $b$ must freely generate $G$. One way to see this is to invoke the fact that a free group of finite rank is Hopfian. Another is to assume there is a nontrivial reduced word in $a$ and $b$ equal to the identity, take a free generating set $x$ and $y$, express $a$ and $b$ in terms of $x$ and $y$, and replace them in the nontrivial word expressing the identity; this will yield a nontrivial word in $x$ and $y$ equal to the identity (some work needs to be done here, of course), contradicting the choice of $x$ and $y$ as a free generating set. And there are other ways, of course.
An easy way to get this is as a corollary to the fact that the free group functor is the left adjoint of the underlying set functor. That is, for every group $G$ and every set $X$,
$$\mathcal{G}roup(\mathbf{F}(X),G) \longleftrightarrow \mathcal{S}et(X,\mathbf{U}(G)),$$
where $\mathbf{F}(X)$ is the free group on the set $X$ and $\mathbf{U}(G)$ is the underlying set of the group $G$.
Because the free group functor is a left adjoint, it sends coproducts to coproducts. That is, the coproduct of two free groups $F(X)$ and $F(Y)$ in the category of groups is the free group on the coproduct of $X$ and $Y$ in the category of sets. The coproduct in the category of groups is the free product, and the coproduct in sets is the disjoint union. Therefore, there is a natural isomorphism
$$F(X\amalg Y) \cong F(X)*F(Y).$$
You can also prove it directly from the universal property: the universal property of the free group on $X\amalg Y$ (the disjoint union) is that for every set-theoretic map $f\colon X\amalg Y\to G$ to a group $G$, there is a unique group homomorphism from $F(X\amalg Y)\to G$ that extends $f$. On the other hand, the universal property of the free product $F(X)*F(Y)$ is that for every pair of group homomorphisms $\varphi\colon F(X)\to G$ and $\psi\colon F(Y)\to G$, there is a unique group homomorphism $\Psi\colon F(X)*F(Y)\to G$ such that $\Psi\circ i_{F(X)}=\varphi$ and $\Psi\circ i_{F(Y)}=\psi$, where $i_{F(X)}\colon F(X)\to F(X)*F(Y)$ and $i_{F(Y)}\colon F(Y)\to F(X)*F(Y)$ are the canonical inclusions.
A set-theoretic map $f\colon X\amalg Y\to G$ is equivalent to a pair of maps $g\colon X\to G$ and $h\colon Y\to G$; the map $g\colon X\to G$ induceds a map $\varphi\colon F(X)\to G$, while the map $h\colon Y\to G$ induces a map $\psi\colon F(Y)\to G$, which in turn induces a map $F(X)*F(Y)\to G$; it is now straightforward to verify that this map extends $f$, and that it is unique, so that $F(X)*F(Y)$ has the universal property of $F(X\amalg Y)$, and therefore they are isomorphic.
Or: the inclusions $X\to F(X)\to F(X)*F(Y)$ and $Y\to F(Y)\to F(X)*F(Y)$ induce an inclusion $X\amalg Y\to F(X)*F(Y)$, which in turn induces a morphism $F(X\amalg Y)\to F(X)*F(Y)$. Conversely, the map $X\to F(X\amalg Y)$ induces a map $F(X)\to F(X\amalg Y)$, and $Y\to F(X\amalg Y)$ induces a map $F(Y)\to F(X\amalg Y)$, and these two maps together induce a map $F(X)*F(Y)\to F(X\amalg Y)$. It is now easy to verify that the induces maps $F(X\amalg Y)\to F(X)*F(Y)$ and $F(X)*F(Y)\to F(X\amalg Y)$ are inverses of each, in the usual abstract nonsense argument about their compositions having the same universal property as the corresponding identity.
P.S. The equality $F(a,b)=\langle a\rangle*\langle b\rangle$ presumably just means that there is a unique isomorphism between the two objects that maps $\{a\}$ and $\{b\}$ to themselves as the identity; this is a consequence of their respective universal properties/adjointness of free group construction.
Best Answer
No, this is not true.
For an extreme class of counter-examples, consider hyperbolic groups, which are an extremely common* class of groups. If $G$ is a hyperbolic group then for all $g,h\in G$ there exists an integer $k\geq1$ such that $\langle g^k, h^k\rangle$ is free. So every non-free two-generated hyperbolic group gives you a counter-example. Often the promised free subgroup is free of rank two. Concrete examples include: \begin{align} \langle a, b\mid abab^2ab^3\cdots ab^n\rangle&\quad\forall\:n\gg1&(1)\\ \langle a, b\mid [a, b]^n\rangle &\quad \forall\:n>1&(2)\\ \langle a, b\mid a^n, [a, b]\rangle&\quad\forall\:n>0&(3)\\ \langle a, b\mid a^p, b^q, (ab)^r\rangle&\quad\forall\:\frac1p+\frac1q+\frac1r<1&(4) \end{align} In (1) and (2) there exists some $k$ such that $\langle a^k, b^k\rangle\cong F_2$. This is the "generic" case, as it happens if both generators have infinite order and don't commute (and a generic group is in fact "non-elementary" torsion-free hyperbolic). Note that the group in (2) is not torsion-free, although the generators have infinite order.
The result still holds for generators of finite order, it just won't be the free group of rank $2$ that you get. In (3), $\langle a^n, b^n\rangle\cong\mathbb{Z}$ which is the free group of rank $1$. In (4), $\langle a^{pq}, b^{pq}\rangle$ is trivial, which is free of rank $0$.
To save myself duplicating work: references to proofs are here. Theorem 5.3.E of Gromov's "Essay" is actually a more powerful result, covering multiple generators.
*Both in the sense that they crop up everywhere, and that they are "generic" within the class of finitely presented groups.