I want to show that if $|G|=pqr$ where $p,g,r$ are primes, then $G$ is not simple.
We have that a group is simple if it doesn't have any non-trivial normal subgroups, right?
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I have done the following:
There are $p$-Sylow, $q$-Sylow and $r$-Sylow in $G$.
$$P\in \text{Syl}_p(G), \ |P|=p, \\ Q\in \text{Syl}_q(G), \ |Q|=q, \\ R\in \text{Syl}_r(G), \ |R|=r$$
$|\text{Syl}_p(G)|=1+kp \mid |G|=pqr$ so $1+kp\mid pqr \Rightarrow 1+kp\mid qr$.
So, $$1+kp=\left\{\begin{matrix} 1 \\ q \\ r \\ qr \end{matrix}\right.$$
Similarily, $|\text{Syl}_q(G)|=1+kq \mid |G|=pqr$ so $1+kq\mid pqr \Rightarrow 1+kq\mid pr$.
So, $$1+kq=\left\{\begin{matrix} 1 \\ p \\ r \\ pr \end{matrix}\right.$$
And, $|\text{Syl}_r(G)|=1+kr \mid |G|=pqr$ so $1+kr\mid pqr \Rightarrow 1+kr\mid pq$.
So, $$1+kr=\left\{\begin{matrix} 1 \\ p \\ q \\ pq \end{matrix}\right.$$
Is this correct?
Suppose that $|\text{Syl}_p(G)|=1$ or $|\text{Syl}_q(G)|=1$ or $|\text{Syl}_r(G)|=1$. That means that there is one Sylow subgroup that has order $p$, $q$, $r$, respectively, right? Is this Sylow subgroup normal?
Suppose that $|\text{Syl}_p(G)|=q$. That means that the number of $p$-Sylow subgroups is $q$, each of which has order $p$, right? How can we check if these subgroups are normal?
Assuming the three primes are different (otherwise you can try to show separatedly), say $\;p<q<r\;$ . Assume $\;1<n_r\;$ , then $\;n_r=pq\;$ , otherwise $\;n_r=p\;$ or $\;n_r=q\;$, which cannot be as also $\;n_r=1\pmod r\;$ and $\;r>q>p\;$ .
Thus, we already have $\;pq(r-1)=pqr-pq\;$ elements of order $\;r\;$ in the group. If we assume that also $\;n_q>1\;$ , then since there are $\;n_q>q>p\;$ Sylow $\;q-$ subgroups, we have at least $\;p(q-1)=pq-p\;$ elements of order $\;q\;$ .
Toghether, we have at least $\;pqr-pq+pq-p=pqr-p\;$ elements of order $\;r\;$ or $\;q\;$ , leaving only $\;p\;$ elements which must be a unique Sylow $\;p$ - subgroup which is then normal.