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if $G$ is a group , $x,y \in G$ and $[x,y]$ is the commutator of $x$ and $y$ so , $[x,y]=x^{-1}y^{-1}xy$

is there a formula to compute $|[x,y]|$ in terms of $|x| , |y|$ ?

FNH
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    no, consider $\mathbb{Z}/2\ast\mathbb{Z}/2$ and $\mathbb{Z}/2\times\mathbb{Z}/2$ for example – yoyo Jun 14 '13 at 18:10
  • $(xy)^n = x^ny^n[y,x]^{n(n-1)/2} $ does not hold in general. In a 2-step nilpotent group you'll have better control on powers versus commutators, but certainly no exact formula. – Jack Schmidt Jun 14 '13 at 18:19
  • @JackSchmidt , yes , it's only true when $x,y$ commute with $[x,y]$ , so i deleted the formula from the question . – FNH Jun 14 '13 at 18:21
  • This is related: http://math.stackexchange.com/questions/41303/examples-and-further-results-about-the-order-of-the-product-of-two-elements-in-a – Jack Schmidt Jun 14 '13 at 18:46

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No. If $|x|=|y|=2$ then $|[x,y]|$ can have any order $1,2,\ldots,\infty$.

Take $G$ dihedral of order $4n$ (with $4\infty = \infty$). Then the commutator of two generating reflections has order $n$.

Take $G = \langle a,b : a^2 = b^2 = (ab)^{2n} = 1 \rangle$ Then $[a,b] = a^{-1} b^{-1} a b = a b a b = (ab)^2$ should have order $n$ if the presentation is not “lying” about the order of $ab$.

However, the group of symmetries of a regular $2n$-gon satisfies this presentation with $a$ and $b$ “adjacent” reflections, and their product $ab$ a “primitive” rotation. Alternatively, you can start with the reflection $a$ and a primitive rotation $r$. Then define $b=ar$. By a fundamental geometric principal, $b$ is also a reflection, and $ab=r$ has order $2n$. At any rate, $[a,b] = [a,ar]$ is a “double” rotation, so only has order $n$.

If $n=\infty$ then $a=(x\mapsto -x)$ is a reflection (over the line $x=0$ if you imagine this in the Euclidean plane) and $b=(x\mapsto 1-x)$ is a reflection (over the line $x=\tfrac12$) and $[a,b]= x \mapsto -x \mapsto 1-(-x) \mapsto -(1-(-x)) \mapsto 1 - ( -(1-(-x))) = x+2$ is a translation by 2, so it has infinite order.

Jack Schmidt
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  • may you clarify your example on dihedral group more plz ? – FNH Jun 14 '13 at 18:35
  • (+1) Is this true that for every group $G$ we have $$[a,b]^r=(ab)^r(a^{-1}b^{-1})^r u$$ wherein $u$ is a product of $r-1, r\ge 1$ commutators? – Mikasa Jun 14 '13 at 18:56
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    @BabakS. I don't think that is quite true. certainly for some product of commutators, but I thought the growth was faster (remember you have to move a not only across b, but also across commutators). In “regular” p-groups it is probably true (at least similar statements are true there). – Jack Schmidt Jun 14 '13 at 19:00
  • @BabakS. if you think you have a proof, I'd like to see it. I'm just saying I don't remember that being true, and if I had to guess, I would guess it was not true. – Jack Schmidt Jun 14 '13 at 19:01
  • @JackSchmidt: Honestly, I was thinking to this problem and meanwhile I had a search in my old notations. Suddenly your neat answer came up and it it's done. But I found a claim as you saw above. That has a proof. I wanted just to note that Jack. Thanks for your time. – Mikasa Jun 14 '13 at 19:05
  • @JackSchmidt , great clarification , but what is $a=(x \rightarrow -x)$ ? also for $b$ and $[a,b]$ ? – FNH Jun 14 '13 at 19:07
  • @BabakS. , do you mean , $[a,b]^r=(ab)^r(a^{-1}b^{-1})^{r(r-1)/2}$ ? because your formula doesn't clarify what is u ? – FNH Jun 14 '13 at 19:09
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    There is a similar formula to the one given above in Leedham-Green and MacKay's book on $p$-groups...but I do not have this book anymore... – user1729 Jun 14 '13 at 19:09
  • @MathsLover: $x\mapsto -x$ is a function from the integers to the integers (or the Euclidean line to the Euclidean line; or the Euclidean plane to the Euclidean plane, but it leaves $y$ alone). This specific function is a reflection (over the line $x=0$ if you think of this as Euclidean plane). The function for $b$ is another reflection (over the parallel line $x=1/2$). $[a,b]$ is a translation by $2$. It just adds 2 to the x-coordinate. – Jack Schmidt Jun 14 '13 at 19:10
  • @BabakS. moved to http://math.stackexchange.com/questions/420503/power-of-commutator-formula – Jack Schmidt Jun 14 '13 at 19:14
  • @user1729 moved to http://math.stackexchange.com/questions/420503/power-of-commutator-formula – Jack Schmidt Jun 14 '13 at 19:16
  • I believe the formula we are after is: $(xy)^m=x^my^m[y, x]^{m/2}$. This holds in a $2$-step nilpotent group. See p141 of Robinson, A Course in the Theory of Groups. – user1729 Jun 14 '13 at 19:16
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    @JackSchmidt , thanx so much :) your example is Great , you motivated me to learn more about the symmetry of the plane and the relation with groups , i think that your arguments in this post can be summarized as follows , despite , $|x| = |y|=2$ but $|[x,y]|$ is $n$ in Dihedral group of order $4n$ so the order of the commutator deponds on the order"hence the structure" of the group which $x,y$ lies on not only the orders of the elements of the commutator , so such formula can't exist . – FNH Jun 14 '13 at 19:42
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    @MathsLover: exactly – Jack Schmidt Jun 14 '13 at 20:20