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    Abstract Algebra: Identity Element of the Group

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    1. Prove that is a is a number in G, a group, and ab = b for some b of G, then a = e, the identity element of the group.

    2. Consider the set of polynomials with real coefficients. Define two elements of this set to be related if their derivatives are equal. Prove that this defines an equivalence relation.

    3. Let H be a subgroup of the group G. Prove that every right coset of H is a left coset of some subgroup of G.

    © BrainMass Inc. brainmass.com December 24, 2021, 10:55 pm ad1c9bdddf
    https://brainmass.com/math/basic-algebra/abstract-algebra-identity-element-group-519362

    SOLUTION This solution is FREE courtesy of BrainMass!

    1. The element b has a unique inverse, b, such that bb=bb=e . Hence
    ab=b
    (ab)b^?1=bb^?1
    a(bb^?1)=bb^?1 (associativity of group multiplication)
    ae=e (definition of inverse)
    a=e (identity property of e )

    2. Consider the set of polynomials with real coefficients. Define two elements of this set to be related if their derivatives are equal. Prove that this defines an equivalence relation. There's not much to do here. The solution boils down to the fact that = is reflexive, symmetric, and transitive. To be more precise let p , q , and r be polynomials. We write p?q if p'=q' , where p' denotes the derivative of p . We want to show that p?p , p?q ==> q?p , and p?q,q?r ==> p?r . We have
    (a) p?p since p'=p' .
    (b) p?q ==> q?p since p'=q' ==> q'=p' .
    (c) p?q,q?r ==> p?r since p'=q',q'=r' ==> p'=r' .
    This proves the assertion.

    3. Let H be a subgroup of the group G. Prove that every right coset of H is a left coset of some subgroup of G.
    Let Hg denote a right right coset of H . Then we can write Hg=gK where K=g?1Hg . To prove that K is a subgroup, we must show that it is closed under
    (i) multiplication and (ii) inversion.
    (i) Suppose k,k' of K . Then, by definition of K , there exist h,h' of H so that
    k=(g^?1)hg, k'=(g^?1)h'g . It follows that
    kk'=((g^?1)hg)((g^?1)h'g)=g?1h(g(g^?1))h'g=g?1heh' g=g?1(hh')g of K .
    (ii) Similarly, k=g?1hg implies that
    k^?1=((g^?1)hg)^?1=(g^?1)(h^?1)(g^?1)^?1=(g^?1)h^?1g of K .
    Thus K is a subgroup. Note that in (ii), we used the basic facts that (ab)^?1=(b^?1)(a^?1), applied to a triple product, and (a^?1)^?1=a .

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 10:55 pm ad1c9bdddf>
    https://brainmass.com/math/basic-algebra/abstract-algebra-identity-element-group-519362

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