Consider the set Z₁0 under multiplication. [0] does not have a multiplicative inverse. So, Z10 is not a group under multiplication. Is Z₁0 - {[0]} a group under multiplication? (Hint: Think about the inverse axiom for a group.)

Consider the set of permutations V = {(1), (1 2) (3 4), (1 3) (2 4), (1 4) (2 3)}. Determine whether V forms a group under the operation of composition. If it is a group, is the group cyclic? If it is a group, is the group abelian?

Write out the Cayley table for the group {1, -1, i, -i} under multiplication where i = sqrt(-1). (In general, we apply the row element on the left of the column element.)

Abstract Algebra

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.

Q1c Kindly answer correctly. Please show the necessary steps.

is it a binary option? is it associative? identity element= inverse elements=