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.

  • $\begingroup$ sorry, mistake and adjusted $\endgroup$ – 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 $x^{\langle g^p\rangle}$.

  • $\begingroup$ Thanks for this argument! $\endgroup$ – Sven Wirsing Apr 16 at 19:04
  • $\begingroup$ @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. $\endgroup$ – Arturo Magidin Apr 16 at 19:10
  • $\begingroup$ Okay, thanks for that solution! $\endgroup$ – Sven Wirsing Apr 16 at 19:14

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.