Let $(G,\cdot)$ be a finite group with $ord(G)=n,\;$ $n\ge3\,$ and $a, b \in G\; \backslash \{e\}$ so that $a^k\neq \; b^p\;\; \forall k, p \in \{1, 2,\ldots,\left[\frac{n}{2}\right]\}$. Show that $a=b^{-1}$.
If $a \in G$ it follows that $\exists \;k=ord(a)$ so that $k\,|\,n$ . Since $k$ is a divisor of $n$, we have $k \in \{2,\ldots,\left[\frac{n}{2}\right],n\}$. Now if we pick $p=\left[\frac{n}{2}\right]$, with $k \in \{2,\ldots,\left[\frac{n}{2}\right]\}$, we have $e=a^k\neq b^{\left[\frac{n}{2}\right]}$ which follows that $ord(b)=n$. It seems like I'm stuck here. Is my approach correct, or do you have another idea?
Let $r=ord(a)$ and $s=ord(b)$. If $r<n$, then $r\le [n/2]$, and the hypothesis implies that $s > [n/2]$. But this means that $s=n$ and $G$ is cyclic generated by $b$.
Write $a=b^k$ with $k>[n/2]$. Let $u=n-k$. Then $a=b^{n-u}$, $a^2=b^{n-2u}$, $a^i = b^{n-iu}$.
Take $i\ge [n/(2u)]$ and $j=n-iu$.
If $u \ge 2$, then $a^i=b^j$, with $i,j \in \{1, 2,\ldots,\left[\frac{n}{2}\right]\}$, which cannot happen.
Therefore, $u=1$ and $a=b^{n-1}=b^{-1}$.