# When does the subgroup generated by a generator of a group admit a complement?

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!

• @IzaakvanDongen, I think I see your point. We just take $H$ to be the subgroup of $\langle y \rangle$ isomorphic to $G / \langle x \rangle$? – 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 $$4$$.

• 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? – u1571372 Jul 15 '20 at 18:30
• It's still not true, let $G$ be cyclic of order $36$, $o(x)=18$, $o(y)=4$. – Derek Holt Jul 16 '20 at 7:56
• 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). – 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 $$G=\langle{x}\rangle{\times}H$$.