Give and example of a non-abelian group $(G,.)$ where $a^2b=ba^2\Rightarrow ab=ba$ for all $a,b\in G$. Can somebody give me some tips, please? Moreover how did you think to get there.

I've found that $C(a^2) \subset C(a)=C(a^{|G|+1}) $

Edit: The answer sheet gives the solution the group of matrices of the form $$\begin{pmatrix} \hat 1 & a & b \\ \hat 0& \hat 1 & c \\ \hat 0 & \hat 0 & \hat 1 \end{pmatrix}\qquad\text{ with }\ a,b,c \in \Bbb{Z}/3\Bbb{Z}.$$ Then $A^3=I_3$ for all such matrices. I wanted to know if there are some easier groups to find. It's pretty hard to find matrices.

  • 3
    $\begingroup$ Where did you find this problem? $\endgroup$ – Aqua Mar 4 '19 at 21:15
  • $\begingroup$ Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers $\endgroup$ – Brian Mar 4 '19 at 21:16
  • 5
    $\begingroup$ A link to the contest page, please. We have a rule against questions from on-going contests, so we need to ascertain that the deadline is already gone. $\endgroup$ – Jyrki Lahtonen Mar 4 '19 at 21:49
  • 5
    $\begingroup$ A link to the page, please. We have Romanian users. And we may be able to spot a deadline date (google translate). $\endgroup$ – Jyrki Lahtonen Mar 4 '19 at 21:54
  • 1
    $\begingroup$ I disagree that $C(a^2)\subseteq C(a)$, assuming that “$C(g)$” stands for the centralizer of $g$, at least in general. For example, in the dihedral group of order $8$, if $a$ is the element of order $2$ that is not central, $a^2=e$ so that $C(a^2)=G$, but $C(a)$ is not all of $G$. Of course, what you are asked to find is a group in which the inclusion holds; and since you always have $C(a)\subseteq C(a^2)$, what you are actually looking for is a group in which $C(a)=C(a^2)$ for all $a\in G$. $\endgroup$ – Arturo Magidin Mar 4 '19 at 22:28

My thought process:

  • The relation $a^2b=ba^2$ can be read as stating that $a^2$ is in the centralizer of $b$, or that $b$ is in the centralizer of $a^2$. Can't tell which is more useful, yet.
  • The relation $ab=ba$ similarly states that $a$ is in the centralizer of $b$, or that $b$ is in the centralizer of $a$.
  • Centralizers of $b$ are involved in both, so the implication can be conveniently rephrased: $$\text{for all $a,b\in G$ we have:}\ a^2\in C_G(b)\implies a\in C_G(b).$$

How to make that implication true in a non-abelian group? Remember that $C_G(b)$ is a subgroup. If it contains the element $a^2$ it will contain all the powers $(a^2)^k=a^{2k}$, $k\in\Bbb{Z}$. Can we make sure that $a$ is among those powers? Yes, we can! Simply insist that for all $a$ we have $a^{2k-1}=1$ for some integer $k$.

Any non-abelian group $G$ of odd order will work. This is because, by Lagrange, every element then has an odd order as well.

See here for an explicit construction of the smallest non-abelian group of odd order.

| cite | improve this answer | |
  • $\begingroup$ $\begin{pmatrix} \hat 1 & a & b \\ \hat 0& \hat 1 & c \\ \hat 0 & \hat 0 & \hat 1 \end{pmatrix} \ ,a,b,c \in \mathbb Z_3. A^3=I_3 $ . This is on the anwer sheet, but I wanted to know if there are some easier groups to find.It's pretty hard to find matrices. $\endgroup$ – Andrei Gabor Mar 5 '19 at 9:57
  • $\begingroup$ That group of order 27 may be the simplest to describe. There is an example of a group of order 21 that can be constructed either as a semidirect product or as a subgroup of $S_7$, or as a group of 2x2 matrices over $\Bbb{Z}_7$. These are all among the simplest to describe groups. What are you looking for? $\endgroup$ – Jyrki Lahtonen Mar 5 '19 at 10:11
  • $\begingroup$ Mind you, Servaes' answer tells how to arrive at that group. You may consider accepting that instead of mine. Your call, of course. You really need to have built up some kind of a library of groups so that you can draw these groups from the shelf in your brain :-) $\endgroup$ – Jyrki Lahtonen Mar 5 '19 at 10:14

One example of such a group is the subgroup of $\operatorname{GL}_3(\Bbb{F}_3)$ of matrices of the form $$\begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix}.$$ My thought process; if $a^2=a^{-1}$ for all $a\in G$ then the implication is immediate. So I'd like a group in which the order of every element divides $3$. Then the order of this group is $3^k$ for some $k$, because I'd like the group to be finite. Now I know that if the order of a group is either $p$ or $p^2$ for a prime $p$, then it is abelian. So I'd like a group of order $p^3$. This is the first one that came to mind.

It isn't hard to check that this group is non-abelian, and if you have a little patience it isn't even that hard to check explicitly that the relation $a^2b=ba^2\implies ab=ba$ holds.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.