By Thomas C. Craven

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8(2). But then aan−1 = an = e, so a−1 = an−1 ∈ H by closure. Example from page 183: H = { f ∈ S5 | f (1) = 1 } consists of all permutations fixing the element 1. Since H is finite, we only need to check that it is closed under the operation: for g, h ∈ H, g(h(1)) = g(1) = 1, so g ◦ h ∈ H. Therefore H is a subgroup of S5 . In fact, we expect that as soon as we define isomorphisms for groups, H should be isomorphic to S4 which we think of as permutations of the set { 2, 3, 4, 5 }. Even though a group G may be nonabelian, it may have some elements that commute with everything in the group (as the identity does).

Thus the homomorphic images are either isomorphic to Z itself (when n = 0) or to Z/(n) = Zn (when n > 0). 3. We showed that I = { f ∈ R | f (r) = 0 } is an ideal in R = C(R, R) for any fixed r ∈ R. The homomorphism φ : R → R defined by φ(f ) = f (r) is surjective [φ(s) = s for any constant function s ∈ R] and has kernel I. Thus R/I is isomorphic to R. 7 Exercise 3, p. 151. Let F be a field, R a nonzero ring and f : F → R a surjective homomorphism. We claim f is actually an isomorphism. 11) that its kernel is zero.

28. Let A G, B G, AB = G and A ∩ B = e . Then A × B ∼ = G. Proof: We −1 −1 first note that ab = ba for every a ∈ A, b ∈ B. Indeed, a b ab = (a−1 b−1 a)b ∈ B and a−1 b−1 ab = a−1 (b−1 ab) ∈ A by normality, so a−1 b−1 ab = e; therefore ab = ba. Define f : A × B → G by f (a, b) = ab. Check that this is a homomorphism (using the commutativity we have have just established). f is surjective by hypothesis. For injectivity, −1 assume f (a1 , b1 ) = f (a2 , b2 ). Then a1 b1 = a2 b2 , hence a−1 ∈ A ∩ B = {e}.