I am currently working on a research paper and im not sure about my arguments.
Let $G$ be a finite group. Then $C_G(H)$ is cyclic for every minimal subgroup $H$ of $G$ if and only if one of the following statements holds:
$(i)$ $G$ is cyclic;
$(ii)$ $G = \left\langle a, b \mid a^m = b^n = 1, b^{-1} a b = a^r \right\rangle$, where $\gcd(m,(r^{t}-1)n) = 1$ for any $1\leq t <n$ and $r^n \equiv 1 \pmod{m}$.
I am currently working on this part of the proof $(\Leftarrow)$ My arguments goes as follows:
If $G$ is cyclic, this is trivial. In the next, suppose that $G = \left\langle a, b \mid a^m = b^n = 1, b^{-1} a b = a^r \right\rangle$, where $\gcd(m,(r^{t}-1)n) = 1$ for any $1\leq t <n$ and $r^n \equiv 1 \pmod{m}$. Let $H$ be a minimal subgroup of $G$, then $|H|$$|$$m$ or $|H|$$|$$n$. Can we assume that $H\leq \langle a \rangle$ or $H\leq \langle b \rangle$? If so why?
Here's the my proof of the case where $H\leq \langle a \rangle$.
Suppose $H\leq \langle a \rangle$. Let $H= \langle a^{k} \rangle$, where $1 \leq k < m$, $k$ is a divisor of $m$. Then $C_G(H)=C_G(H)\cap G= C_G(H) \cap(\langle a \rangle \rtimes\langle b \rangle) = \langle a \rangle \rtimes C_G(H) \cap \langle b \rangle$. Let $C_G(H) \cap \langle b \rangle= \langle b^{\ell} \rangle$, where $1\leq\ell\leq n$. Since $b^{-\ell}a^{k}b^{-\ell}=a^{r^\ell k}$, we have $(r^{\ell}-1)k \equiv 0 \pmod{m}$. If $1\leq\ell< n$, we have $((r^{\ell}-1)k,m)=1$ by the hypothesis $\ell = n$ and then $C_G(H)= \langle a \rangle$.