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Fix $\epsilon >0$ and set $d=2/\epsilon$. Suppose, towards a contradiction, thatwhere $\lambda$ will absorb constants depending only on $\epsilon$. Put $n=\lfloor |G{|}^{\epsilon }\rfloor$. Once $\lambda$ is large enough, $n\ge N(d)$, $n<D$, andPart (a) yields $H⊴C<G$ with $[G:C]=O_{\epsilon }(1)$, $H\subseteq S^{\lfloor n/2\rfloor }$, and $C/H$ nilpotent of class $O_{\epsilon }(1)$.

The subgroup core ${\bigcap }_{g\in G}gC{g}^{-1}$ is normal in $G$ and has index at most $[G:C]!$. Since $G$ is a simple group, the core is either $\{1\}$ or $G$. In the first case $|G|\le [G:C]!=O_{\epsilon }(1)$, which is excluded by increasing $\lambda$. Hence the core is $G$, so $C=G$.

Now $H⊴G$. Since $n/2<D$, the ball $S^{\lfloor n/2\rfloor }$ is not all of $G$, so $H\ne G$. Simplicity gives $H=\{1\}$, and therefore $G=C/H$ is a nilpotent group. A nontrivial finite nilpotent group has nontrivial center of a group; simplicity would force that center to be all of $G$, making $G$ Abelian. This contradicts the assumption that $G$ is non-abelian. Consequently

For $n\ge 2$, the commutator subgroup of $S_n$ is $A_n$. The group $A_5$ is nonabelian and simple, so its commutator subgroup is a nontrivial normal subgroup and must equal $A_5$ itself. Hence the derived series of $S_5$ isand never reaches the identity. Therefore

Let $G$ act by left multiplication on the set $G/H$ of $n$ left cosets. This gives a group action homomorphismIts kernelis a normal subgroup of $G$, and the first isomorphism theorem gives $G/K\cong \mathrm{im}\phi$. By Lagrange's theorem, $|G/K|$ divides $n!$. The image acts transitively on the $n$ cosets, so the orbit-stabilizer theorem shows that $n$ divides $|G/K|$; in particular,

Now suppose $G$ is nonabelian and simple. Since $H$ is proper, the coset action is nontrivial, so $K\ne G$. Simplicity gives $K=\{e\}$, and $\phi$ embeds $G$ into $S_n$. Composing with the sign homomorphism givesIts kernel is normal. A nontrivial map would embed the simple group $G$ into the abelian group of order two, which is impossible because $G$ is nonabelian. The sign is therefore always $+1$, so

A Sylow subgroup of a finite group $G$ is a subgroup whose order is the largest power of a prime $p$ dividing $|G|$. The Sylow theorems say that such subgroups exist, that every $p$-subgroup lies in one, that all Sylow $p$-subgroups are conjugate, and that their number $n_p$ divides $|G|/p^a$ and obeys $n_p\equiv 1(\mathrm{mod}p)$.

For the last congruence, let one Sylow $p$-subgroup $P$ act by conjugation on the set of all Sylow $p$-subgroups. Every orbit other than a fixed point has size divisible by $p$. If $Q$ is fixed, then $P,Q\le N_G(Q)$. They are Sylow subgroups of this normalizer, so they are conjugate within $N_G(Q)$; because every element of $N_G(Q)$ fixes $Q$ under conjugation, this forces $P=Q$. Thus there is exactly one fixed point and $n_p\equiv 1(\mathrm{mod}p)$.

Now let $H<A_n$ have index $m$ with $1<m<n$. The coset action gives a homomorphismIts kernel is a normal subgroup of the simple group $A_n$. It cannot be all of $A_n$, since the action is transitive and nontrivial, so it is trivial. This would embed $A_n$ into $S_m$, contrary to $|A_n|=n!/2>m!=|S_m|$. Hence no such subgroup exists.

Suppose finally that a group $G$ of order $90$ were simple. The Sylow count satisfiesand simplicity excludes $n_5=1$, so $n_5=6$. Conjugation on these six subgroups gives a nontrivial homomorphism $G\to S_6$, which simplicity makes injective. Its image lies in $A_6$, because the composite with the sign homomorphism $S_6\to \{\pm 1\}$ must be trivial. It would therefore be a subgroup of $A_6$ of index $360/90=4$, contradicting the result just proved. Thus

A nonabelian simple group has no nonidentity conjugacy class of prime-power size. Coprime-degree irreducible characters vanish on such an element: the conjugacy-class sum makes the character-to-degree ratio an algebraic integer, and the Kronecker theorem on algebraic integers in the unit disk makes a nonzero ratio a root of unity, forcing a scalar in a faithful group representation. Column character orthogonality then makes $1/p$ an algebraic integer, impossible. This is the character-theoretic ingredient in Burnside's theorem.

A normal subgroup of $A_6$ has order dividing 360 and consists of the identity plus whole conjugacy classes of the alternating group on six letters. If it omits the odd-sized class of size 45, its order is an odd divisor and no nontrivial class sum fits. If it includes that class, the possible class sums not exceeding 180 are $46,86,118,126,136,158,176$, none dividing 360. Every proper subgroup has order at most 180; therefore $A_6$ is a simple group.