# Index of cyclic subgroup in its normal closure

Let $G$ be finite group, $x, y \in G$. Let $m=|\langle x \rangle ^G: \langle x \rangle|$, $n=|\langle y \rangle^G: \langle y \rangle|$. In general, we can not get that $$(*) \ \ \ \ mn=|\langle xy \rangle^G:\langle xy \rangle|.$$

So we need additional condition to make $(*)$ hold.

If we require that $\langle x \rangle^G \cap \langle y \rangle^G=1$, does $(*)$ hold? If not, which conditons can make it hold?

We also interested with the case that $G$ is a infinite group.

Any reference about the normal closure of the product of two element will be welcomed.

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Do you denote with $\langle x \rangle^G$ the smallest normal subgroup of $G$ containing $x$? I ask, since I'm used to the notation $\langle x^G\rangle$. If yes, then $\langle x^G\rangle\cap\langle y^G\rangle = 1$ implies (*), as you can work in the direct product of $\langle x^G\rangle$ and $\langle y^G\rangle$. –  j.p. Jun 6 '12 at 6:18
@jb: This is not true. Let's say that $<x^G>$ and $<y^G>$ are both $\mathbb Z_2 \times \mathbb Z_2$. Then $m=2$, $n=2$. But $<xy^G>$ could be the whole of $<x^G><y^G>$, so the index of $<xy>$ would be $8$. This is if the group acts by conjugation on $x$ and $y$ totally separately, so the conjugacy class has $9$ elements. It could also act on them identically, for a conjugacy class of $3$ elements and an index of $2$. –  Will Sawin Jun 6 '12 at 6:30
@jb: Yes, $\langle x \rangle ^G$ means the smallest normal subgroup of $G$ containing $x$. @Will: Thank you for the counterexample. –  Wei Zhou Jun 6 '12 at 11:35
@Wei: If the orders of $x$ and $y$ are coprime and the normal closures intersect trivially, (*) holds, since $xy$ has order $|\langle x \rangle|\cdot|\langle y \rangle|$ due to both conditions. –  j.p. Jun 6 '12 at 12:06
@Wei: $X := \langle x^G\rangle$ and $Y := \langle y^G\rangle$ are normal subgroups of $G$ with $X\cap Y = 1$, hence the (normal) subgroup $Z:=XY$ of $G$ is the direct product of $X$ and $Y$. And here my proof breaks down ...(Sorry!). A counterexample to my claim you don't understand is $(Z_2^2\times Z_2^2)\rtimes Z_3$ where $Z_3$ fixes both components $Z_2^2$ and acts nontrivially on both components, and $x$ rsp. $y$ are nontrivial elements of the first rsp. second component. –  j.p. Jun 6 '12 at 20:16