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Suppose I have a finite group with elements $x,y$ such that $\langle x \rangle \cap \langle y \rangle = \{1\}$. Is it true that $|\langle x,y \rangle|=|\langle x \rangle|\cdot|\langle y\rangle|$?

I know that the trivial intersection implies that the product map is injective: that is, $\langle x \rangle \langle y \rangle = \{x^i y^j\}$ has size $|\langle x \rangle|\cdot|\langle y\rangle|$, but I am not sure if I can make the extra step to say that the group generated by $x$ and $y$ also has the same size.

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  • $\begingroup$ Do you mean $|\langle xy \rangle|=|\langle x \rangle|\cdot|\langle y\rangle|$ - i.e. the group generated by the product of $x$ and $y$ or $|\langle x, y \rangle|=|\langle x \rangle|\cdot|\langle y\rangle|$ - i.e. the group generated by $x$ and $y$ (which is what you say later). $\endgroup$ – Christopher Nov 20 '13 at 14:58
  • $\begingroup$ @user73985 Thanks for catching the typo; edited. $\endgroup$ – angryavian Nov 20 '13 at 14:59
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The following well-known equality is related to your question ($H$ and $K$ are arbitrary finite subgroups of a group $G$). $$|HK|=\frac{|H|\cdot |K|}{|H\cap K|}$$ See here for a proof. Note that $HK$ is not necessarily a subgroup, so does not always equal $\langle H, K\rangle$. On the other hand, if $HK$ is a subgroup then your result holds. Note that $HK$ is a subgroup if one of $H$ or $K$ is normal, and proving this is a nice exercise.

If $HK$ is a subgroup and $H\cap K=1$ then $HK$ is called the Zappa-Szep product of $H$ and $K$, written $H\bowtie K$. This is a generalisation of semidirect products (when either $H$ or $K$ is normal) and direct products (when both $H$ and $K$ are normal). See this question for more details. The answer mentions the specific case you are talking about here - the case when $H$ and $K$ are both finite cyclic. Such Zappa-Szep products were classified by Jesse Douglas, was one of two winners of the first Fields Medals (although he didn't win his medal for this...).

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It is false for an arbitrary finite group; in particular, a non-abelian finite group.

Example: Take $G=S_3$, $x=(12)$, and $y=(23)$. Then clearly their subgroups intersect trivially, but $$|\langle x\rangle |=|\langle y\rangle |=2,\quad |\langle x,y\rangle | = |G|=6\neq 4=|\langle x\rangle ||\langle y\rangle |.$$

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  • $\begingroup$ Ah ok, thanks. So, the best we can say for non-abelian groups is that $|\langle x,y\rangle| \ge |\langle x\rangle| \cdot |\langle y \rangle|$? $\endgroup$ – angryavian Nov 20 '13 at 15:07
  • $\begingroup$ It is true if $\langle x\rangle$ and $\langle y\rangle$ are normal subgroups, though, as then the elements in the first group commute with the elements in the second group. $\endgroup$ – Stefan Hamcke Nov 20 '13 at 15:07
  • $\begingroup$ @StefanH Because it is the direct product in that case? $\endgroup$ – angryavian Nov 20 '13 at 15:08
  • $\begingroup$ @blf: Yes, $⟨x⟩\cdot⟨y⟩$ is the direct product then. $\endgroup$ – Stefan Hamcke Nov 20 '13 at 15:10
  • $\begingroup$ @StefanH What if one of them is normal? Consider the dihedral group $D_n$ with $x$ being a rotation and $y$ being a reflection. It still holds here [I think] because of the $(yx)^2=1$ relation. $\endgroup$ – angryavian Nov 20 '13 at 15:19

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