Edit: This is what the OP is trying to ask, I think.
Let $G$ be a group of order $360$. Suppose that there are ten Sylow $3$-subgroups, 36 Sylow $5$-subgroups and 45 Sylow $2$-subgroups. Show that this implies that $G$ is a simple group.
My idea is to use Lemma 1 Lemma2 and proof by contradiction to rule out all possible order of $H$($H$ is a minimal normal group in $G$).
Lemma 1 : G is finite group , $n \mid |G|$ , $L_{n}=\{g:g^{n}=1\}$ , then $n\mid |L_{n}|$.
Lemma 2 : Group of order 9 are either isomorphic to $Z_{9}$ or $Z_{3}\times Z_{3}$
$n_{p}(G)$: the number of Sylow p-group in $G$
$s_{n}(G)$: the number of elements of order $n$ in $G$
If $G$ is not simple group and $H$ is a minimal normal group.
If $5\mid |H|$ , then $H$ has all elements of order $5$ so $|H|\geq n_{5}\phi(5)=144$ thus $|H|=2^{2}3^{2}5=180$. It also follows that $H$ has all elements of order $3^{r}$. Let $x\in H$, $o(x)=3^{r}(r=1,2)$. If $r=2$, hence $s_{9}(H)=n_{3}(G)\phi(9)=60$. But $|N|=180<144+60$, so $r=1$. And then I couldn't go on
Based on a hint form David A. Craven and Derek Holt, I conducted the following proof
It is easy to rule out $60$,$5$, $2^3$,$3^2$
If $|H|=2^{2}$ or $3$, then $H\cong C_{2}\times C_{2}$ or $C_3$
hence
$$
N_{G}(H)/C_{G}(H)\lesssim Aut(H),|Aut(H)|=2.
$$
So $｜C_{G}(H)｜=180$.
It follows that $H< N_{G}(R) $ where $R\in Syl_{5}(C_{G}(H))$.
But $|H|\nmid |N_{G}(R)|=10$.
If $|H|=2$, consider the minimal normal group of $G/N$ which order could be $2,4,,3,9,5,60$. So there is a normal subgroup group $M(\supseteq H)$ of $G$ which order could be $4,8,6,18,120,10$. It is easy to rule out $4,8,120,10$.
If $|M|=18$, because of $H\lhd M$, So $|Syl_{2}(M)|=1$. Contradiction.
If $|M|=6$, $M$($M\supseteq H$ and $H\subseteq Z(G)$) is 6-cyclic, so $|C_{G}(M)|=180$. It follows that
$C_{G}(M)\lhd G$.
Similarly, exists $K\lhd C_{G}(M)$, $K\supseteq H$. $|K|$ could be $4,6,18,10$.
Just consider $|K|=6$. If $|K|=6$. Similarly, $C_{M}(K)\lhd M$, $|C_{M}(K)|=90$ contradiction.
Is my proof correct? If I not, how can we prove it?
Best Answer
Let me throw in an alternative argument. Suppose $G$ is solvable. Then by Hall's theorem, there exists a subgroup $H\le G$ of order $45$. By Sylow's theorem, $H$ has a normal Sylow $3$-subgroup $P$. Hence, $5$ divides $|N_G(P)|$. But since $P$ is also a Sylow of $G$, the number of Sylow $3$-subgroups of $G$ cannot be $10$. The same argument works with $5$ instead of $3$.
EDIT: A complete argument on request: Let $P$ be a Sylow $p$-subgroup of $G$. Let $N$ be a non-trivial normal subgroup of $G$. Clearly, $N$ or $G/N$ is solvable. Morever, $36=n_5(G)=n_5(G/N)n_5(NP)$. By Hall's theorem mentioned above, $n_5(G/N)=1$ or $n_5(NP)=1$. In the first case, $|N|\ge 36*5=180$, thus $|N|=180$. Now $P$ is self-normalizing in $N$. By Burnside's theorem, $N$ has a normal $5$-complement. Then $N$ is solvable. Contradiction. Similarly, in the second case $|N|=2$ and $G/N$ has a normal $5$-complement. So $G/N$ is solvable. Contradiction.