Abstract Algebra I - download pdf or read online

By Randall R. Holmes

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Let x ∈ 3Z. We have x = 3m for some m ∈ Z. So −x = −(3m) = 3(−m) ∈ 3Z. By the Subgroup Theorem, 3Z is a subgroup of Z. More generally, nZ = {nm | m ∈ Z} is a subgroup of Z for every integer n (same proof with n replacing 3). 2 Example Define Let X be a set and let x0 be a fixed element of X. H = {σ ∈ SX | σ(x0 ) = x0 }. Prove that H is a subgroup of SX . 2). The binary operation of SX is composition ◦ (but we have decided to write σ ◦ τ simply as στ ) and the role of the identity e is played by the identity function 1X .

The reader can check that this is indeed the case here. Therefore, ϕ is an isomorphism. Because of the existence of the isomorphism ϕ from Z4 to U4 , it follows that these two groups are essentially the same. One can think of ϕ as a renaming function: it gives to 0 the new name 1, to 1 the new name i, to 2 the new name −1, and to 3 the new name −i. Moreover, because of the homomorphism property, this renaming is compatible with the binary 34 operations of the two groups. In short, the only difference between the two groups is the names we give their elements and the symbol we use for their binary operations (+ for Z4 and · for U4 ).

The binary operation of SX is composition ◦ (but we have decided to write σ ◦ τ simply as στ ) and the role of the identity e is played by the identity function 1X . ) We have 1X (x0 ) = x0 , so 1X ∈ H. ) Let σ, τ ∈ H. We have (στ )(x0 ) = σ(τ (x0 )) = σ(x0 ) (τ ∈ H) = x0 (σ ∈ H). Therefore, στ ∈ H. ) Let σ ∈ H. We have σ −1 (x0 ) = σ −1 σ(x0 ) (σ ∈ H) = x0 (definition of inverse function). Therefore, σ −1 ∈ H. By the Subgroup Theorem, H is a subgroup of SX . 3 Example Let G be an abelian group and let H = {x ∈ G | x2 = e}.

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Abstract Algebra I by Randall R. Holmes


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