Find a group $G$ with $a\in G$ such that $|a|=6$ but $C_G(a)\neq C_G(a^3)$.

Question

This is part of Exercise 46 of Chapter 3 of Gallian's "Contemporary Abstract Algebra".

Notation 1: The centraliser of $g$ in a group $G$ is denoted $C_G(g)$.

Notation 2: The dihedral group $D_n$ has order $2n$.

I'm interested in how one would answer the following question using the tools available in the textbook prior to it.

Find a group $G$ with $a\in G$ such that $|a|=6$ but $C_G(a)\neq C_G(a^3)$.

NOTE: The textbook doesn't cover permutation groups prior to this.

Thoughts:

• I understand that the group $H$ given by the presentation $$\langle a, x\mid a^6, a^3x=xa^3\rangle$$ would suffice as $x\in C_H(a^3)\setminus C_H(a)$ by construction; however, presentations aren't covered in the preceding material. It's not clear to me how I would exploit my understanding of this perspective to construct/name a group in the spirit of the question.

• Perhaps a dihedral group $D_n$ would do for some $n$ s.t. $6\mid n$ but I'm not sure. These groups are defined in terms of reflections & rotations in the text, but the way I see them is, again, as in terms of a few different presentations for them. From my comment below: The problem is that it's defined in terms of a regular $n$-gon in this textbook. I'm used to just using a presentation. If I'm to use $D_6$ here, then I'm to argue from the perspective of the book; I'm not so sure how to proceed. (Emphasis added.)

Please help :)

Answer 1

Have you already covered the symmetric groups $S_n$ in your course? If you can use them, then an idea could be to employ the fact that non-overlapping cycles commute with each other. For example, if $a$ is a product of two independent cycles of lengths $2$ and $3$, then the latter one will disappear in $a^3$, and thus its centralizer will be different.

Answer 2

Let $D_{12}$ be the dihedral group, presented as follows $$D_{12}=\left\langle x,y\;:\;x^2=y^6=1,\quad xyx=y^{-1}\right\rangle.$$ Consider $a=y$, it is easy to see that its order is $6$. Now, we have that $xyx=y^{-1}\neq y$ but $xy^3x=y^{-3}=y^3$ (note that $x=x^{-1}$). Therefore, $x$ lies on the centrlizer of $y^3$ but not in the centralizer of $y$.

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Not covered in the preceding material but:

Consider $G=S_8$ the symmetric group. Let $\sigma=(1\ 2\ 3\ 4\ 5\ 6)$ be a permutation, it is clear that $|\sigma|=6$, note that $\sigma^3=(1 4)$. We have that $(5\ 6)$ lies in the centralizer of $\sigma^3$ but does not lie on the centralizer of $\sigma$.