The Alternating Group A4 of Even Permutations of {1, 2, 3, 4}
In this table, the permutations of A4 are designated as ε, α2,..., α12. ◦ ε α2 α3 α4 α5 α6 α7 α8 α9 α10 α11 α12
(1) = ε εα 2 α3 α4 α5 α6 α7 α8 α9 α10 α11 α12
(12)(34) = α2 α2 εα 4 α3 α6 α5 α8 α7 α10 α9 α12 α11
(13)(24) = α3 α3 α4 εα 2 α7 α8 α5 α6 α11 α12 α9 α10
(14)(23) = α4 α4 α3 α2 εα 8 α7 α6 α5 α12 α11 α10 α9
(123) = α5 α5 α8 α6 α7 α9 α12 α10 α11 ε α4 α2 α3
(243) = α6 α6 α7 α5 α8 α10 α11 α9 α12 α2 α3 ε α4
(142) = α7 α7 α6 α8 α5 α11 α10 α12 α9 α3 α2 α4 ε
(134) = α8 α8 α5 α7 α6 α12 α9 α11 α10 α4 ε α3 α2
(132) = α9 α9 α11 α12 α10 ε α3 α4 α2 α5 α7 α8 α6
(143) = α10 α10 α12 α11 α9 α2 α4 α3 ε α6 α8 α7 α5
(234) = α11 α11 α9 α10 α12 α3 ε α2 α4 α7 α5 α6 α8
(124) = α12 α12 α10 α9 α11 α4 α2 ε α3 α8 α6 α5 α7
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 1 / 6 1. Let G = Z/20Z and H = 4Z/20Z. List the elements of G/H. (How do you know G/H is a group?)(You may assume that a factor group of an Abelian group is Abelian, which you will be showing for PS 8)
2. Determine the order of (Z ⊕ Z)/ < (4, 2) >. Is the group cyclic?
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 2 / 6 Now we know: G = {0¯, 1¯, 2¯, 3¯, 4¯,..., 19} and H = {0¯, 4¯, 8¯, 12, 16}.
The elements of G/H are: 0¯ + H = H = {0¯, 4¯, 8¯, 12, 16} = 4¯ + H = 8¯ + H = 12 + H = 16 + H 1¯ + H = {1¯, 5¯, 9¯, 13, 17} = 5¯ + H = 9¯ + H = 13 + H = 17 + H 2¯ + H = {2¯, 6¯, 10, 14, 18} = 6¯ + H = 10 + H = 14 + H = 18 + H 3¯ + H = {3¯, 7¯, 11, 15, 19} = 7¯ + H = 11 + H = 15 + H = 19 + H
Note: Since |G/H| = 4, we know that G/H is either cyclic and isomorphic to Z4 or else it is not cyclic, in which case it is isomorphic to Z2 ⊕ Z2.
1. Let G = Z/20Z and H = 4Z/20Z. List the elements of G/H.
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 3 / 6 Now we know: G = {0¯, 1¯, 2¯, 3¯, 4¯,..., 19} and H = {0¯, 4¯, 8¯, 12, 16}.
The elements of G/H are: 0¯ + H = H = {0¯, 4¯, 8¯, 12, 16} = 4¯ + H = 8¯ + H = 12 + H = 16 + H 1¯ + H = {1¯, 5¯, 9¯, 13, 17} = 5¯ + H = 9¯ + H = 13 + H = 17 + H 2¯ + H = {2¯, 6¯, 10, 14, 18} = 6¯ + H = 10 + H = 14 + H = 18 + H 3¯ + H = {3¯, 7¯, 11, 15, 19} = 7¯ + H = 11 + H = 15 + H = 19 + H
Note: Since |G/H| = 4, we know that G/H is either cyclic and isomorphic to Z4 or else it is not cyclic, in which case it is isomorphic to Z2 ⊕ Z2.
1. Let G = Z/20Z and H = 4Z/20Z. List the elements of G/H. Helpful shorthand for cosets (just remember the elements are really sets!): Write a + 20Z as ¯a.
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 3 / 6 Now we know: G = {0¯, 1¯, 2¯, 3¯, 4¯,..., 19} and H = {0¯, 4¯, 8¯, 12, 16}.
The elements of G/H are: 0¯ + H = H = {0¯, 4¯, 8¯, 12, 16} = 4¯ + H = 8¯ + H = 12 + H = 16 + H 1¯ + H = {1¯, 5¯, 9¯, 13, 17} = 5¯ + H = 9¯ + H = 13 + H = 17 + H 2¯ + H = {2¯, 6¯, 10, 14, 18} = 6¯ + H = 10 + H = 14 + H = 18 + H 3¯ + H = {3¯, 7¯, 11, 15, 19} = 7¯ + H = 11 + H = 15 + H = 19 + H
Note: Since |G/H| = 4, we know that G/H is either cyclic and isomorphic to Z4 or else it is not cyclic, in which case it is isomorphic to Z2 ⊕ Z2.
1. Let G = Z/20Z and H = 4Z/20Z. List the elements of G/H. Helpful shorthand for cosets (just remember the elements are really sets!): Write a + 20Z as ¯a. With this notation, the elements of G are the cosets G = {0¯, 1¯, 2¯, 3¯, 4¯,..., 19}.
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 3 / 6 Now we know: G = {0¯, 1¯, 2¯, 3¯, 4¯,..., 19} and H = {0¯, 4¯, 8¯, 12, 16}.
The elements of G/H are: 0¯ + H = H = {0¯, 4¯, 8¯, 12, 16} = 4¯ + H = 8¯ + H = 12 + H = 16 + H 1¯ + H = {1¯, 5¯, 9¯, 13, 17} = 5¯ + H = 9¯ + H = 13 + H = 17 + H 2¯ + H = {2¯, 6¯, 10, 14, 18} = 6¯ + H = 10 + H = 14 + H = 18 + H 3¯ + H = {3¯, 7¯, 11, 15, 19} = 7¯ + H = 11 + H = 15 + H = 19 + H
Note: Since |G/H| = 4, we know that G/H is either cyclic and isomorphic to Z4 or else it is not cyclic, in which case it is isomorphic to Z2 ⊕ Z2.
1. Let G = Z/20Z and H = 4Z/20Z. List the elements of G/H.
G = {0¯, 1¯, 2¯, 3¯, 4¯,..., 19}.
By definition, H = 4Z/20Z = {¯a ∈ Z/20Z|a ∈ 4Z}.
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 3 / 6 Now we know: G = {0¯, 1¯, 2¯, 3¯, 4¯,..., 19} and H = {0¯, 4¯, 8¯, 12, 16}.
The elements of G/H are: 0¯ + H = H = {0¯, 4¯, 8¯, 12, 16} = 4¯ + H = 8¯ + H = 12 + H = 16 + H 1¯ + H = {1¯, 5¯, 9¯, 13, 17} = 5¯ + H = 9¯ + H = 13 + H = 17 + H 2¯ + H = {2¯, 6¯, 10, 14, 18} = 6¯ + H = 10 + H = 14 + H = 18 + H 3¯ + H = {3¯, 7¯, 11, 15, 19} = 7¯ + H = 11 + H = 15 + H = 19 + H
Note: Since |G/H| = 4, we know that G/H is either cyclic and isomorphic to Z4 or else it is not cyclic, in which case it is isomorphic to Z2 ⊕ Z2.
1. Let G = Z/20Z and H = 4Z/20Z. List the elements of G/H.
G = {0¯, 1¯, 2¯, 3¯, 4¯,..., 19}.
By definition, H = 4Z/20Z = {¯a ∈ Z/20Z|a ∈ 4Z}. Hence H = 4Z/20Z = {0¯, 4¯, 8¯, 12, 16}.
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 3 / 6 The elements of G/H are: 0¯ + H = H = {0¯, 4¯, 8¯, 12, 16} = 4¯ + H = 8¯ + H = 12 + H = 16 + H 1¯ + H = {1¯, 5¯, 9¯, 13, 17} = 5¯ + H = 9¯ + H = 13 + H = 17 + H 2¯ + H = {2¯, 6¯, 10, 14, 18} = 6¯ + H = 10 + H = 14 + H = 18 + H 3¯ + H = {3¯, 7¯, 11, 15, 19} = 7¯ + H = 11 + H = 15 + H = 19 + H
Note: Since |G/H| = 4, we know that G/H is either cyclic and isomorphic to Z4 or else it is not cyclic, in which case it is isomorphic to Z2 ⊕ Z2.
1. Let G = Z/20Z and H = 4Z/20Z. List the elements of G/H.
Now we know: G = {0¯, 1¯, 2¯, 3¯, 4¯,..., 19} and H = {0¯, 4¯, 8¯, 12, 16}.
Why is G/H a group?
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 3 / 6 The elements of G/H are: 0¯ + H = H = {0¯, 4¯, 8¯, 12, 16} = 4¯ + H = 8¯ + H = 12 + H = 16 + H 1¯ + H = {1¯, 5¯, 9¯, 13, 17} = 5¯ + H = 9¯ + H = 13 + H = 17 + H 2¯ + H = {2¯, 6¯, 10, 14, 18} = 6¯ + H = 10 + H = 14 + H = 18 + H 3¯ + H = {3¯, 7¯, 11, 15, 19} = 7¯ + H = 11 + H = 15 + H = 19 + H
Note: Since |G/H| = 4, we know that G/H is either cyclic and isomorphic to Z4 or else it is not cyclic, in which case it is isomorphic to Z2 ⊕ Z2.
1. Let G = Z/20Z and H = 4Z/20Z. List the elements of G/H.
Now we know: G = {0¯, 1¯, 2¯, 3¯, 4¯,..., 19} and H = {0¯, 4¯, 8¯, 12, 16}.
Why is G/H a group?
Using that a factor group of an Abelian group is Abelian, every subgroup of G is normal. In particular, H / G, and so G/H is a group.
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 3 / 6 The elements of G/H are: 0¯ + H = H = {0¯, 4¯, 8¯, 12, 16} = 4¯ + H = 8¯ + H = 12 + H = 16 + H 1¯ + H = {1¯, 5¯, 9¯, 13, 17} = 5¯ + H = 9¯ + H = 13 + H = 17 + H 2¯ + H = {2¯, 6¯, 10, 14, 18} = 6¯ + H = 10 + H = 14 + H = 18 + H 3¯ + H = {3¯, 7¯, 11, 15, 19} = 7¯ + H = 11 + H = 15 + H = 19 + H
Note: Since |G/H| = 4, we know that G/H is either cyclic and isomorphic to Z4 or else it is not cyclic, in which case it is isomorphic to Z2 ⊕ Z2.
1. Let G = Z/20Z and H = 4Z/20Z. List the elements of G/H.
Now we know: G = {0¯, 1¯, 2¯, 3¯, 4¯,..., 19} and H = {0¯, 4¯, 8¯, 12, 16}.
List the elements of G/H. Since |G| = 20 and |H| = 5, from Lagrange’s Theorem, we know that |G/H| = 4, so we know how many distinct cosets we’re looking for.
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 3 / 6 Note: Since |G/H| = 4, we know that G/H is either cyclic and isomorphic to Z4 or else it is not cyclic, in which case it is isomorphic to Z2 ⊕ Z2.
1. Let G = Z/20Z and H = 4Z/20Z. List the elements of G/H.
Now we know: G = {0¯, 1¯, 2¯, 3¯, 4¯,..., 19} and H = {0¯, 4¯, 8¯, 12, 16}.
The elements of G/H are: 0¯ + H = H = {0¯, 4¯, 8¯, 12, 16} = 4¯ + H = 8¯ + H = 12 + H = 16 + H 1¯ + H = {1¯, 5¯, 9¯, 13, 17} = 5¯ + H = 9¯ + H = 13 + H = 17 + H 2¯ + H = {2¯, 6¯, 10, 14, 18} = 6¯ + H = 10 + H = 14 + H = 18 + H 3¯ + H = {3¯, 7¯, 11, 15, 19} = 7¯ + H = 11 + H = 15 + H = 19 + H
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 3 / 6 1. Let G = Z/20Z and H = 4Z/20Z. List the elements of G/H.
Now we know: G = {0¯, 1¯, 2¯, 3¯, 4¯,..., 19} and H = {0¯, 4¯, 8¯, 12, 16}.
The elements of G/H are: 0¯ + H = H = {0¯, 4¯, 8¯, 12, 16} = 4¯ + H = 8¯ + H = 12 + H = 16 + H 1¯ + H = {1¯, 5¯, 9¯, 13, 17} = 5¯ + H = 9¯ + H = 13 + H = 17 + H 2¯ + H = {2¯, 6¯, 10, 14, 18} = 6¯ + H = 10 + H = 14 + H = 18 + H 3¯ + H = {3¯, 7¯, 11, 15, 19} = 7¯ + H = 11 + H = 15 + H = 19 + H
Note: Since |G/H| = 4, we know that G/H is either cyclic and isomorphic to 4 or else it is not cyclic, in which case it is isomorphic to 2 ⊕ 2. MathZ 321-Abstracti (Sklensky) In-Class Work NovemberZ 15, 2010Z 3 / 6 In that case, the order would be 8. Is this true?
< (4, 2) > = {(0, 0), (±4, ±2), (±8, ±4), (±12, ±6),...} = {(4n, 2n)|n ∈ Z}
A few elts (all of the form (n, 0)+ < (4, 2) >) of Z ⊕ Z/ < (4, 2) >: (0, 0)+ < (4, 2) >, (1, 0)+ < (4, 2) >, (2, 0)+ < (4, 2) >, (3, 0)+ < (4, 2) >, (4, 0)+ < (4, 2) > . . .. These are all distinct. In general, (n, 0)+ < (4, 2) >6= (m, 0)+ < (4, 2) > as long as n 6= m. (To see this, remember that aH = bH ⇐⇒ a−1b ∈ H.) Thus there are an infinite number of distinct cosets, and therefore |Z ⊕ Z/ < (4, 2) > | = ∞.
2. Determine the order of (Z ⊕ Z)/ < (4, 2) >. Is the group cyclic? We have seen that Z/nZ ≈ Zn. So .... First Guess: (Z ⊕ Z) < (4, 2) >≈ Z/4Z ⊕ Z/2Z
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 4 / 6 < (4, 2) > = {(0, 0), (±4, ±2), (±8, ±4), (±12, ±6),...} = {(4n, 2n)|n ∈ Z}
A few elts (all of the form (n, 0)+ < (4, 2) >) of Z ⊕ Z/ < (4, 2) >: (0, 0)+ < (4, 2) >, (1, 0)+ < (4, 2) >, (2, 0)+ < (4, 2) >, (3, 0)+ < (4, 2) >, (4, 0)+ < (4, 2) > . . .. These are all distinct. In general, (n, 0)+ < (4, 2) >6= (m, 0)+ < (4, 2) > as long as n 6= m. (To see this, remember that aH = bH ⇐⇒ a−1b ∈ H.) Thus there are an infinite number of distinct cosets, and therefore |Z ⊕ Z/ < (4, 2) > | = ∞.
2. Determine the order of (Z ⊕ Z)/ < (4, 2) >. Is the group cyclic? We have seen that Z/nZ ≈ Zn. So .... First Guess: (Z ⊕ Z) < (4, 2) >≈ Z/4Z ⊕ Z/2Z In that case, the order would be 8. Is this true?
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 4 / 6 A few elts (all of the form (n, 0)+ < (4, 2) >) of Z ⊕ Z/ < (4, 2) >: (0, 0)+ < (4, 2) >, (1, 0)+ < (4, 2) >, (2, 0)+ < (4, 2) >, (3, 0)+ < (4, 2) >, (4, 0)+ < (4, 2) > . . .. These are all distinct. In general, (n, 0)+ < (4, 2) >6= (m, 0)+ < (4, 2) > as long as n 6= m. (To see this, remember that aH = bH ⇐⇒ a−1b ∈ H.) Thus there are an infinite number of distinct cosets, and therefore |Z ⊕ Z/ < (4, 2) > | = ∞.
2. Determine the order of (Z ⊕ Z)/ < (4, 2) >. Is the group cyclic? We have seen that Z/nZ ≈ Zn. So .... First Guess: (Z ⊕ Z) < (4, 2) >≈ Z/4Z ⊕ Z/2Z In that case, the order would be 8. Is this true?
< (4, 2) > = {(0, 0), (±4, ±2), (±8, ±4), (±12, ±6),...} = {(4n, 2n)|n ∈ Z}
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 4 / 6 In general, (n, 0)+ < (4, 2) >6= (m, 0)+ < (4, 2) > as long as n 6= m. (To see this, remember that aH = bH ⇐⇒ a−1b ∈ H.) Thus there are an infinite number of distinct cosets, and therefore |Z ⊕ Z/ < (4, 2) > | = ∞.
2. Determine the order of (Z ⊕ Z)/ < (4, 2) >. Is the group cyclic? We have seen that Z/nZ ≈ Zn. So .... First Guess: (Z ⊕ Z) < (4, 2) >≈ Z/4Z ⊕ Z/2Z In that case, the order would be 8. Is this true?
< (4, 2) > = {(0, 0), (±4, ±2), (±8, ±4), (±12, ±6),...} = {(4n, 2n)|n ∈ Z}
A few elts (all of the form (n, 0)+ < (4, 2) >) of Z ⊕ Z/ < (4, 2) >: (0, 0)+ < (4, 2) >, (1, 0)+ < (4, 2) >, (2, 0)+ < (4, 2) >, (3, 0)+ < (4, 2) >, (4, 0)+ < (4, 2) > . . .. These are all distinct.
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 4 / 6 Thus there are an infinite number of distinct cosets, and therefore |Z ⊕ Z/ < (4, 2) > | = ∞.
2. Determine the order of (Z ⊕ Z)/ < (4, 2) >. Is the group cyclic? We have seen that Z/nZ ≈ Zn. So .... First Guess: (Z ⊕ Z) < (4, 2) >≈ Z/4Z ⊕ Z/2Z In that case, the order would be 8. Is this true?
< (4, 2) > = {(0, 0), (±4, ±2), (±8, ±4), (±12, ±6),...} = {(4n, 2n)|n ∈ Z}
A few elts (all of the form (n, 0)+ < (4, 2) >) of Z ⊕ Z/ < (4, 2) >: (0, 0)+ < (4, 2) >, (1, 0)+ < (4, 2) >, (2, 0)+ < (4, 2) >, (3, 0)+ < (4, 2) >, (4, 0)+ < (4, 2) > . . .. These are all distinct. In general, (n, 0)+ < (4, 2) >6= (m, 0)+ < (4, 2) > as long as n 6= m. (To see this, remember that aH = bH ⇐⇒ a−1b ∈ H.)
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 4 / 6 2. Determine the order of (Z ⊕ Z)/ < (4, 2) >. Is the group cyclic? We have seen that Z/nZ ≈ Zn. So .... First Guess: (Z ⊕ Z) < (4, 2) >≈ Z/4Z ⊕ Z/2Z In that case, the order would be 8. Is this true?
< (4, 2) > = {(0, 0), (±4, ±2), (±8, ±4), (±12, ±6),...} = {(4n, 2n)|n ∈ Z}
A few elts (all of the form (n, 0)+ < (4, 2) >) of Z ⊕ Z/ < (4, 2) >: (0, 0)+ < (4, 2) >, (1, 0)+ < (4, 2) >, (2, 0)+ < (4, 2) >, (3, 0)+ < (4, 2) >, (4, 0)+ < (4, 2) > . . .. These are all distinct. In general, (n, 0)+ < (4, 2) >6= (m, 0)+ < (4, 2) > as long as n 6= m. (To see this, remember that aH = bH ⇐⇒ a−1b ∈ H.) Thus there are an infinite number of distinct cosets, and therefore |Z ⊕ Z/ < (4, 2) > | = ∞. Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 4 / 6 Goal: find (a, b) ∈ Z ⊕ Z but not in < (4, 2) > such that
2((a, b)+ < (4, 2) >= (0, 0)+ < (4, 2) > .
Want: (a, b) ∈ Z ⊕ Z but not in < (4, 2) > such that (2a, 2b) ∈< (4, 2) >. (2, 1) seems like it would work. Check:
2((2, 1)+ < (4, 2) >) = (4, 2)+ < (4, 2) >=< 4, 2 > .
2. Determine the order of (Z ⊕ Z)/ < (4, 2) >. Is the group cyclic? Is (Z ⊕ Z)/ < (4, 2) > cyclic? Since the order of the group is infinite, if it is cyclic then every non-identity element must have infinite order.
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 5 / 6 Goal: find (a, b) ∈ Z ⊕ Z but not in < (4, 2) > such that
2((a, b)+ < (4, 2) >= (0, 0)+ < (4, 2) > .
Want: (a, b) ∈ Z ⊕ Z but not in < (4, 2) > such that (2a, 2b) ∈< (4, 2) >. (2, 1) seems like it would work. Check:
2((2, 1)+ < (4, 2) >) = (4, 2)+ < (4, 2) >=< 4, 2 > .
2. Determine the order of (Z ⊕ Z)/ < (4, 2) >. Is the group cyclic? Is (Z ⊕ Z)/ < (4, 2) > cyclic? Since the order of the group is infinite, if it is cyclic then every non-identity element must have infinite order. Goal: find (a, b) ∈ Z ⊕ Z but not in < (4, 2) > such that (a, b)+ < (4, 2) > has finite order.
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 5 / 6 Goal: find (a, b) ∈ Z ⊕ Z but not in < (4, 2) > such that
2((a, b)+ < (4, 2) >= (0, 0)+ < (4, 2) > .
Want: (a, b) ∈ Z ⊕ Z but not in < (4, 2) > such that (2a, 2b) ∈< (4, 2) >. (2, 1) seems like it would work. Check: 2((2, 1)+ < (4, 2) >) = (4, 2)+ < (4, 2) >=< 4, 2 > .
2. Determine the order of (Z ⊕ Z)/ < (4, 2) >. Is the group cyclic? Is (Z ⊕ Z)/ < (4, 2) > cyclic? Since the order of the group is infinite, if it is cyclic then every non-identity element must have infinite order. Goal: find (a, b) ∈ Z ⊕ Z but not in < (4, 2) > such that (a, b)+ < (4, 2) > has finite order. The quickest order to find is order 2, so try to find an elt with 2(a, b)+ < (4, 2) > = (0, 0)+ < (4, 2) >
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 5 / 6 (2, 1) seems like it would work. Check:
2((2, 1)+ < (4, 2) >) = (4, 2)+ < (4, 2) >=< 4, 2 > .
2. Determine the order of (Z ⊕ Z)/ < (4, 2) >. Is the group cyclic? Is (Z ⊕ Z)/ < (4, 2) > cyclic? Since the order of the group is infinite, if it is cyclic then every non-identity element must have infinite order. Goal: find (a, b) ∈ Z ⊕ Z but not in < (4, 2) > such that
2((a, b)+ < (4, 2) >= (0, 0)+ < (4, 2) > .
Want: (a, b) ∈ Z ⊕ Z but not in < (4, 2) > such that (2a, 2b) ∈< (4, 2) >.
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 5 / 6 2. Determine the order of (Z ⊕ Z)/ < (4, 2) >. Is the group cyclic? Is (Z ⊕ Z)/ < (4, 2) > cyclic? Since the order of the group is infinite, if it is cyclic then every non-identity element must have infinite order. Goal: find (a, b) ∈ Z ⊕ Z but not in < (4, 2) > such that
2((a, b)+ < (4, 2) >= (0, 0)+ < (4, 2) > .
Want: (a, b) ∈ Z ⊕ Z but not in < (4, 2) > such that (2a, 2b) ∈< (4, 2) >. (2, 1) seems like it would work. Check:
2((2, 1)+ < (4, 2) >) = (4, 2)+ < (4, 2) >=< 4, 2 > .
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 5 / 6 2. Determine the order of (Z ⊕ Z)/ < (4, 2) >. Is the group cyclic? Is (Z ⊕ Z)/ < (4, 2) > cyclic? Since the order of the group is infinite, if it were cyclic, every non-identity element would have infinite order.
(2, 1)+ < (4, 2) >6= (0, 0)+ < (4, 2) > and (2, 1)+ < (4, 2) >”2” = (0, 0)+ < (4, 2) >,
so we have found an element of order 2 in (Z ⊕ Z)/ < (4, 2) >. Thus this group is not cyclic.
Math 321-Abstracti (Sklensky) In-Class Work November 15, 2010 6 / 6