Burnside proved, by use of character theory, that if a finite group $G$ has a conjugacy class $C$ such that $\vert C \vert$ is a prime power $> 1$, then $G$ is not simple. Let us call this statement "Burnside's non-simplicity theorem".
Let $G$ be a group of order $4400$. Let $n_{5}(G)$ denote the number of Sylow $5$-subgroups of $G$. In a proof that G is not simple, W.R. Scott (12.3.5, p. 335) uses Burnside's non-simplicity theorem, and thus character theory (since Scott proves Burnside's non-simplicity theorem by use of character theory), in the case where $n_{5}(G) = 11$. He settles the other cases without character theory.
In fact, if I'm not wrong, it is not very difficult to avoid character theory, by use of the following lemma:
(Edit : the following lemma is the result of an improvement made by j.p. to my initial formulation.)
Lemma. Let $H$ be a finite group, let $p$ be a prime divisor of $\vert H \vert$. Let $n_{p}(H)$ denote the number of Sylow $p$-subgroups of $H$. Assume that
1° $n_{p}(H)$ is the least divisor $d$ of $\vert H \vert$ such that $d > 1$ and $d \equiv 1 \pmod{p}$;
2° $n_{p}(H) \not\equiv 1 \pmod{p^{r}}$, where $p^{r}$ denotes the greatest power of $p$ that divides $\vert H \vert$ (in other words, $p^{r}$ is the order of the Sylow $p$-subgroups of $H$).
Then $H$ is not simple.
Proof. In view of a strong form of the Sylow congruence theorem, we conclude from hypothesis 2° that we can choose two distinct Sylow $p$-subgroups of $H$, say $P$ and $Q$, such that $P \cap Q > 1$. We may additionally assume that $P\cap Q$ is maximal among the intersection of two Sylow $p$-subgroups. Let $N:=N_H(P\cap Q)$ be the normalizer of this intersection in $H$. As $P$ and $Q$ are (finite) $p$-groups, $N\cap P$ and $N\cap Q$ both properly contain $P\cap Q$. $N\cap P$ and $N\cap Q$ are contained in Sylow $p$-subgroups $P'$ rsp. $Q'$ of $N$, which in turn are contained in Sylow $p$-subgroups $P''$ rsp. $Q''$ of $H$. By maximality of $P\cap Q$ we get $P=P''$ and $Q=Q''$ and hence that $N\cap P = P'$ and $N\cap Q = Q'$ are (different) Sylow $p$-subgroups of $N$, thus $N$ has more than one Sylow $p$-subgroup. On the other hand, by Sylow theorems, the number of Sylow $p$-subgroups of $N$ is $\equiv 1 \pmod{p}$ and divides $\vert N\vert$, which divides $\vert H \vert$. Thus the number of Sylow $p$-subgroups of $N$ is $> 1$, is $\equiv 1 \pmod{p}$ and divides $\vert H \vert$. By minimality of $n_{p}(H)$ (hypothesis 1°),
(*) the number of Sylow $p$-subgroups of $N$ is $\geq n_{p}(H)$.
As $P\cap Q$ is normal in $N$, every Sylow $p$-subgroup of $N$ contains $P\cap Q$. But, as by the maximality of $P\cap Q$ every Sylow $p$-subgroup of $N$ is contained in a unique different Sylow $p$-subgroup of $H$, by (*) every Sylow $p$-subgroup of $H$ contains some Sylow $p$-subgroup of $N$, and hence $P\cap Q$. Thus the intersection of all Sylow $p$-subgroups of $H$, which is a normal subgroup, is non-trivial, and thus $H$ is not simple.
Applying this lemma with $H = G$ (where $\vert G \vert = 4400$) and $p = 5$, we find that if $n_{5}(G) = 11$, then $G$ is not simple. Thus, in each case, it can be proved without character theory that a group of order $4400$ is not simple.
In exerc. 12.3.10, p. 337, W.R. Scott asks for a proof that there are no simple groups of order $1200$, $2240$ or $2800$. I presume that use of Burnside's non-simplicity theorem is expected, but here again, if I'm not wrong, character theory can be skipped.
My question is : do you know a nonzero natural number $n$ with at least $3$ distinct prime factors such that
(i) it is relatively easy to prove that there is no simple group of order $n$ with use of Burnside's non-simplicity theorem;
(ii) it is not easy to prove it without Burnside's non-simplicity theorem and without character theory ?
Thanks in advance for the answers.
