By Mark Steinberger
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Extra resources for Algebra [Lecture notes]
Proof Since (g, h) = (g, e) · (e, h), f (g, h) = f ◦ ι1 (g) · f ◦ ι2 (h). Thus, f is determined by the f ◦ ιi , which are determined by the restriction of f to the stated subgroups. The commutativity assertion follows since if two elements commute, so must their images under any homomorphism. Given the gi as above, deﬁne f : G × H → K by f (g, h) = g1 (g) · g2 (h). Then f (g, h) · f (g , h ) = g1 (g)g2 (h)g1 (g )g2 (h ) = g1 (g)g1 (g )g2 (h)g2 (h ) = g1 (gg )g2 (hh ), with the key step being given by the fact that g1 (g ) commutes with g2 (h).
17. Show that every submonoid of a ﬁnite group is a group. 3 The Subgroups of the Integers One of the simplest yet most powerful results in mathematics is the Euclidean Algorithm. We shall use it here to identify all subgroups of Z and to derive the properties of prime decomposition in Z. 1. (The Euclidean Algorithm2 ) Let m and n be integers, with n > 0. Then there are integers q and r, with 0 ≤ r < n, such that m = qn + r. Proof First, we assume that m ≥ 0, and argue by induction on m. If m < n, we may take q = 0 and r = m.
Thus, (h, k) = (e, e), and hence μ is injective. Since H and K generate G, any element in G may be written as a product g1 . . gn where gi is an element of either H or K for i = 1, . . , n. Since elements of H commute with those of K, we can rewrite this as a product h1 . . hr k1 . . ks with hi ∈ H for i = 1, . . , r and ki ∈ K for i = 1, . . , s, by moving the elements in H past those in K. But this is just μ(h1 . . hr , k1 . . ks ), so μ is onto. 7) of G. Now let us generalize the material in this section to the study of products of more than two groups.
Algebra [Lecture notes] by Mark Steinberger