1. ## So difficult: groups.

Hi there,
3 Weeks ago, I started learning about Groups. I have so much troubles with it, but I try and try, and most of it I can work out on my own, if I work hard enough (which is really really hard).

Exact copy of website: "
6.3 Show that the following are equivalent. (A,~) is a normal group. Let x,y in set A.
a) xyx'y = i
b) (xy)^2 = a^2 b^2
c) A is commutative
"

2. Show that the following are equivalent. (A,~) is a normal group. Let x,y in set A.
a) xyx'y = i
b) (xy)^2 = a^2 b^2
c) A is commutative
This is a very strangely formulated problem. E.g., it goes from x, y to some unrelated a, b...

First: what is a "normal group"? Is "normal" used in the usual sense or is it a technical term?

The only way I can interpret the problem is the following. Please correct me if I am wrong.

Given a group $A$, prove that the following are equivalent.
(1) For all $x,y\in A$, $xyx^{-1}y^{-1}=i$ (group unit)
(2) For all $x,y\in A$, $xyxy=xxyy$
(3) $A$ is commutative.

If this is the case, then this claim is proved by multiplying both sides of an equation on the right or on the left. In general, if $a,b,c$ are elements of some group $G$, then the following three facts are equivalent:
(a) $a=b$
(b) $ac=bc$
(c) $ca=cb$
For example, if $a=b$, then multiplying both sides on the right we have $ac=bc$. Conversely, if $ac=bc$, then multiplying both sides on the right by $c^{-1}$ we get $a=b$. It is very important to multiply either both sides on the left or both sides on the right.

So, for your problem, suppose that $xyx^{-1}y^{-1}=i$. Multiply this equation by $yx$ on the right. Note that to deduce that $A$ is commutative, you must know that $xyx^{-1}y^{-1}=i$ for all $x$ and $y$.

3. Originally Posted by MaryB
Hi there,
3 Weeks ago, I started learning about Groups. I have so much troubles with it, but I try and try, and most of it I can work out on my own, if I work hard enough (which is really really hard).

Exact copy of website: "
6.3 Show that the following are equivalent. (A,~) is a normal group. Let x,y in set A.
a) xyx'y = i
b) (xy)^2 = a^2 b^2
c) A is commutative
"
I suspect you have misread the problem. There is no such thing as a "normal group". There are "normal subgroups" of a group. Perhaps you just meant "a group". And what does "(A,~)" mean? Normally we represent a group by something like (A, *) with A a set and "*" the operation but you don't use "~" in your operations. Perhaps it is intended to indicate "nothing" because you are using just juxtaposition to mean the operation. xyx'y= i makes no sense because you haven't said what "x' " is. Did you mean xyx-1y-1= i with i being the identity? Saying that "(xy)^2= a^2b^2" makes no sense because you haven't said what a and b are. Did you mean (xy)^2= x^2y^2?
And, finally, it doesn't make sense to say that "A is commutative" because A is a set, not a group. Perhaps you meant "(A, ~) is commutative"

4. Hi there, I saw there was an edit on the site yesterday; b) must be: $(xy)^{2} = x^{2} y^{2}$. And a) must be: $xyx^{'}y{'}=i$. By "normal" he means just a group, so it's not commutative.

@emakarov; I don't understand what you did with a), because it must be:
$xyx^{'}y{'}=i$. Why does ' mean -1 ?
@HoI: I didn't misread the problem... I copied it directly from the website.
I did say what x is: "Let x,y in set A." so x is an element in A.
Maybe it's easier if I say A is abelian? But that's the same as A is commutative.

5. I assume that you denote the inverse of some $x\in A$ by $x'$. Usually the notation for the inverse of $x$ is $x^{-1}$. This is to make it look like an inverse of a number, e.g., $5^{-1}=1/5$; $5\cdot 5^{-1}=1$.