2
$\begingroup$

I would like to know necessary and sufficient conditions on $x$ and $y$ to have $\langle x,y\rangle=\langle x\rangle\langle xy\rangle$.

Sure that:

$\bullet$ $\langle x\rangle$ and $\langle xy\rangle$ must commute (otherwise the product wouldn't be a group).

$\bullet$ $x\notin\langle xy\rangle$, otherwise $\langle xy\rangle=\langle x,y\rangle$ (in fact $x\in\langle xy\rangle\Rightarrow x^{-1}\in\langle xy\rangle\Rightarrow y\in\langle xy\rangle$) hence we would have that a group generated by two elements is cyclic, a contradiction (it's implicit that 2 is the lowest number of generators). Similarly $y\notin\langle xy\rangle$.

But I can't extrapolate more informations. If someone help me it would be great. Thanks

$\endgroup$

1 Answer 1

4
$\begingroup$

Suppose $x,y \in G$ and $G$ is some group. Then $\langle x,y\rangle$ is always a group, so if $\langle x \rangle \langle xy \rangle$ is a group then $\langle x \rangle \langle xy \rangle = \langle xy \rangle\langle x \rangle$ (that is $\langle x \rangle$ and $\langle xy \rangle$ permute in the modern English language). Conversely suppose $\langle x \rangle$ and $\langle xy \rangle$ permute. Then $x = x (xy)^0 \in \langle x \rangle \langle xy \rangle$ and $y = x^{-1} (xy)^1 \in \langle x \rangle \langle xy \rangle$, so $\langle x,y \rangle \leq \langle x \rangle \langle xy \rangle$. Of course $x^i (xy)^j \in \langle x,y\rangle$ so $\langle x \rangle \langle xy \rangle \subset \langle x,y \rangle$.

Hence the necessary and sufficient condition for $\langle x,y \rangle = \langle x \rangle \langle xy \rangle$ is that $\langle x \rangle$ and $\langle xy \rangle$ permute.

I'm not sure what you want to say as far as $\langle x,y \rangle$ not being cyclic.

$\endgroup$
1
  • $\begingroup$ Thank you Jack! However my definition of cylcic group is a group that is generated by one element. Hence $\langle x,y\rangle$ couldn't be cyclic by definition. $\endgroup$
    – Joe
    Apr 10, 2014 at 17:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .