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Let $G = \langle x,y \rangle$ be a finite bicyclic group generated by the two elements $x,y \in G$ and assume that $x \not\in \langle y \rangle,y \not\in \langle x \rangle.$

Is it true that $G = \langle x \rangle \times H$ for some subgroup $H \leq G$, i.e. does $\langle x \rangle$ admit a complement in $G$?

This seemed intuitive to me, but I am struggling to show it. Thanks in advance!

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  • $\begingroup$ @IzaakvanDongen, I think I see your point. We just take $H$ to be the subgroup of $\langle y \rangle$ isomorphic to $G / \langle x \rangle$? $\endgroup$ – u1571372 Jul 14 '20 at 14:22
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Not necessarily. Let $G = {\mathbb Z}/3{\mathbb Z}\oplus {\mathbb Z}/12{\mathbb Z}$, $x = (1,6)$, $y=(0,1)$.

Note that $\langle x \rangle$ has order 6, and it cannot have a complement, because the $2$-part of $G$ is cyclic of order $4$.

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  • $\begingroup$ Thanks for the answer, Prof. Holt! If you don't mind a follow-up question: for my particular application it suffices that $x$ is contained in some (proper) subgroup of $G$ admitting a complement. I believe this weaker result is true, correct? $\endgroup$ – u1571372 Jul 15 '20 at 18:30
  • $\begingroup$ It's still not true, let $G$ be cyclic of order $36$, $o(x)=18$, $o(y)=4$. $\endgroup$ – Derek Holt Jul 16 '20 at 7:56
  • $\begingroup$ Thanks, but I've forgot to point out that $G$ is not a cyclic group in my application (I added that info in the original question). $\endgroup$ – u1571372 Jul 16 '20 at 20:26
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Let $G=\mathbb{Z}_{12}$, represented as $\{1,a,a^2,...,a^{11}\}$.

Then letting $x=a^2$ and $y=a^3$, we get $G=\langle{x,y}\rangle$.

It's easily verified that $x\not\in\langle{y}\rangle$ and $y\not\in\langle{x}\rangle$.

Suppose $G=\langle{x}\rangle{\times}H$ for some subgroup $H$ of $G$.

Since $\langle{x}\rangle$ has order $6$, $H$ must have order $2$.

Noting that $\langle{a^6}\rangle$ is the only subgroup of $G$ of order $2$, it follows that $H=\langle{a^6}\rangle$.

But then $H\subset\langle{x}\rangle$, so $\langle{x}\rangle\cap H$ is nontrivial, contrary to the assumption that $G=\langle{x}\rangle{\times}H$.

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