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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?

Greetings!

Shaun
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Soberbio P. Pacheco
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3 Answers3

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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

  • $a_i < 3$
  • $p_i^k \not \equiv 1 \bmod{p_j}$ for all $1 \leq k \leq a_i$

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.

lhf
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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.

Nicky Hekster
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  • hey great solution! But how to show that $PQ \cong P \times Q$? I can't imagine the isomorphism between them. – Levon Minasian Mar 13 '21 at 16:46
  • @Levon Minasian: The subgroup $PQ$ has two normal subgroups $P$ and $Q$ of order $7$ and $11$ respectively. This implies $P \cap Q=1$, $[P,Q]=1$ ($P$ and $Q$ commute which each other) and every element of $PQ$ can be uniquely expressed as $xy$, $x \in P, y \in Q$. Define a map $\varphi : PQ \rightarrow P \times Q$ by $\varphi(xy)=(x,y)$, where $x \in P$ and $y \in Q$. You need to show that this map is well-defined, a homomorphism and bijective. Well, I have paved the way, so up to you to fill in the details and finish the proof. – Nicky Hekster Mar 14 '21 at 13:38
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    Wow! Got it, thanks! – Levon Minasian Mar 14 '21 at 15:43
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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.