Let be a finite group $ G $ that has an element $ a\neq 1 $ for which exists a prime number $ p $ such that $ x^{1+p}=a^{-1}xa, $ for all $ x\in G. $ a) Prove that the order of $ G $ is a power of $ p. $ b) Show that $ H:=\{x\in G|\text{ord} (x)=p\}\le G $ and $ \text{ord}^2(H)>\text{ord}(G). $
Problem
Source: Romania National Olympiad 2014, Grade XII, Problem 4
Tags: number theory, prime numbers, group theory, abstract algebra
Doru2718
26.10.2021 14:01
Beautiful problem.
a) Since $x\to a^{-1}xa$ is an automorphism of $G$, $x\to x^{p+1}$ will be one too.
Claim: $x^p\in Z(G)$ for any $x\in G$
Proof: Click to reveal hidden textLet $x\in G$.Then for any $y\in G$ there is a $z$ s.t. $z^{p+1}=y$. Hence $x^py=x^pz^{p+1}=x^{-1}x^{p+1}y^{p+1}=x^{-1}(xy)^{p+1}=(yx)^{p+1}x^{-1}=y^{p+1}x^{p+1}x^{-1}=yx^p$, so $x^p\in Z(G)$ $\square$.
Notice that $x^{p(p+1)}=a^{-1}x^pa=x^p\Rightarrow x^{p^2}=1$, for any $x\in G$. By Cauchy's theorem we get that $|G|$ is a power of $p$. $\square$
b) We have $x,y\in H\Rightarrow (xy)^p=y^{-1}x^{-1}(xy)^{p+1}=y^{-1}x^{-1}x^{p+1}y^{p+1}=y^{-1}x^{p}y^{p+1}=e$, so $H\leq G$.
Now, notice that $x\in H\iff x^{p+1}=x\iff a^{-1}xa=x\iff x\in C(a)$, therefore $H=C(a)$, so in particular $a^p=e$. Take $M=\{gag^{-1}|\text{ }g\in G\}$ the conjugacy class of $a$. Since $(gag^{-1})^p=ga^pg^{-1}=e\Rightarrow M\subseteq H$, and $e\in H-M$, as $a\neq e$.
Hence $|H|>|M|=[G:C(a)]=\frac{|G|}{|H|}\Rightarrow |H|^2>|G|$. $\square$
KevinYang2.71
11.07.2024 05:39
Let $\pi\in\mathrm{Inn}(G)$ be the inner automorphism given by $x\mapsto a^{-1}xa=x^{1+p}$. a) Notice that $g\in C_G(a)\iff\pi(g)=g\iff g^p=e$ so \[ C_G(a)=\{g\in G\mid\pi(g)=g\}=\{g\in G\mid g^p=e\}=H. \]Note that $a\in H$. For $g\in G$ and $h\in H$, we have \[ \pi(g^{-1}hg)=(g^{-1}hg)^{1+p}=g^{-1}h^{1+p}g=g^{-1}hg \]so $g^{-1}hg\in H$. Thus $H\trianglelefteq G$. For $g\in G$, we have $g^{1+p}=a^{-1}ga\in gH$ so $g^p\in H$. Hence $g^{p^2}=e$ for all $g\in G$, so we are done by Cauchy's theorem. $\square$ b) Since $g^{p^2}=e$ for all $g\in G$, it follows that $\pi^{-1}$ is given by $x\mapsto x^{1-p}$. Since $\pi^{-1}$ is an automorphism, we have \[ y^{1-p}x^{1-p}=(yx)^{1-p}=y(xy)^{-p}x\implies y^{-p}x^{-p}=(xy)^{-p}\implies x^py^p=(xy)^p \]for all $x,y\in G$. Let homomorphism $\varphi:G\mapsto H$ be given by $x\mapsto x^p$. Since $\ker\varphi=H$, it follows that $|G|=|H|\cdot|\mathrm{im}\ \varphi|$ so it suffices to prove that $\mathrm{im}\ \varphi<H$. We claim that $a\not\in\mathrm{im}\ \varphi$, which finishes. Suppose FTSOC there exists $g\in G$ such that $g^p=a$. Then $ga=g^{1+p}=ag$ implies that $g\in H$. It follows that $g^p=e\neq a$, a contradiction. $\square$