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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!
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 $4$.
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 $G=\langle{x}\rangle{\times}H$.
