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Here's how I solve the problems. Thanks for pointing out what might be the weakness of my solutions. Actually, what I want are other ways of solving this kind of problems, appart from counting the elements. Is there any reference by which I can be fully instructed? Any help is sincerely appreciated.

PS: I have also applied this method to the proof that any simple group of order $60$ is isomorphic to $A_5$.

I will apply the well-known Sylow theorem, strong Cayley theorem, $N/C$ theorem, Burnside's transfer theorem and some basic theories in group-theory.(All by contradiction)

1. Groups of order $180$ are not simple.

First of all, we assume group $G$ of order $180$ to be simple.

$180=2^2\cdot 3^2\cdot 5$. So a Sylow $5$-subgroup would have order 5.

By Sylow theorem, $n_5 \mid 2^2\cdot 3^2$, and $n_5\equiv 1$ $($mod $5)$. There are three possibilities: $n_5= 1,\,6,\,36.$ And we denote one of the Sylow $5$-subgroups by $P_5$.

$1)$ If $n_5=1$, then $P_5$ will be a normal subgroup of $G$, and we're done.

$2)$ If $n_5=36$, then $|G:N_G(P_5)|=36$, thereby $|N_G(P_5)|=\dfrac{180}{36}=5$. Because every group of order $5$ is abelian (cyclic), here we have $P_5\trianglelefteq C_G(P_5)\trianglelefteq N_G(P_5)$. Whereas $|P_5|=|N_G(P_5)|$, $C_G(P_5)=N_G(P_5)$. By Burnside's transfer theorem, there exists $K\triangleleft G$, and we are done.

$3)^{[1]}$ If $n_5=6$, then $|G:N_G(P_5)|=6$, thereby $|N_G(P_5)|=\dfrac{180}{6}=30.$ By $N/C$ theorem, $N_G(P_5)/C_G(P_5)\lesssim$ $ \text{Aut}(P_5)$, therefore we will know that $\dfrac{|N_G(P_5)|}{|C_G(P_5)|}$ divides $|\text{Aut}(P_5)|$. Since $P_5$ is cyclic, which leads to $|\text{Aut}(P_5)|=\varphi(5)=4$, $\dfrac{|N_G(P_5)|}{|C_G(P_5)|}=1$, $2$, or $4$. If $1$, then by Burnside's transfer theorem, we 're done. $4$ is impossible, by which $|C_G(P_5)|=\dfrac{|N_G(P_5)|}{4}$ will be a fraction. Then $\dfrac{|N_G(P_5)|}{|C_G(P_5)|}$ must be $2$, thereby, $|C_G(P_5)|=15$. It is known that every group of order $15$ is cyclic, so $C_G(P_5)$ is a cyclic group of order $15$. Then $G$ has an element of order $15$. $~~~$[ strong Cayley theorem ] $G$ is simple, hence $\ker\Phi $ must be the trivial subgroup$\{e\}$ $($if $\ker \Phi=G$, then $G\leqslant N_G(P_5)$, which is ridiculous $)^{[2]}$. Hence, $G=G/\{e\}=G/\ker\Phi\lesssim S_6$. However, it contradicts the statement that $G$ has element of order $15$, $2$ also impossible.

This completes the proof for $180$.

$~$

2. Groups of order $540$ are not simple.

First of all, we assume group $G$ of order $540$ to be simple.

$540=2^2\cdot 3^3\cdot 5$. So a Sylow $5$-subgroup would have order 5.

By Sylow theorem, $n_5 \mid 2^2\cdot 3^3$, and $n_5\equiv 1$ $($mod $5)$. There are three possibilities: $n_5= 1,\,6,\,36.$ And we denote one of the Sylow $5$-subgroups by $P_5$.

$1)$ If $n_5=1$, then $P_5$ will be a normal subgroup of $G$, and we're done.

$2)$ If $n_5=6$, then $|G:N_G(P_5)|=6$. By strong Cayley theorem, $G\lesssim S_6$. However, $|G|=540\not\mid 720=|S_6|$, contradicting our assumption, and we're done.

$3)$ If $n_5=36$, then $|G:N_G(P_5)|=36$, thereby $|N_G(P_5)|=\dfrac{540}{36}=15$. By $N/C$ theorem, $\dfrac{|N_G(P_5)|}{|C_G(P_5)|}$ divides $|\text{Aut}(P_5)|=\varphi(5)=4$, therefore it must be $1$. By Burnside's transfer theorem, there exists $K\triangleleft G$. We 're done.

This completes the proof for $540$.

$~$

3. Groups of order $1080$ are not simple.

First of all, we assume group $G$ of order $1080$ to be simple.

$1080=2^3\cdot 3^3\cdot 5$. So a Sylow $5$-subgroup would have order 5.

By Sylow theorem, $n_5 \mid 2^3\cdot 3^3$, and $n_5\equiv 1$ $($mod $5)$. There are four possibilities: $n_5= 1,\,6,\,36,\,216.$ And we denote one of the Sylow $5$-subgroups by $P_5$.

$1)$ If $n_5=1$, then $P_5$ will be a normal subgroup of $G$, and we're done.

$2)$ If $n_5=6$, then $G\lesssim S_6$, contradicting the fact that $|G|=1080>720=|S_6|$, impossible.

$3)$ If $n_5=216$, then $|G:N_G(P_5)|=216$, thereby $|N_G(P_5)|=\dfrac{1080}{216}=5$. Because every group of order $5$ is abelian (cyclic), here we have $P_5\trianglelefteq C_G(P_5)\trianglelefteq N_G(P_5)$. Whereas $|P_5|=|N_G(P_5)|$, $C_G(P_5)=N_G(P_5)$. By Burnside's transfer theorem, there exists $K\triangleleft G$, and we are done.

$4)$ If $n_5=36$, then $|G:N_G(P_5)|=36$, thereby $|N_G(P_5)|=\dfrac{1080}{36}=30.$ By $N/C$ theorem, $\dfrac{|N_G(P_5)|}{|C_G(P_5)|}$ divides $|\text{Aut}(P_5)|=\varphi(5)=4$, $1$ an $4$ obviously impossible, and it must be $2$. $C_G(P_5)$ is an cyclic group of order $15$, hence $C_G(P_5)$ has an unique subgroup of order $3$ denoted by $H_3$, $H_3\triangleleft C_G(P_5)$. $\forall z\in N_G(P_5)$,$z^{-1}H_3z\triangleleft z^{-1}C_G(P_5)z=C_G(P_5)$, is an subgroup of $C_G(P_5)$ with order $3$, therefore $z^{-1}H_3z=H_3$, $H_3\triangleleft N_G(P_5)$, furthermore, $N_G(H_3)\geqslant N_G(P_5)$. To be exact, $H_3$ is also a subgroup of $G$ with order $3$, by Sylow's first theorem, there exists a subgroup $H_9$ of $G$, which has order $9$, s.t. $H_3\triangleleft H_9$, thereby this $H_9$ is a subgroup of $N_G(H_3)$; note that $9\not\mid 30$, $H_9\not\subset N_G(P_5)$, we'll get $N_G(H_3)> N_G(P_5)$ ie. $|N_G(H_3):N_G(P_5)|>1$. About $36=|G:N_G(P_5)|=|G:N_G(H_3)|\cdot|N_G(H_3):N_G(P_5)|$: firstly, $|N_G(H_3):N_G(P_5)|$ is a divisor of $36$$(\text{not 1})$; secondly, $|N_G(H_3):N_G(P_5)|=|N_G(H_3):N_{N_G{(H_3)}}(P_5)|\equiv 1~(\text{mod}~5)^{[3]}$. Therefore $|N_G(H_3):N_G(P_5)|=6$ or $36$, and $|G:N_G(H_3)|=1$ or $6$. If $1$, then we're done. If $6$, it will contradict the fact that $G$ cannot embed in $S_6$.

This completes the proof for $1080$.

Note:

[1] In the third step of $180$, we can actually do with it quickly. By a corollary from strong Cayley theorem, simple group $G$ embeds not only $S_6$, but also $A_6$; however, it will have index 2, hence normal, contradicting the simplicity of $ A_6$. We are done.

[2] $\Phi$ is a map introduced in strong Cayley theorem ; $G$ is simple, hence $\ker\Phi $, as a normal subgroup, must be $G$ itself or the trivial subgroup$\{e\}$. If $\ker \Phi=G $, then according to strong Cayley theorem, $$G=\ker\Phi=\bigcap\limits_{x\in G} x^{-1}N_G{(P_5)}x\leqslant N_G(P_5).$$ Whereas, $|G:N_G{(P_5)}|>1$, it will be ridiculous.

[3] Since $N_G(P_5)$ is contained in $N_G(H_3)$, looking for such an element in $G$ will be equivlient to looking for it in $N_G(H_3)$, that is to say, $N_G(P_5)=N_{N_G(H_3)}(P_5)$. Furthermore, $|N_G(H_3):N_G(P_5)|=|N_G(H_3):N_{N_G{(H_3)}}(P_5)|$, which represents the number of Sylow $5$-subgroups contained in $N_G(H_3)$, satisfies Sylow’s third theorem.

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