Topics of Algebra Herstein Chapter 2.7 problem 18,19

Question

In Herstein chapter 2.7 problem 17
Let G be a group in which,for some integer n>1 $(ab)^n=a^nb^n$ for all a,b$\in G$.Show that
a) $G^n$={$x^n|x\in G$} is normal subgroup
b) $G^{n-1}$={$x^{n-1}|x\in G$} is normal subgroup
I can solve this problem as$x^n,y^n \in G^n$ and from given condition $x^ny^n=(xy)^n$ $(x^n)^{-1}=(x^{-1})^n \in G^n$ Therefore 2 step test$G^n$ is subgroup $tx^nt^{-1}$= $(txt^{-1})^n$ for every $t\in G$ This implies $txt^{-1} \in G^n$
Similarly for b)
Is there any institution behind defining this kind of normal subgroup?
I thought using this type of normal subgroup by fixing any integer as power and taking that power of all element we can obtain normal subgroup in G .
Problem 18: From above show that
a) $a^{n-1}b^n=b^na^{n-1}$ for all a,b$\in G$
b) $(aba^{-1}b^{-1})^{n(n-1)}=e$ for all $a,b\in G$.
I had tried but not able to solve.Any help will be appreciated.
Thanks a Lot

Answer 1

I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
These problems are Problem 18 and Problem 19 on p.54 in Herstein's book.
I solved Problem 19 as follows:

(a) $a^{n-1}b^n=a^{-1}(a^nb^n)=a^{-1}(ab)^n=(ba)^{n-1}b=(ba)^na^{-1}=(b^na^n)a^{-1}=b^na^{n-1}$.

(b) Note that $(ab)^{n-1}=b^{n-1}a^{n-1}$ and $(a^kba^{-k})^n=a^kb^na^{-k}$ hold for all $a,b\in G$.

$(aba^{-1}b^{-1})^{n(n-1)}=[(aba^{-1})^n(b^{-1})^n]^{n-1}=[(ab^na^{-1})(b^{-n})]^{n-1}=[a(b^na^{-1}b^{-n})]^{n-1}=(b^na^{-1}b^{-n})^{n-1}a^{n-1}=[b^n(a^{-1})^{n-1}b^{-n}]a^{n-1}=[(a^{-1})^{n-1}b^nb^{-n}]a^{n-1}=e.$