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I have seen two statements of Burnside's Theorem and they are as follows.

Statement 1: Let $p, q$ be distinct prime numbers and $a,b \in \mathbb{Z}_{\geq 0}$. There does not exist a non-abelian simple group $G$ of order $p^aq^b$.

Statement 2. Let $p,q$ be distinct prime numbers and $a,b \in \mathbb{Z}_{\geq 0}$. Then any group $G$ of order $p^aq^b$ is solvable.

So as these statements are equivalent does this mean that every solvable group is not non-abelian and simple? Why is this the case?

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up vote 5 down vote accepted

Strange way of phrasing it (the negation should be "not (non-abelian and simple)", otherwise it can be mis-parsed), but yes: a solvable group must either be abelian, or else it is not simple. More than that, if $G$ is solvable, then there cannot exist subgroups $H$ and $K$ of $G$, such that $H\triangleleft K$ and $K/H$ is nonabelian and simple.

However, it is possible for a group to be non-simple and yet not be solvable (e.g., $S_n$ with $n\gt 4$).

A group $G$ is solvable if and only if there exists a normal series $$\{1\} = H_n \triangleleft H_{n-1}\triangleleft\cdots\triangleleft H_1\triangleleft H_0 = G$$ such that $H_{i+1}\triangleleft H_i$ and $H_i/H_{i+1}$ is abelian.

For every finite group $G$, there is a normal series as above in which $H_i/H_{i+1}$ is simple, and $G$ will be solvable if and only if in that series all those quotients are cyclic groups of prime order. Thus, in particular, if there are no nonabelian simple groups of order $p^aq^b$, then any such series for a group of order $p^{\alpha}q^{\beta}$ cannot have $H_i/H_{i+1}$ nonabelian and simple, hence they will all be abelian and simple, so $G$ will be solvable. This proves that Statement $2$ implies statement $1$. To see that statement $1$ implies statement $2$, note that a nonabelian simple group is not solvable.

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