Whether a group containing a free group is also a free group Suppose $G$ is a group generated by two elements $s,t$.
Suppose $H$ is a subgroup of $G$ such that $H= \langle s^k,t^k \rangle$ is a free group where $k$ is some integer not equal to $1$.
Does it imply that $G$ is also a free group?
Note: I know there are examples of non free group with a free subgroup. But in those examples, the subgroup had different generators.
But I want to know about this particular case where the generators of the subgroup has some power of the generators of the mother group.
 A: 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.
