How to use induction to prove Cauchy's theorem? This is an exercise from "A book of abstract algebra" by Charles C. Pinter called "Cauchy's Theorem" (If $G$ is a group and $p$ is any prime divisor of $\vert G \vert$ then $G$ has at least one element of order $p$).
It requires some prerequisites which I managed to prove in previous exercises (Cauchy's theorem for abelian groups, the class equation and some properties about conjugacy classes). Here is the exercise:
The proof proceeds by induction. Assume that $G$ is not an abelian group. Let $\vert G\vert=k$ (and $p$ a prime dividing $k$), and assume Cauchy's theorem for every group of order less than $k$. Let $\mathbf C$ be the center of $G$, let $C_a=\{x\in G\:;xax^{-1}=a\}$ be the centralizer of $a$ for each $a\in G$, and let $k=c+k_s+k_{s+1}+\cdots+k_t$ be the (conjugacy) class equation of $G$


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*Prove: If $p$ is a factor of $\vert C_a\vert$ for any $a\in G$, where $a\notin \mathbf C$, we are done.

*Prove that for any $a\notin \mathbf C$, if $p$ is not a factor of $\vert C_a\vert$, then $p$ is a factor of $(G:C_a)$

*Solving the class equation $k=c+k_s+k_{s+1}+\cdots+k_t$ for $c$ (the size of $\mathbf C$), explain why $p$ is a factor of $c$. We are now done. (Explain why)


Here is my attempt:
Knowing that $C_a$ is a subgroup of $G$ and that its order is $$\vert C_a\vert=p\cdot n=\frac{\vert G\vert}{(G:C_a)}\lt k\:\text{for some $n\in\Bbb N$}$$ then $C_a$ has at least (by induction hypothesis) one element of order $p$ therefore $G$ has an element of order $p$ and part $1.$ is proved (is it right? why does the restriction $a\notin \mathbf C$ matter?)
If $p$ is not a factor of $\vert C_a\vert$ then $$(G:C_a)=\frac{\vert G\vert}{\vert C_a\vert}$$ is, because $p$ isn't factored out of $\vert G\vert$ (again the restriction $a\notin \mathbf C$ doesn't make sense to me)
Now, using $(G:C_a)=\vert [a]\vert$, where $[a]$ is the conjugacy class of $a$ (a prerequisite) and $k=c+k_s+k_{s+1}+\cdots+k_t$ where $c$ is the size of the center and $k_i$ are the sizes of every conjugacy class, if both $k$ and every $k_i$ are multiples of $p$ then $c$ must also be a multiple of $p$
Using Cauchy's theorem for abelian groups (already proven as a prerequisite), $\mathbf C$ (of order $c$) is abelian, so $3.$ is proven (also it would satisfy the inductive hypothesis).
I know that the centralizer of some element in the center is the whole group, but why exclude it from the question if its size is also a multiple of $p$?

I think that the restriction is necessary for 2. If $a\in C$, you have $C_{a}$ it's all $G$, so the hypothesis of "$p$ is not a factor of $|C_{a}|$" contradicts the hypothesis of the "$p$ is a factor of $|G|$".
