Answer to Question #271038 in Abstract Algebra for Asmitha

Question

Let G be a group such that (ab)^p = a^p b^p for all a,b belongs to G, where p is a prime number. Let S= {x belongs to G /x^p^m = e for some m depending on x} . Prove S is a normal subgroup of G

Answer 1

Let s be any element in S and g another element in G

Since s is in S we have $x^{p m}=e$ , where p is prime and for x in X

$g s g^{-1}=g x g^{-1}$

To prove that $\mathrm{gxg}^{-1}$ is in S, it is enough to prove that $\left(\mathrm{gxg}^{-1}\right) ^{\mathrm{pm}}=\mathrm{e}$

$\begin{aligned} &\left(g x g^{-1}\right)^{p m}=g^{p m} x^{p m} g^{-p m} \\ &=g^{p m} e g^{-p m} \\ &=e \end{aligned}$

Hence S is normal subgroup