301 Abstract Algebra 2:00 - 2:50 MWF Hentzel 432 Carver hentzel@iastate.edu Web Page http://www.math.iastate.edu/hentzel/class.301 Text: A First Course in Abstract Algebra, sixth Edition by John B. Fraleigh Course: Chapter 1,2,3,4,5 The material we are covering today is on Page 128 to 135 Section 2.4 beginning with "Direct Products and Finitely Generated Abelian Groups." Monday, September 17, 2001 ********************************** * * * TEST MONDAY SEPTEMBER 24 * * * ********************************** The test will cover pages 51 through 135 and 197 through 208. Main Idea: Everything is known about the structure of commutative groups. Key Words: Direct Product, Cross Product. sigma(p^e), prime power Objective: Learn how figure out the structure of Finite commutative groups. We first introduce the construction of the direct product of two or more groups. These groups do not have to be commutative. If G and H are groups, then G x H = { (g,h) | g in G and h in H } under coordinatewise multiplication is a group. Proof of Associativity, Identity, Inverses. (a) [(g1,h1)*(g2,h2)]*(g3,h3) = ( (g1g2)g3,(h1h2)h3) || || (g1,h1)*[(g2,h2)*(g3,h3)] = ( g1(g2g3),h1(h2h3)) (b) The identity is (e , e ) since G H (e ,e )(g,h) = (e g, e h) = (g,h) and G H G H (g,h)(e ,e ) = (g e , h e ) = (g e , h e ) G H G H G H -1 (c) (g,h) = (g',h') since (g,h)(g',h') = (gg',hh') = (e , e ) G H (g',h')(g,h) = (g'g,h'h) = (e , e ) G H Z2 X Z3 is a group with 6 elements. if x = (1,1) then x = (1,1) x^2 = (0,2) x^3 = (1,0) x^4 = (0,1) x^5 = (1,2) x^6 = (0,0) Notice that Z2 x Z3 is cyclic. Whenever r and s are relatively prime, Zr x Zs is cyclic and is generated by (1,1). Notice that rs (1,1) = (rs,rs) = (0,0) so the order of (1,1) is not n more than rs. Also, if (0,0) = (1,1) = (n,n), then r divides n and s divides n. When r and s are relatively prime, then rs divides n so n >= rs. Thus if we have a number which factors into relative prime factors, then Zrs = Zr x Zs and essentially they are just the same group, the cyclic group of order rs. In thinking about the group, writing Zrs is somewhat preferred because it is more compact. But in the decomposition, we usually write Zr x Zs because the pieces are smaller. Example: List all the commutative groups possible of order 10. Answer: Z2 x Z5. We can break the pieces up using the relative prime result until the factors are all the same prime. Example: Z100 = Z4 x Z25. Example: List all the commutative groups possible of order 16. Z2xZ2xZ2xZ2 Z2xZ2xZ4 Z2xZ8 Z4xZ4 Z16 Only one of these is cyclic. Which one? Example: List all the commutative groups of order 12. Z4 x Z3 Z2 x Z2 x Z3 Only one of these is cyclic. Which one? In Zn, the numbers relatively prime to n form a group under multiplication. What is the structure of this group. 1 2 3 4 5 6 7 8 9 10 11 12 x x x x There are four elements. It is either Z2 x Z2 or Z4. Which is it. 5^1 = 5 5^2 = 25 = 1 7^1 = 7 7^2 = 49 = 1 11^1 = 11 11^1 = 121 = 1 The group is Z2 x Z2 It is also the direct sum of two of its own subgroups. This is called the "internal direct sum". {1,5} x {1,7} The elements of the group are 1*1 = 1 1*7 = 7 5*1 = 5 5*7 = 11 Theorem: If G is a finite commutative group, then G is the direct sum of cyclic groups of prime power order. Question: What is K4. Assignment: (1) Find the order of the element (1,1) in Z9 x Z12 (2) How many subgroups of Z4 x Z25 are there? (3) Decompose the multiplicative group of the numbers relatively prime to 21 into an internal direct sum of cyclic groups. (4) List the possible commutative groups of order 36. Previous Problems: 1(a). How many 6 bead necklaces are there using exactly 2 red, 2 blue, and 2 green beads? 1 I 90 2 (******) 0 2 (***)(***) 0 1 (**)(**)(**) 6 3 (*)(*)(**)(**) 18 3 (**)(**)(**) 18 ------- 132/12 = 11 <== ans (b). How many 8 bead necklaces are there using exactly 3 red and 5 blue beads? 1 I 1 56 = 56 4 (********) 4 0 2 (****)(****) 2 0 1 (**)(**)(**)(**) 1 0 4 (**)(**)(**)(**) 4 0 4 (*)(*)(**)(**)(**) 4 6 = 24 -- ------ 16 80/16 = 5 <====ans (c). How many 8 bead necklaces are there using exactly 4 red and 4 blue beads? 1 I 1 70 = 70 4 (********) 4 0 2 (****)(****) 2 2 = 4 1 (**)(**)(**)(**) 1 6 = 6 4 (**)(**)(**)(**) 4 6 = 24 4 (*)(*)(**)(**)(**) 4 6 = 24 -- ------ 16 128/16 = 8 2. Suppose that G is a finite group. Let H be a subgroup of G. -1 For fixed element g of G, prove that gHg is also a subgroup of G. Proof: We have to show closure and inverses. If g h1 g' and g h2 g' are in gHg', then (g h1 g')(g h2 g') = g h1 h2 g' which is in gHg' since h1 h2 is in H. (g h g')' = g'' h' g' = g h' g' since g'' = g. Thus the inverse of g h g' is also in gHg' so gHg' is a subgroup. 3. Let S be the collection of all subgroups of G. Show that the action * of G acting of S where if g is in G and H is in S, -1 then g*H = g H g satisfies the requirements of a group acting -1 -1 -1 on a set. Hint: Show (ab) = b a . Proof: There are two things we have to show. (1) e*s = s (2) (g1 g2)*s = g1*(g2*s)). e*H = e' H e = e H e = H. (g1 g2)*H = (g1 g2)' H (g1 g2) = g2' g1' H g1 g2 = g2'(g1' H g1) g2 = g2*(g1*h). Actually, this does not work correctly. I should have said -1 g*H = g H g Now (g1 g2)*H = (g1 g2) H (g1 g2)' = (g1 g2) H (g2' g1') = g1(g2 H g2')g1' = g1*(g2*H). 4. How many elements are there in the group of symmetries of: (a) the equilateral triangle, ans 6 (b) the square, ans 8 (c) the pentagon, ans 10 (d) the octahedron, ans 24 (e) the dodecahedron. ans 60