Abstract Algebra I, Edition: version 14 Apr 2015 by Randall R. Holmes

By Randall R. Holmes

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2, −1, 0, 1, 2, . . } by putting bars over each element of Z and by using a binary operation that acts just like + acts on the unadorned integers (so, for instance, 2 3 = 5), then the group (Z, ) is essentially the same as the original group (Z, +). We say that these two groups are “isomorphic” (the precise definition is given below). One often encounters two groups like this that are essentially the same in the sense that the only difference between them is the symbols used for their elements and the symbols used for their binary operations.

Let σ ∈ Sn . The algorithm illustrated above produces a product σ1 σ2 · · · σm of disjoint cycles. The proof that σ actually equals this product is Exercise 7–6. The proof of the uniqueness statement in the theorem is omitted. In the statement of the theorem the word “product” has a broad meaning including any number of factors, with the case of a product of one factor meaning that element itself. 3 Permutation is product of transpositions Recall that a cycle of length two, like (3, 5), is called a transposition.

In the example given at the first of the section Z is isomorphic to Z. Indeed, an isomorphism ϕ : Z → Z is given by ϕ(x) = x. Note that the homomorphism property is satisfied by the way we defined : ϕ(x + y) = x + y = x y = ϕ(x) ϕ(y). 2 Example Let R be viewed as a group under addition and let + R (positive reals) be viewed as a group under multiplication. Prove that R∼ = R+ . Solution Define ϕ : R → R+ by ϕ(x) = ex . Since ex is defined and positive for every x ∈ R, the function ϕ is well defined.

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