Question on a theorem implying a group is cyclic 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.
 A: *

*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.


*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 $1001$.
