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Question: I was working on the problem below, and I have a proof, but I can't seem to remember why a certain part of the proof implies a group is cyclic. Here's the problem:

Show a group of order $2002$ has an abelian subgroup of index $2$.

Proof. Using Sylow, $n_{13}=1$, $n_{11}=1$, and $n_7=1$. Let $P\in Syl_{13}(G)$, $Q\in Syl_{11}(G)$, and $R\in Syl_7(G)$. As $P, Q, R\trianglelefteq G$, we have $PQR\trianglelefteq G$. Since $(|P|, |Q|, |R|)=1$, we have $P\cap Q\cap R=\langle 1\rangle$, so $|PQR|=7\cdot11\cdot13$. Thus, $|G:PQR|=2$ (so we have our subgroup of index $2$), and to show $PQR$ is abelian we show that it is cyclic since $13\not\equiv 1\bmod11$, $13\not\equiv 1\bmod 7$, $11\not\equiv 1\bmod13$, and so on.

But, I can't seem to remember what theorem or why we can claim a group is cyclic based on the last line above. Can anyone point me in the right direction?

Thank you.

user26857
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User7238
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    slightly related : https://mathoverflow.net/questions/148731/for-which-n-is-there-only-one-group-of-order-n – Asinomás Jul 14 '21 at 20:24
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    Careful; it is not enough for the triple intersection to be trivial: you need $P\cap Q$ trivial, and you need $PQ\cap R$ trivial (or some such). For example, in $C_2\times C_2$, the three subgroups of order $2$ have pairwise trivial intersection, but the size of their product is not the product of their sizes (which would be $8$, larger than the whole group). – Arturo Magidin Jul 14 '21 at 20:28
  • @ArturoMagidin Thank you! Here, though, I would need all three, correct? $P\cap Q$, $PQ\cap R$, and $P\cap Q\cap R$ to all be trivial, right? I don't believe anymore work would need shown to claim $P\cap Q$ is trivial, but would anything else need done to show $PQ\cap R$ is trivial? I wouldn't think so, but I am curious if I am missing something.. – User7238 Jul 14 '21 at 20:35
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    The fact that $P\cap Q\cap R$ is trivial is irrelevant. $P\cap Q$ is easy to show to be trivial, and then an easy argument shows $PQ\cap R$ is trivial (e.g., consider the orders). But you need to argue that way, you can't argue as you did. – Arturo Magidin Jul 14 '21 at 20:37
  • $1001$ is a cyclic number – lhf Jul 15 '21 at 00:09

1 Answers1

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  1. To conclude that $|PQR|=7\times 11\times 13$, it is insufficient to have $P\cap Q\cap R = \{e\}$. To see this, consider the case of the Klein $4$-group, $C_2\times C_2$, and the subgroups $A=\langle (x,e)\rangle$, $B=\langle (e,x)\rangle$, and $C=\langle (x,x)\rangle$. We have $A\cap B\cap C=\{(e,e)\}$ (in fact, the pairwise intersections are trivial), but $4=|ABC|\neq |A|\times|B|\times |C|=8$. You are trying to invoke the result that says that if $A$ and $B$ are subgroups, then $|AB||A\cap B|=|A|\,|B|$; that holds for pairwise products, not triple products. You need to first note that $P\cap Q=\{e\}$, so $|PQ|=7\times 11$; then show that $PQ\cap R=\{e\}$, so that $|PQR| = |(PQ)R| = |PQ|\,|R|=(7\times 11)\times 13$. This is not hard to do, but it still needs to be done like that, and not as you do.

  2. The argument you use in the final line seems to miss the point. Note that in any group, if $K$ and $N$ are normal and $K\cap N=\{e\}$, then $kn=nk$ for each $k\in K$ and $n\in N$ (Proof: $knk^{-1}n^{-1}\in K\cap N$, so $kn=nk$). Thus, $PQ$ is abelian and normal, of order $77$, and isomorphic to $P\times Q$. This is a product of two cyclic groups of coprime order, so it is cyclic. Thus, $PQ$ is cyclic of order $77$. And hence $PQR$ is abelian, normal, of order $1001$, and isomorphic to the product of a cyclic group of order $77$ and one of order $13$, and since $\gcd(77,13)=1$, again we have that the product is cyclic of order $1001$.

Arturo Magidin
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