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MA 541 Modern Algebra ISolutions to HW #1

1. Which of the following are groups under the given operation? Justify youranswers.

(a) The collection of odd integers under +.

This is not a group, since it is not closed. Consider any elementm. Thenm +m= 2m, which is even.

(b) The collection of even integers under +.

This is a group.

Consider any pair of elements m, n 2Z. Then m = 2x andn = 2y for some pair of integers x and y. Therefore, m+n =(2x) + ( 2y) = 2(x + y), which is even, so the operation is binary.

Consider elements a, b, and c in 2Z. Then, by associativity ofaddition in the integers, (a + b) + c= a + (b + c), so the operationis associative.

Consider any element a 2Z. Then 0 +a= a+ 0 =a, so thereis an identity, namely 0.

Consider any element a 2Z. Thena has an inverse, namelya, sincea+(a) = 0, which we showed above to be the identityelement.

(c) Z7 under multiplication.

This is not a group, since the element 0 does not have an inverse.For anya (Z7, ), 0a= 0= 1.

(d) Z7 {0} under multiplication (that is, Z7 with zero removed).

This is a group.

Consider elementsa,b, andc in Z7 {0}. Then, by associativityof multiplication in the integers, (a b) c = a (b c), so theoperation is associative.

Consider any element a Z7 {0}. Then 1a = a 1 = a, sothere is an identity, namely 1.

Consider any element a Z7 {0}. Thena has an inverse. Thejustification follows: 11 = 1 1 (mod 7); 24 = 8 1 (mod7); 35 = 15 1 (mod 7); 42 = 8 1 (mod 7); 53 = 15 1(mod 7); 66 = 36 1 (mod 7).

(e) The collection of positive real numbers under multiplication.

This is a group. For any pair of elements a, b R+, we havea b > 0, sincea >0andb >0. So the operation is binary.

Consider elements a, b, and c in R+. By the associativity ofmultiplication in the reals, (a b) c= a (b c), so the operationis associative.

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Consider any element a R+. Then 1a= a1 =a, so there isan identity, namely 1.

Consider any element a R+. Since a > 0, we have 1/a > 0.Sincea (1/a) = 1 each element ofR+ has an inverse in R+.

(f) The collection of positive real numbers under division.

This is not a group, since the operation is not associative. Forexample, (1/3)/2 = 1/6= 1/(3/2) = 2/3.

2. Forb Zm, we say thatb is a unit if there exists an elementc Zmsuchthatb c1 (mod m). For example, 2 is a unit in Z5 since 2 3 1 (mod5).

(a) For each of the following m determine all of the units in Zm.

m= 3; 4; 5; 6; 7; 8; 9

m = 3. The units are 1 and 2.


m = 5. The units are 1, 2, 3, and 4.


m = 7. The units are 1, 2, 3, 4, 5, and 6.

m = 8. The units are 1, 3, 5, and 7.

m = 9. The units are 1, 2, 4, 5, 7, and 8.

(b) Many observations to make!

(c) In any of the above examples, do the non-zero elements ofZm forma group?

(Z3 {0}, ), (Z5 {0}, ), and (Z7 {0}, ) are all groups.

3. (a) For eacha Z7, compute a, a + a, a + a + a, etc. until the patternis clear. Any observations?

a= 0. We have 0, 0, 0, and the pattern repeats.

a= 1. We have 1, 2, 3, 4, 5, 6, 0, and the pattern repeats.






(b) For eacha Z8, compute a, a + a, a + a + a, etc. until the patternis clear. Any observations?



a= 2. We have 2, 4, 6, 0, and the pattern repeats.

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a= 3. We have 3, 6, 1, 4, 7, 2, 5, 0, and the pattern repeats.





(c) For eacha Z7, compute a, aa, aaa, etc. until the pattern isclear. Any observations?




a= 3. We have 3, 2, 6, 4, 5, 1, and the pattern repeats.


a= 5. We have 5, 4, 6, 2, 3, 1, and the pattern repeats. a= 6. We have 6, 1, 6, 1, and the pattern repeats.

(d) For eacha Z8, compute a, aa, aaa, etc. until the pattern isclear. Any observations?



a= 2. We have 2, 4, 0, 0, 0, 0, and the pattern continues.


a= 4. We have 4, 0, 0, 0, 0 and the pattern continues.


a= 6. We have 6, 4, 0, 0, 0, 0, and the pattern continues.

a= 7. We have 7, 1, 7, 1, and the pattern repeats.(e) Any overall observations?

Many observations!

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