I am doing the following exercise for an assignment:
Assume that $G$ is any finite group with non-trivial elements such that $bab^{-1}=a^{-1}$. Let $k$ be a natural number and use induction to show that $$(ab)^{k}=\begin{cases} \\b^k\,\,\,\,\,\,\,\text{ if }k=2m \\ab^k\,\,\,\,\text{ if }k=2m+1 \end{cases} $$
If $b$ is its own inverse, then the induction proof is easy, in fact I don't see how it's possible to prove this without $b$ being its own inverse.
So $G$ is just any group, which means that I can't be sure that $b\ast b=e$ in $G$, but I know that $G$ is isometric to a subgroup of $S_G$, where the permutation $\rho_b \in S_G$ is its own inverse.
Am I correct in thinking that I can just do the proof as if $b\ast b=e$? Because of the isomorphism I could do the entire proof for: $$(\rho_a\rho_b)^{k}=\begin{cases} \\(\rho_b)^k\,\,\,\,\,\,\,\text{ if }k=2m \\\rho_a(\rho_b)^k\,\,\text{ if }k=2m+1 \end{cases} $$
and then just map it back into $G$ anyway, right?
If I want an explicit map, can I for example say:
Let $\phi:G\rightarrow S_G$ be an isomorphism between $G$ and a subgroup of $S_G$, such that $\phi(i)=\rho_i$ for all $i\in G$, where $\rho_i$ is the permutation corresponding to the column in G's multiplication table such that $\rho_i (e)=i$.
If you have a finite group, you can write out a multiplication table, and you can create permutations that correspond to that table to get the actual permutations if you need them.
In the case where I'm not completely off base: What do I have to justify and what do I not have to justify here? How can I summarize this in a mathematically concise and precise way?
I think you're really over-thinking this. Since you're given $bab^{-1} = a^{-1}$, you know that $abab^{-1} = e$, so that $ab = ba^{-1}$.
Since you want to prove something about powers, note that $(ab)^2 = (ab)(ab) = (ba^{-1})(ab) = b^2$. This is your key to proving the statement. Break it up into two cases, even and odd power, use the fact that you can turn adjacent pairs of $(ab)$s into $b^2$s, and formalize it with induction.