If $G$ is a nonabelian simple group then $G$ has at least $7$ non-normal subgroups of distinct order If $G$ is (any) nonabelian simple group, then why has $G$ at least $7$ subgroups of distinct orders?
My question is a lemma in an article.
The proof of lemma in the article is:

Noticing the subgroups of distinct orders are not conjugate in $G$, we calculate the number of subgroups of distinct orders of prime powers of $p$ for every $p\in\pi(G)$.
  One more fact is that $G$ cannot have all Sylow subgroups self-normalizing, otherwise $G$ is a $p$-group for some prime $p$ by Corollary in other article, a contradiction. By checking the number of prime power divisors of $|G|$ and noticing that there is at least one Sylow subgroup not self-normalizing, one can see that $G$ has at least $7$ subgroups of distinct orders.

Now I want to know, why is the number of this subgroups greater than $7$?
 A: I did not understand autors reasoninig but it can be proved by Burnside theorems.
Theorem(Burnside) Any finite simple group has at least $3$ prime divisior.
Theorem(Burnside) If $N_G(P)=C_G(P)$ for $P\in Syl_p(G)$, $G$ is $p$ nilpotent.
Theorem(Burnside) If $P\in Syl_p(G)$ is cyclic for a smallest $p$ dividing order of $G$,$G$ is $p$ nilpotent.
Claim: If $G$ is a finite abelian group then $G$ has at leats $7$ subgroups of different orders.
We know that $G$ has at least $3$ prime divisior. Let they are $p,q,r$. W.L.G assume $r$ is the smallest prime dividing order of $G$.
By, Burnside theorem we know that sylow $r$ subgroup can not be  cyclic. we have at least $2$ subgroup of order distinc power of $r$.
Let $H$ be a subgroup of order $p$. Notice that $H<N_G(H)$ if $H$ is not sylow subgroup.(normalizer will grows in sylow-$p$ subgroups). If $H$ is a Sylow subgroup then again we must have $H<N_G(H)$ otherwise $G$ is $p$ nilpotent By Burnside.(we can assume that $H$ and $N_G(H)$ are different groups)
An let $K$ be a subgroup of order $q$.
we have $2+2+1=5$ subgroup. The left $2$ is not so hard !
