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.

  • $\begingroup$ 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

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...).


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 $$|\langle x\rangle |=|\langle y\rangle |=2,\quad |\langle x,y\rangle | = |G|=6\neq 4=|\langle x\rangle ||\langle y\rangle |.$$

  • $\begingroup$ 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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