Question

Which of the following is a group?
{\text{A}}{\text{. }}\left\\{ {1,2,4,8} \right\\}{\text{ under multiplication}} \\\ {\text{B}}{\text{. }}\left\\{ {0, \pm 2, \pm 4, \pm 6...} \right\\}{\text{ under addition}} \\\ {\text{C}}{\text{. }}\left\\{ {1, - 1} \right\\}{\text{ under addition}} \\\ {\text{D}}{\text{. }}\left\\{ {0,1,2,3,4} \right\\}{\text{ under multiplication modulo 5}} \\\

Answer 1

Hint: To find out which amongst the given options is a group, we verify each of the options to find out if it holds the given condition or not. Hence, we determine the answer.

Complete step-by-step answer:

A. Let G ≡ {1, 2, 4, 8} under multiplication.
If we multiply elements of the above set, then it is not necessary that it will lie in the same set.
1 × 2 = 2 ∈ G
2 × 4 = 8 ∈ G
4 × 8 = 32 $\notin$ G
Hence, not a group.

B. Let H ≡ {0, ±2, ±4, ±6…}
If we add any elements of this set, then we will get the result in the same set.
0 + 2 = 2 ∈ H
0 – 2 + 4 – 6 = -4 ∈ H and so on,
Hence it is a group

C. Let F ≡ {1, -1} under addition
If we add any elements of this set, then it is not necessary that it will lie in the same set.
1 + -1 = 0 $\notin$F
Hence it is not a group.

D. Let P ≡ {0, 1, 2, 3, 4} under multiplication modulo 5.
Now we check if this group satisfies the conditions under multiplication modulo, which are as follows.

Step -1: Check the operation defined in the question is a binary operation:
Making Cayley table, we get:

0 1 2 3 4  

0 0 0 0 0 0
1 0 1 2 3 4
2 0 2 4 1 3
3 0 3 1 4 2
4 0 4 3 2 1

Calculations above show that the operation defined in the question assigns to each ordered pair of elements of {0, 1, 2, 3, 4} an element in {0, 1, 2, 3, 4}. Hence, it is a binary operation.

Step -2: Check the existence of identity:
If we look at the Cayley table above, we see that 1.a = a.1 = a for all a in {0, 1, 2, 3, 4}. So, 1 is an identity.

Step -3: Check the existence of inverses:
The Cayley table shows that for each element a in {0, 1, 2, 3, 4}, there is an element in b in {0, 1, 2, 3, 4} such that a.b = b.a = e.
Except element 0
1.2= 1.1 = 1
2.3 = 3.2 = 1
3.2 = 2.3 = 1
4.4 = 4.4 = 1
But 0.b = b.0 ≠ 1
Hence, it does not satisfy inverse property.
Therefore it is not a group.

Hence, Option B is the correct answer.

Note: In order to solve questions of this type the key is to verify if the group holds its given condition. Conditions like multiplication, addition are straightforward where we just perform the calculation. Whereas for multiplication under modulo we use a Cayley table and check if the group holds the required conditions. Hence, we determine the answer.