When two elements of a group generate the same subgroup

Question

Let $G$ be a group and $x,y$ two elements from $G$. Suppose that $\langle xy \rangle = \langle yx \rangle$ , i.e. they generate the same subgroup of $G$. I want to find a counterexample in which $xy \neq yx$ and $xyyx \neq 1$. Maybe in a permutation group such a counterexample exists, but I haven't found one yet.

A counterexample for $xy \neq yx$ can be easily found in dihedral group $D_6$, for example $x=\rho^2\sigma$, $y=\rho\sigma$, where $\sigma^2=\rho^6=1$, but, in this case, $xyyx=1$.

Thank you in advance!

Answer 1

Say $xy = z$. Then asking $\langle z \rangle = \langle xy \rangle = \langle yx \rangle$ is the same thing as asking $yx = z^k$, where $k$ had better be coprime to the order of $z$ (do you see why?).

Then you are asking for a group with elements $x$ and $z$ so that

• $x^{-1} z x = z^k$
• $z^n = 1$ for $n$ coprime to $k$.

Given such elements, we can set $y = x^{-1}z$, as then $xy = z$ is obvious, and $yx = x^{-1}zx = z^k$.

If $k \neq 1$ then we'll know $xy = z \neq z^k = yx$.

If $k \neq n-1$, then we'll know $xy = z \neq (z^k)^{-1} = yx$, and so $xyyx \neq 1$.

This leads us to a group $G$ with presentation

$$\langle x, z ~|~ x^{-1}zx = z^k, z^n = 1 \rangle$$

Ideally we'd get a finite group at the end of the day, so let's assert $x$ has finite order too.

$$\langle x, z ~|~ x^{-1}zx = z^k, z^n = 1, x^m = 1 \rangle$$

This group looks like two cyclic groups $\langle x ~|~ x^m = 1 \rangle$ and $\langle z ~|~ z^n = 1 \rangle$ that have been "glued together" in such a way that $x^{-1}zx = z^k$.... Luckily there's a construction called the semidirect product which exists precisely for gluing groups together in this way!

Then our group is $C_m \ltimes_k C_n$, for (almost) any $m$ and $n$ we like. For technical reasons we need $m \mid n-1$ if we want to get a real group out the other side. Here the action is given by $x^{-1}zx = z^k$ for any $k \neq n-1$ (this requirement ensures $(xy)(yx) = zz^k \neq 1$).

By $C_m \cong \mathbb{Z}/m$, I mean the cyclic group of order $m$ where we write our elements as $x^i$ (or $z^i$ for $C_n$) instead of integers.

This may not be a super satisfactory answer (since it's not the kind of group that one meets in an introductory group theory class), but hopefully this shows the kind of strategies which one might use to build these examples for yourself. As a (fun?) exercise, you should chase through the definition of a semidirect product and make sure you understand the multiplication table for some small choices of $m$, $n$, and $k$.

For instance, what are the group elements of $C_3 \ltimes_3 C_7$? It's a group of size $21$, so it's reasonable to get your hands on it. What is the group multiplication? Good luck!

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I hope this helps ^_^

Answer 2

Suppose $3<n<p$ and $\Bbb F_p^+$ is the additive group of the finite field $\Bbb F_p:=\frac{\Bbb Z}{p\Bbb Z}=\{\mathbf 0,\mathbf 1\,\dots,\mathbf{p-1}\}$ of size $p$. Let $$x=(u,\mathbf 1)\;\;\;\;\;y=(v,\mathbf 1)$$ in the product group $D_{n}\times\Bbb F_p^+$ where $u,v\in D_n$ are reflections and $\langle uv\rangle$ is the entire $D_n$ subgroup $R_n$ of rotations. Now, $$\langle yx\rangle=\langle xy\rangle=\langle (uv,\mathbf 2)\rangle =R_n\times\Bbb F_p^+$$ Since $\text{ord}(\mathbf 2)=p$ is odd we must have $$(uv,\mathbf 2)=xy\neq (yx)^{-1}=(uv,\mathbf 2^{-1})$$ and finally $xy\neq yx$ because $uv\neq vu$.

Answer 3

Finally, I have come up with my own (theoretical) example, based on Oliver's concrete examples. First, some preparations:

Let $G,H$ be groups, $a\in G,b\in H$ of finite order. If their orders are coprime, then: $\langle(a,b)\rangle=\langle a\rangle\times\langle b\rangle$. In general we only have $\subseteq$. For the second inclusion, pick $n,m\in\mathbb{Z}\Rightarrow (a^n,b^m)\in\langle a\rangle\times\langle b\rangle$. Let $p:=\text{ord}_G(a),q:=\text{ord}_H(b)$. We can suppose WLOG that $n\in\overline{0,p-1},m\in\overline{0,q-1}$. We want to find $r\in\mathbb{Z}$, such that $(a^n,b^m)=(a,b)^r=(a^r,b^r)$. This means: $r\equiv_p m\land r\equiv_q n$. But this is solved via Chinese Remainder Theorem, because $(p,q)=1$.

Another (simpler) way to prove the other inclusion is to realize that $|\langle(a,b)\rangle|=LCM(p,q)=pq=|\langle a\rangle\times\langle b\rangle|$.

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Now, suppose $x,y\in G$ such that $\langle xy\rangle=\langle yx\rangle$ and $xy\neq yx$ (just like in my attempted example with $D_6$). Pick $a\in H$ such that $a^2\neq a^{-2}\iff a^4\neq 1_H$, with order of $a^2$ coprime with the order of $xy$ (or $yx$, they are the same, but required to be finite). Then $(x,a),(y,a)\in G\times H$ meet the required conditions, i.e.: $$\langle(x,a)(y,a)\rangle=\langle(xy,a^2)\rangle=\langle xy\rangle\times\langle a^2\rangle=\langle yx\rangle\times\langle a^2\rangle=\langle(yx,a^2)\rangle=\langle(y,a)(x,a)\rangle;$$

$$(x,a)(y,a)=(xy,a^2)\neq(yx,a^2)=(y,a)(x,a);$$

$$(x,a)(y,a)=(xy,a^2)\neq((yx)^{-1},a^{-2})=(yx,a^2)^{-1}=((y,a)(x,a))^{-1}.$$