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Math Help - 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 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.
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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: (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.
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  5. #5
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    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.
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