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Math Help - A few Abstract Algebra questions

  1. #1
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    A few Abstract Algebra questions

    I have a few problems I can't figure out, I'm sure they're not too complex, I just can't grasp how I'm supposed to tackle these.... (* means binary operation in these)

    1. Let (G, *) be a group and let g be some fixed element of G. Show that G = {g*x such that x is in G}

    2. Let G be a nonempty set and let * be an associative binary operation on G. Assume that both the left and right cancellation laws hold in (G, *). Assume moreover that G is finite. Show that (G, *) is a group. (Obviously, * being associative is a part of what makes it a group. However, I don't know how to show how an identity exists and an inverse. I don't get how the group being finite has anything to do with this.)

    3. Let G be a group and let x in G be an element of order 18. Find the orders of x^2, x^3, x^4, x^5, x^12.

    4. Show that if G is a finite group, then every element of G is of finite order.

    5. Give an example of an infinite group G such that every element of G has finite order.
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  2. #2
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    Anyone? For the first one, I talked to my professor, and said I need to prove each was a subset of the other. So I think I can get that one. But if someone can help me just one or a couple of the questions, that would be a big help? I'm pretty confused.
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  3. #3
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    Quote Originally Posted by Janu42 View Post
    2. Let G be a nonempty set and let * be an associative binary operation on G. Assume that both the left and right cancellation laws hold in (G, *). Assume moreover that G is finite. Show that (G, *) is a group. (Obviously, * being associative is a part of what makes it a group. However, I don't know how to show how an identity exists and an inverse. I don't get how the group being finite has anything to do with this.)
    Let G = \{x_1,...,x_n\} and consider the function \phi_g: G\to G defined by \phi_g(x) = gx. This function is one-to-one, because if \phi_g(x) = \phi_g(y) \implies gx = gy \implies x=y by the left-cancellation law. But since G is finite any injective map to itself must automatically be a surjective map (pigeonhole principle). Therefore, for any b\in G there exists x\in G so that \phi_g(x) = b i.e. gx = b. We have established that the equation gx = b is always sovable (for x) for any g,b\in G. Similarly, xg=b is always solvable. Pick any g\in G. We have established that there exists e\in G so that ge=g. We will argue now that e is a right identity element. Let b\in G so there is x such that xg=b. Thus, be = (xg)e = x(ge) = xg = b. Great, we have found a left identity element. Now find a right identity element, show they are the same and fill in all the steps.
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    Quote Originally Posted by Janu42 View Post
    3. Let G be a group and let x in G be an element of order 18. Find the orders of x^2, x^3, x^4, x^5, x^12.
    If x has order 18 then x^k has order \tfrac{18}{(k,18)}.

    4. Show that if G is a finite group, then every element of G is of finite order.
    Let G = \{x_1,x_2,...,x_n\}.
    If x = e then proof complete.
    If x\not = e and x^2=e then proof complete.
    If x,x^2\not = e and x^3 = e then proof complete.
    And keep on going.

    Say that x,x^2,x^3,...,x^k are all non-identity elements. Then all of these must be distinct. Because otherwise x^i = x^j \implies x^{i-j} = e (assuming i>j) and this is a contradiction. The point here is if x is a non-identity element all exponents of x that are not idenitity elements (i.e. x,x^2,...,x^k) must be distinct. However, G is finite and so we eventually reach a point when the number of exponents will exceede the number of elements in the group. In this case eventually the exponent produces the identity element.

    5. Give an example of an infinite group G such that every element of G has finite order.
    Let G= \{(x_1,x_2,x_3,... ) : x_i = 0 \text{ or } 1\}.
    Define (x_1,x_2,...) + (y_1,y_2,...) = (z_1,z_2,...) where z_i = x_1 + x_2 (\bmod 2).
    Check that G is a group.
    Now prove that all non-trivial elements have order 2.
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  5. #5
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    Thank you so much. That was a big help and I really appreciate it.
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