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Prove that T(G) is subgroup of G

  1. Nov 4, 2015 #1
    1. The problem statement, all variables and given/known data
    lin2.png

    2. Relevant equations
    subgroup axioms:

    1. a, b in T(G), then ab in T(G)
    2. existence of identity element.
    3. a in T(G), then a^-1 in T(G)

    3. The attempt at a solution

    1.

    let a be in T(G), then a^n = e.
    let b be in T(G), then b^n = e

    (ab)^n = (a^n)(b^n) = (e)(e) = e

    axiom 1 holds.

    2.

    let e be in T(G), then e^n = e

    conclusion: identity element, e, of G, must be 1.

    3.

    let a be in T(G), then a^n = e.

    show that (a^-1)^n = e

    (a^-1)^n = (a^n)^-1 = 1 / a^n = 1 / e

    therefore

    (a^-1)^n = 1 / e

    we know from axiom 2 that e = 1, so:

    1 / e = e.

    Done.

    Is my proof correct? there's not markscheme, can't check answer,
     
  2. jcsd
  3. Nov 4, 2015 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    There is an error in showing that T is closed under the operation. If x is in T then there exist some number, n, such that x^n= e. If y is in T then there exist some integer m such that y^m= e. You CANNOT assume that m and n are the same. So look at (xy)^{mn}= (x^n)^m(x^m)^n.

    To show that there is an identity element you start by saying "Let e be in T(G) then e^n= e and then say "conclusion: identity element of G, e, is 1". That is meaningless!
    First, there is no "1" mentioned in the problem. Further, you simply assert that e is in T(G), not that it is an identity element. Nothing you say there has any connection to there being an identity element in T(G).

    Instead start by saying "Let e be the identity of G (NOT T(G)). Then, because e^n= e (for all n, not just one), e is in T(G). And since T(G) is a subgroup of G it is also true that ex= xe= x for any x in G. Therefore e, the identity for G, is also the identity for T(G).

    Whether you can simply assert that (a^-1)^n= (a^n)^-1 depends upon previous theorems you have proved. Of course, it is true that, since G is abelian, (a^n)(a^-1)^n= e (do you see why "abelian is important?) so that (a^-1)^n= (a^n)^-1.
     
  4. Nov 4, 2015 #3

    1.

    a in T, there exists n such that a^n = e
    b in T, there exists m such that b^m = e

    (ab)^mn = (a^n)^m (b^m)^n = e^m e^n = e

    axiom 1 holds.

    2.

    let e be identiy in G
    e^n = e ==> e in T(G)

    T(G) subgroup of G, it's also true that ea = ae =a for all a in G

    conclusion: idenity of G, e, is also identity for T(G).

    Axiom 2 holds.

    3.

    this is where im confused:

    Prove that : a in T(G) ====> a^-1 in T(G)

    proof:

    (a^-1)^n = e
    (a^n) (a^-1)^n = (a^n) e
    (a)^n (a^-1)^n = (a^n)
    (a a^-1)^n = a^n
    e^n = a^n

    and a^n = e, so:

    e^n = e

    LHS = RHS

    axiom 3 holds.

    Am i going along the right lines?
     
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