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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

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

  2. $G^{(n−1)}=\{x^{n−1}|x\in G\}$ is a normal subgroup of $G$.

II. Let $G$ be as in the problem above. Show that

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

  2. $(aba^{−1}b^{−1})^{n(n−1)}=e$ for all $a,b\in G$.

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Welcome to MSE. Please tell us what have you tried. Consider rewriting your question in a polite language. –  user21436 Feb 8 '12 at 15:29
    
Such as asking for help instead of just writing down the question. And, if this is homework, add the homework tag. –  Graphth Feb 8 '12 at 15:30
    
These two exercises are from I.N. Herstein's Topics in Algebra (2.ed, page 54, ex 18 and ex 19). –  Mikko Korhonen Feb 8 '12 at 15:49
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yes the problem is from i.n.herstein ch.2 section on normal subgroups. i copied it as it is out of laziness. –  ajinkya Feb 9 '12 at 4:44

2 Answers 2

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.

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the two parts you have proved above are indeed quite easy. it is the proving normal part where i got stuck. If it is true that the map f:x->$x^{n-1}$ is onto on G, then we are done. please try if you can prove that. –  ajinkya Feb 9 '12 at 4:50
    
@noob The "normal" should be the easy part: try proving that in any group $G$ and for any integer $n$, the set of $n$th powers is closed under conjugation. You don't need the given condition on $G$. –  Ted Feb 9 '12 at 6:38

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.

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