I am a math autodidact learning abstract algebra in high school. I have limited resources at my school, and only one busy teacher to help me out, so I thought I would ask for help here.
I use Gallian's book for the exercises, and I was asked to prove: $$(a^{-1}ba)^n = a^{-1}b^na$$ for $$n \in Z$$ (integers) and $$(a,b) \in G$$ Where G is a group. When $$n = 0 $$ then $$ e = a^{-1}ea=e$$ So with this base case we have:
$$(a^{-1}ba)^n(a^{-1}ba)=(a^{-1}b^na)(a^{-1}ba) =(a^{-1}b^n)(e)(ba)=a^{-1}b^{n+1}a$$ And for negative integers where $n \in Z+$ $$(a^{-1}ba)^{-n} = ((a^{-1}ba)^n)^{-1} = (a^{-1}b^na)^{-1} = (a^{-1}b^{-n}a)$$
Is this correct? My instructor has really been bagging on my proofs and I don't want to ask him yet.