problems in group theory I. Let $G$ be a group in which, for some integer $n\gt 1$, $(ab)^n=a^n b^n$ for all $a,b\in G$. Show that 


*

*$G^{(n)}=\{x^{n}|x\in G\}$ is a normal subgroup of $G$. 

*$G^{(n−1)}=\{x^{n−1}|x\in G\}$ is a normal subgroup of $G$.
II. Let $G$ be as in the problem above. Show that


*

*$a^{n−1}b^n=b^n a^{n−1}$ for all $a,b\in G$.

*$(aba^{−1}b^{−1})^{n(n−1)}=e$ for all $a,b\in G$.
 A: Firstly, you don't seem to want to be doing us the favour of writing your question in a manner we'd like to see it! 
Secondly, your notation is cumbersome for me to type it over here. So, set $G^{(n)}=G_n$ and similarly for $n-1$.
This is the way to go about it:
1) Note that identity element in $G$ belongs to $G_n$. If $g,h \in G_n$, then, there exists $x, y \in G$ such that $g=x^n$ and $h=y^n$. So, $gh^{-1}=x^ny^{-n}=(xy^{-1})^n$ Since, $gh^{-1}$ can be written in the form of $l^n$ for some $l \in G$,we have that $gh^{-1} \in G_n$. Hence, $G_n$ is a subgroup. 
To prove it is normal, note that $n^{th}$ power map is a homomorphism from $G$ to $G$. 
Here's where you need to think, I don't want to kill the purpose of the home work, or so, you tell yourself, I don't know how to prove it!
For the next exercise, here's what I'll do:
Given, $$\begin{align*}(ab)^n&=a^nb^n\\ab \cdot ab \cdots ab&=aa^{n-1}b^{n-1}b\\(ba)^{n-1}&=a^{n-1}b^{n-1}\\(ba)^n&=a^{n-1}b^{n}a\\b^na^n&=a^{n-1}b^{n}a\\b^na^{n-1}&=a^{n-1}b^n\end{align*}$$
Make sure you justify each step above. I will elaborate if you seem to get lost somewhere.
A: Just got how to solve this one. In fact it is quite easy. As you have shown above, $G_n$ is a group. To show it is normal, all we need is to make use of the following: $gx^{n}g^{-1}=(gxg^{-1})^n$ and similarly for n-1. To show $G_{n-1}$ is a group, we can use the following: $a^{n-1}b^{n-1}=(ba)^{n-1}$. You have already proved the first part of the second question. For the second part we just need this, $(xy)^{n(n-1)}=(yx)^{n(n-1)}$ and some little manipulation.
