Proof of Burnside's theorem Let $|G| = p^{a}q^{b}$ where $p\neq q$ and $a,b$ are positive integers (i.e. excluding the case where $G$ is a $p$-group). 
In preparation for this proof, I have shown that if $Z(G) = 1$ there exists a proper nontrivial normal subgroup of $G$.
My attempted proof:
Suppose that if $|G| = p^{a'}q^{b'}$ where $a'\leq a$ and $b'\leq b$, not both equal, then $G$ is solvable. We proceed the induction on $|G|$ by considering the two cases $Z(G) = 1$ and $Z(G) \neq 1$.
Assume that $Z(G)=1$, then by the preparatory result we know there exists a proper nontrivial normal subgroup $N$ of $G$. Since $|N|$ divides $|G|$, $|N| < |G|$ and $|G/N| < |G|$ we have by the induction hypothesis that $N$ and $G/N$ are solvable, thus $G$ is also solvable.
Assume that $Z(G) \neq 1$, moreover assume that $Z(G) \neq G$ because then $G$ is abelian and trivially solvable. Then $Z(G)$ is a proper nontrivial normal subgroup of $G$, and and $G$ is solvable by a similar argument to the one above.
Is my proof correct?
 A: This is not an answer to my question, but my proof of the preparatory result stated in the beginning of the post.
The setup is the same as in my post.
$\textbf{Lemma}$: If $Z(G)=1$ then there exists a proper nontrivial normal subgroup of $G$.
Suppose $Z(G) = 1$, and let $C$ be a set of representatives for each conjugacy class of $G$. We have the class equation 
                        \begin{equation}\label{eq:class}
       |G| =\sum_{x\in C}[G:N(x)]
      \end{equation} 
                        where $N(x) = \{a \in G|axa^{-1} = x\}$ and $|C(x)| = [G:N(x)]$. The equation above can be modified by letting $C' \subset C$ be the subset containing representatives for each conjugacy class having more than one element (note that $1\notin C'$), yielding
                        \begin{equation}
       |G| =|Z(G)| + \sum_{x\in C'}[G:N(x)] = 1 + \sum_{x\in C'}[G:N(x)]
      \end{equation}
                        Now, because $|G|=p^{q}q^{b}$ we have that $q$ divides $\big(1 + \sum_{x\in C'}[G:N(x)]\big)$. Clearly, $q$ cannot divide $[G:N(x)]$ for all $x\in C'$ because then it would divide $1$ as well. Therefore there exists an $x\in G\backslash \{1\}$ such that $q$ does not divide $$|C(x)| = [G:N(x)] = \frac{p^{a}q^{b}}{|N(x)|}$$ which implies that $|C(x)|=p^{k}$ for $k\leq a$. $\textbf{From a previous problem:}$ We know that there exists a nontrivial irreducible representation $\rho :G \rightarrow GL(V)$ whose kernel $N \triangleleft G$ is a proper normal subgroup. Moreover, the image of $x$ in $G/N$, i.e. $xN$, lies in $Z(G/N)$. Suppose that $N$ is the trivial subgroup, then $x \in Z(G) = 1$, which is a contradiction. Therefore, $N$ is a proper nontrivial normal subgroup of $G$.
EDIT: The statement of the previous problem:
Let $g\in G\backslash\{1\}$ and suppose $|\mathcal{C}(g)|$ is a $p$-power for some prime $p$.
(a) Show that there exists a nontrivial irreducible representation $\rho : G \rightarrow GL(V)$ with character $\chi$ such that $\chi(g) \neq 0$ and $p$ does not divide $\chi(1)$.
(b) Let $\rho$ be as in (a) and explain why $\rho_{g}$ is multiplication by a complex number.
(c) Let $N \leq G$ be the kernel of $\rho$. Verify that $N \neq G$ and that the image of $g$ in $G/N$ lies in the center of $G/N$.
