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Thread: So difficult: groups.

  1. #1
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    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).

    But this assignment is just to hard for me. Can someone please help me?
    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
    "
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  2. #2
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    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 $\displaystyle A$, prove that the following are equivalent.
    (1) For all $\displaystyle x,y\in A$, $\displaystyle xyx^{-1}y^{-1}=i$ (group unit)
    (2) For all $\displaystyle x,y\in A$, $\displaystyle xyxy=xxyy$
    (3) $\displaystyle A$ 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 $\displaystyle a,b,c$ are elements of some group $\displaystyle G$, then the following three facts are equivalent:
    (a) $\displaystyle a=b$
    (b) $\displaystyle ac=bc$
    (c) $\displaystyle ca=cb$
    For example, if $\displaystyle a=b$, then multiplying both sides on the right we have $\displaystyle ac=bc$. Conversely, if $\displaystyle ac=bc$, then multiplying both sides on the right by $\displaystyle c^{-1}$ we get $\displaystyle a=b$. It is very important to multiply either both sides on the left or both sides on the right.

    So, for your problem, suppose that $\displaystyle xyx^{-1}y^{-1}=i$. Multiply this equation by $\displaystyle yx$ on the right. Note that to deduce that $\displaystyle A$ is commutative, you must know that $\displaystyle xyx^{-1}y^{-1}=i$ for all $\displaystyle x$ and $\displaystyle y$.
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  3. #3
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    Quote Originally Posted by MaryB View Post
    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).

    But this assignment is just to hard for me. Can someone please help me?
    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"
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  4. #4
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    Hi there, I saw there was an edit on the site yesterday; b) must be: $\displaystyle (xy)^{2} = x^{2} y^{2}$. And a) must be: $\displaystyle xyx^{'}y{'}=i$. 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:
    $\displaystyle xyx^{'}y{'}=i$. 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.
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  5. #5
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    I assume that you denote the inverse of some $\displaystyle x\in A$ by $\displaystyle x'$. Usually the notation for the inverse of $\displaystyle x$ is $\displaystyle x^{-1}$. This is to make it look like an inverse of a number, e.g., $\displaystyle 5^{-1}=1/5$; $\displaystyle 5\cdot 5^{-1}=1$.
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