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!

  • $\begingroup$ @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

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 $4$.

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.