# Centralizers in p-groups

Let $$G$$ be a finite p-group and $$g\in G$$ such that $$C_G(g)=C_G(g^p)$$. Is it possible that for all $$x\in G\setminus C_G(g)$$ the identity $$x^{\langle g^p \rangle}=x^{\langle g \rangle}$$ is valid?

E.g., in the dihedral Group $$D_{16}$$ there are elements $$a,b$$ such that $$o(a)=8$$, $$o(b)=2$$, $$a^b=a^{-1}$$ are valid and $$G$$ is generated by $$a,b$$. Here, $$C_G(a)=C_G(a^2)$$ is true but $$b^{\langle a^2 \rangle}$$ is not identical to $$b^{\langle a \rangle}$$. In other words, the set $$b^{\langle a \rangle}$$ (containing 4 elements) decomposes under $$\langle a^2 \rangle$$ by conjugation into two orbits of length two.

• sorry, mistake and adjusted – Sven Wirsing Apr 16 at 18:42

Trivially, $$x^{\langle g^p\rangle} \subseteq x^{\langle g\rangle}$$.

In order to get equality, we must have that $$x^g\in x^{\langle g^p\rangle}$$, and hence that there exists $$k$$ such that $$g^{-1}xg = g^{-kp}xg^{kp}$$. That would require $$x = g^{-kp+1}xg^{kp-1}$$, hence $$x\in C_G(g^{kp-1}) = C_G(g)=C_G(g^p)$$ (since $$kp-1$$ is relatively prime to $$p$$, and so $$g^{kp-1}$$ generates the same cyclic subgroup as $$g$$). But that means that $$x = x^g = x^{g^{kp}}$$.

In other words, if we have equality, then $$x\in C_G(g)$$. The converse implication is of course trivial. So if $$x\notin C_G(g)$$, then we cannot have equality of the two conjugacy orbits. In short, the equality will hold for all $$x\in G\setminus C_G(g)$$ if and only if that set is empty, if and only if $$g\in Z(G)$$.

Alternatively, we can do a counting argument. Let $$i\geq 0$$ be the smallest nonnegative integer such that $$g^{p^i}\in C_G(x)$$. That is, $$\langle g^{p^i}\rangle = C_{\langle g\rangle}(x)$$. Then the number of distinct elements in the conjugacy orbit $$x^{\langle g\rangle}$$ is $$p^i$$. If $$i=0$$, then $$x\in C_{G}(g)$$, so we may assume that $$i\gt 0$$. But in that situation, $$i-1$$ is the smallest nonnegative integer such that $$(g^p)^{p^{i-1}}\in C_G(x)$$, so that $$\langle (g^p)^{p^{i-1}}\rangle = C_{\langle g^p\rangle}(x)$$; hence the conjugacy orbit $$x^{\langle g^p\rangle}$$ must have $$p^{i-1}$$ elements. That is, in this situation, you cannot have equality.

In particular, we also show that if $$x\in G\setminus C_G(g)$$, then $$x^{\langle g\rangle}$$ has exactly $$p$$ times as many elements as $$x^{\langle g^p\rangle}$$.

• Thanks for this argument! – Sven Wirsing Apr 16 at 19:04
• @SvenWirsing: See the edit for an alternative argument using counting, that shows that in fact when $x$ does not commute with $g$, then the orbit under powers of $g$ has exactly $p$ times the number of elements as the orbit under powers of $g^p$, as might be expected. – Arturo Magidin Apr 16 at 19:10
• Okay, thanks for that solution! – Sven Wirsing Apr 16 at 19:14