please try to show a detailed solution , i will be memorizing and possibly repli
ID: 3076364 • Letter: P
Question
please try to show a detailed solution , i will be memorizing and possibly replicating the steps, Only do the part asked in the title there is a separate question for the other parts. Answers will be rated according to the amount of detailed provided thank you!
If H and G are groups, define a binary operation on the product H x G by (hi,g1)(h2,g2) J h1h2,9192), where h1h2 and gig2 refer to the group operations on H and G, respectively. (a) Show that this binary operation defines a group structure on H x G. (b) Find all subgroups of Z2 x Z2. (c) Find all subgroups of Z2 x Z3.Explanation / Answer
(i) Let a = (h1,g1) and b = (h2,g2) be arbitrary elements of H x G. Then ab = (h1,g1)(h2,g2) = (h1h2,g1g2) is an element of H x G because both H and G are closed under their respective operations, so h1h2 is an element of H and g1g2 is an element of G. So this binary operation is closed in H x G.
(ii) Let a = (h1,g1), b = (h2,g2), c = (h3,g3) be arbitrary elements of H x G. Then (ab)c = ((h1,g1)(h2,g2))(h3,g3) = (h1h2,g1g2)(h3,g3) = ((h1h2)h3,(g1g2)g3). By the associativity of the binary operations in groups H and G, (h1h2)h3 = h1(h2h3) and (g1g2)g3 = g1(g2g3), so (ab)c = ((h1h2)h3,(g1g2)g3) = (h1(h2h3),g1(g2g3)) = (h1,g1)(h2h3,g2g3) = (h1,g1)((h2,g2)(h3,g3)) = a(bc), so the binary operation is associative in H x G.
(iii) H and G are groups, so they each have their own identities. Let p and q be the identity elements of H and G respectively, so that for any h in H and g in G, ph = hp = p and qg = gq = g. Then the identity element exists in H x G and is e = (p,q). To show this, let a = (h,g) be an arbitrary element of H x G. Then ea = (p,q)(h,g) = (ph,qg) = (h,g) = a, and ae = (h,g)(p,q) = (hp,gq) = (h,g) = a, so that ea = ae = a for all elements a of H x G.
(iv) H x G also inherits the inverse elements of H and G. If we let a = (h,g) be an arbitrary element in H x G, then a-1 = (h-1,g-1) because aa-1 = (h,g)(h-1,g-1) = (hh-1,gg-1) = (p,q) = e, and a-1a = (h-1,g-1)(h,g) = (h-1h,g-1g) = (p,q) = e. Thus there exists some element a-1 in H x G for every element a in H x G such that aa-1 = a-1a = e; note that a-1 = (h-1,g-1) is in H x G because h-1 is in H and g-1 is in G by the existence of inverse elements in H and G.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.