By Randall R. Holmes

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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 ). 10 Isomorphism Consider the group Z = {. . , −2, −1, 0, 1, 2, . . } under addition. If we create a new group Z = {. . , −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, +).

The center of G, denoted Z(G), is the set of those elements of G that commute with every element of G: Z(G) = {z ∈ G | zx = xz for all x ∈ G}. Prove that Z(G) is a subgroup of G. 5–2 Fix n ∈ N. Let SLn (R) be the set of all n × n matrices over R having determinant 1: SLn (R) = {A ∈ Matn (R) | det(A) = 1}. Prove that SLn (R) is a subgroup of GLn (R) (= invertible n × n matrices over R). ) Hint: From linear algebra, we know that a square matrix is invertible if and only if its determinant is nonzero.

Then m = km and n = kn for some integers m and n . We have km n (a, b) = (km n a, km n b) = (n ma, m nb) = (0, 0). Since (a, b) has order mn and km n > 0 it follows that mn ≤ km n . So mn ≤ km n ≤ km kn = mn. Since the ends of this string are the same, the intermediate ≤ signs must actually be equalities. Thus km n = km kn which implies k = 1 as desired. (⇐) Assume that gcd(m, n) = 1. Let k be the order of the element (1, 1) of Zm ⊕ Zn . Then (k1, k1) = k(1, 1) = (0, 0). 3, m divides k and n divides k.

### Abstract Algebra I by Randall R. Holmes

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