If $\left|G\right| = 1001= 7\cdot11\cdot13$, we want to prove that all possible subgroups exists, and all of them are normal.
My first idea is to apply Sylows' Theorems. For subgroups with order $7,11$ and $13$, we can apply the 1st Theorem.
But, how about subgroups with order $77,91$ and $143$? And how to prove they are normal?
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On
Let $P \in Syl_7(G)$, $Q \in Syl_{11}(G)$ and $R \in Syl_{13}(G)$. With the help of the Sylow Theorems and working out the congruences that need to be satisfied, it is easy to see that $P,Q,R \lhd G$. Since $P,Q$ are normal in $G$, $PQ$ is a subgroup and also normal. In fact one can show $PQ \cong P \times Q \cong C_{77}$. Now repeat the argument, looking at the normal subroups $PQ$ and $R$ and observe that $|PQ \cdot R|=\frac{|PQ| \cdot |R|}{|PQ \cap R|}=1001$. So $G=PQR \cong PQ \times R \cong C_{77} \times C_{13} \cong C_{1001}$. Hence $G$ is even cyclic and all subgroups are normal, even characteristic.
On
By Sylow III, $n_7=1$ and hence $P_7\unlhd G$. As such, $P_7$ is the union of conjugacy classes; but the least size of the nontrivial conjugacy classes is $7$ and $|P_7\cap Z(G)|=1$ or $7$; so, the only possibility is $P_7$ being central. Moreover, $G/P_7$ is cyclic, because $11\nmid(13-1)$. Therefore, $G$ is Abelian (and hence cyclic), all whose subgroups are then normal.
Every group of order $1001$ is abelian because $1001$ is an abelian number. Therefore, every subgroup of a group of order $1001$ is normal.
$n$ is an abelian number when every group of order $n$ is abelian. This happens iff $n$ is a cubefree nilpotent number, that is, if $n = p_1^{a_1} \cdots p_r^{a_r}$, then
This is easily checked to be the case for $n=1001$.
Actually, $1001$ is a cyclic number and so every group of order $1001$ is cyclic. $n$ is an cyclic number when every group of order $n$ is cyclic. This happens iff $n$ is a squarefree nilpotent number.