Burnside Theorems … I have read that Burnside's theorem implies that a group with order $p^aq^b$ cannot be simple.
Well  how can I prove that: From  Burnside's conjugation theorem (If G has a conjugacy class whose order is divisible  nontrivial power of a prime, then G is not simple) deduce the Burside solubility theorem, i.e., for p and q primes, every group of order $p^aq^b$ is solvable.
Proof.
I proceed by induction on $a+b$. If $a+b ≤ 1$ then G is cyclic of prime order, hence abelian, and therefore solvable since its commutator subgroup is trivial. Now if G is of order $p^aq^b$ then, by Burnside’s conjugation theorem, there is a normal subgroup H of G. By the inductive hypothesis, both H and G/H are solvable since they are groups whose order is of the form $p^a′q^b′$ where $a′ + b′ < a + b$, so there are abelian normal towers
and
$H =A_0 ≥A_1 ≥···≥A_s ={1}$ $G/H =B_0 ≥B_1 ≥···≥B_t ={1}$.
By the Lattice Isomorphism Theorem the $B_i$ correspond to subgroups $C_i$ of G containing H such that $C_i/H = B_i$ for each i. It is easy to see that $C_i+1$ is normal in $C_i$ for each i since $B_i+1$ is normal in $B_i$, and the Second Isomorprism Theorem asserts that $C_i/C_i+1 ∼= (C_i/H)/(C_i+1/H) ∼= B_i/B_i+1$ for each i, so each quotient $C_i/C_i+1$ is abelian. Finally, $B_t = {1}$ corresponds to $C_t = H$, so we have that $C_t/A_0 = {1}$ is abelian. Therefore
$G=C_0 ≥C_1 ≥···≥C_t ≥A_0 ≥A_1 ≥···≥A_s ={1}$ is an abelian normal tower for G, and G is solvable. 
Is it correct?
Thanks for you help and time.
 A: If you know the following:
THEOREM: If $G$ has a conjugacy class of size $p^\alpha$ for a prime $p$, then $G$ is not simple.
Then Burnside's theorem is easy: take a section of the composition series of a group $G$ - with $|G|=p^aq^b$ - and if this section is not cyclic, then look at the conjugacy class of a central element inside a Sylow subgroup.
MORE DETAILS: If a section $H$ of the composition series is not cyclic, it has order divisible by both $p$ and $q$; let's say it has order $p^nq^m$.  Then the Sylow $p$-subgroup has a non-trivial center, with an element $x$ in it.  Since $|C(x)|=p^nq^k$ for some $k < m$ (why?), the conjugacy class of $x$ has size $[H:C(x)]=q^{m-k}$, and you can apply the theorem above to reach a contradiction.
A: If i take you comment  @Steve D the if i proceed by induction on $a+b$. If $a+b ≤ 1$ then G is cyclic of prime order, hence abelian, and therefore solvable since its commutator subgroup is trivial. Now if G is of order $p^aq^b$ then, by Burnside’s conjugation, there is a normal subgroup H of G. By the inductive hypothesis, both H and G/H are solvable since they are groups of order of the form $p^a′q^b′$ where $a′ + b′ < a + b$, so there are abelian normal towers
and
$H =A_0 ≥A_1 ≥···≥A_s ={1}$ $G/H =B_0 ≥B_1 ≥···≥B_t ={1}$.
By the Lattice Isomorphism Theorem the $B_i$ correspond to subgroups $C_i$ of G containing H such that $C_i/H = B_i$ for each i. It is easy to see that $C_i+1$ is normal in $C_i$ for each i since $B_i+1$ is normal in $B_i$, and the Second Isomorprism Theorem asserts that $C_i/C_i+1 ∼= (C_i/H)/(C_i+1/H) ∼= B_i/B_i+1$ for each i, so each quotient $C_i/C_i+1$ is abelian. Finally, $B_t = {1}$ corresponds to $C_t = H$, so we have that $C_t/A_0 = {1}$ is abelian. Therefore
$G=C_0 ≥C_1 ≥···≥C_t ≥A_0 ≥A_1 ≥···≥A_s ={1}$ is an abelian normal tower for G, and G is solvable. 
Is it correct?
