-{[a b]₁ b1 a, b, c, d € R, | a, b, c, d € R, and ad — bc and ad-bc=1}. Is = d Let G = = 1}. Is G a group?
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![6. Let G
= {[a b]1 a,b,c,d € R,
| a, b, c, d € R, and ad — bc =
-
d
= 1}. Is
Is G a group?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F76043b90-1f33-46dd-a5c8-708bcf994f14%2Fbf19e0ab-b257-4fa0-b8fc-863d96305276%2F3vpc4lz_processed.jpeg&w=3840&q=75)
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- In Exercises 15 and 16, the given table defines an operation of multiplication on the set S={ e,a,b,c }. In each case, find a condition in Definition 3.1 that fails to hold, and thereby show that S is not a group. See Figure 3.7 e a b c e e a b c a e a b c b e a b c c e a b cIn Exercises and, the given table defines an operation of multiplication on the set. In each case, find a condition in Definition that fails to hold, and thereby show that is not a group. 15. See Figure.If a is an element of order m in a group G and ak=e, prove that m divides k.
- Let n be appositive integer, n1. Prove by induction that the set of transpositions (1,2),(1,3),...,(1,n) generates the entire group Sn.42. For an arbitrary set , the power set was defined in Section by , and addition in was defined by Prove that is a group with respect to this operation of addition. If has distinct elements, state the order of .In Exercises 114, decide whether each of the given sets is a group with respect to the indicated operation. If it is not a group, state a condition in Definition 3.1 that fails to hold. The set of all multiples of a positive integer n is group with operation multiplication.
- Let a,b,c, and d be elements of a group G. Find an expression for (abcd)1 in terms of a1,b1,c1, and d1.Let a and b belong to a group. If |a| = 10 and |b| = 21, show that <a> n <b> = {e} (n means intersects)Create steps along with justifications to verify that in a group system (G, +) the following property holds:For any two elements from G, ‘a’ and ‘b’, -(a + b) = (-b) + (-a). In other words, the claim is that (-b) + (-a) plays the role of the inverse of a + b. Create steps to show that the element (-b) + (-a) does, in fact, play the role of an inverse to the element a + b, i.e., show that:i. (a + b) + ( (-b) + (-a) ) = e, where e represents the identity in the group; andii. ( (-b) + (-a) ) + (a + b) = e.