By Randall R. Holmes

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**Example text**

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.

We claim that (x−1 1 , x2 ) is an inverse of (x1 , x2 ). We have −1 −1 −1 (x−1 1 , x2 )(x1 , x2 ) = (x1 x1 , x2 x2 ) = (e1 , e2 ) −1 −1 −1 and similarly (x1 , x2 )(x−1 1 , x2 ) = (e1 , e2 ). Therefore, (x1 , x2 ) is an inverse of (x1 , x2 ). This finishes the proof that G1 × G2 is a group under componentwise multiplication. 32 G1 × G2 is the direct product of the groups G1 and G2 . , the binary operations are both +), then the direct product is called the direct sum and it is denoted G1 ⊕ G2 . In this case, the operation is denoted + and it is called componentwise addition: (x1 , x2 ) + (y1 , y2 ) = (x1 + y1 , x2 + y2 ).

Prove that H is a subgroup of G. 2). ) We have e2 = ee = e, so e ∈ H. ) Let x, y ∈ H. We have (xy)2 = xyxy = xxyy (G is abelian) = x2 y 2 (x, y ∈ H) = ee = e. Therefore, xy ∈ H. ) Let x ∈ H. We have x−1 2 = x−2 = x =e (law of exponents) 2 −1 −1 (law of exponents) (x ∈ H) = e. Therefore, x−1 ∈ H. By the Subgroup Theorem, H is a subgroup of G. 3 Cyclic subgroup Let G be a group and let a be a fixed element of G. Put a = {am | m ∈ Z}. 1 Theorem. a is a subgroup of G. Proof. 2). ) We have e = a0 ∈ a .

### Abstract Algebra I by Randall R. Holmes

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