My Algebra lecturer is a very strict about proofs(w.r.t Completeness , correctness and format ) more so than I have encountered in the past or any of my lecturers of the courses I am take concurrent. I do appreciate this as Mathematics along with other scientific disciplines are found on being as precise as possible. It does make it rather difficult to judge what constitutes a proof, Which brings me onto my main topic
The Question: Suppose $f \in S_n$ and consists of a single cycle $( a_1 , a_2 ,...,a_r).$ Let $g\in S_n$ and show that $gfg^{-1} = (g(a_1) , g(a_2) ,... ,g(a_r)) $ and furthermore show that $f$ and $gfg^{-1}$ has the same disjoint cycle structure
My Proposed Proof:
Suppose $a_i \in \{a_1, a_2,...,a_r\}$
Let $h = gfg^{-1}$. Since $f(a_i) = a_{i+1}$
then
$$ h(g(a_i) = gfg^{-1} g(a_i)$$ $$ = g(f(a_i)) = g(a_{i+1})$$
So $gfg^{-1} $ has a cycle $(g(a_1) , g(a_2) ,... ,g(a_r)) $
otherwise $a_i \in \{ a_{r+1} ,.., a_n\}$
Since $f(a_i) =a_i$
then $$ h(g(a_i) = gfg^{-1} g(a_i)$$ $$ = g(f(a_i)) = g(a_{i})$$
So $f$ and $gfg^{-1}$ has the same disjoint cycle structure
If this proof correct, and if so any suggestions/comments about additions and changes that should be made in order to make the proof as complete , rigorous and properly formatted.