Test: Group Theory - 9 - Mathematics MCQ

(21) = {gcd(24, 21)} = (3) = {0, 3, 6, 9, 12, 15, 18, 21}
(10) = {gcd(24, 10)} = h2i = {0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22}
(21) ∩ (10) = {0, 6, 12, 18} = (6)
(a²¹) = (a^gcd(24,21)) = (a3) = {e, a³, a⁶, a⁹, a¹², a¹⁵, a¹⁸, a²¹}
(a10) = (a^gcd(24,10)) = (a²) = {e, a², a⁴, a⁶, a⁸, a¹⁰, a¹², a¹⁴, a¹⁶, a¹⁸, a²⁰, a²²}
(a²¹) ∩ (a¹⁰)
= (a³) ∩ (a²)
= (a⁶) = {e, a⁶, a¹², a¹⁸}

THEOREM. (Primitive Root Theorem) U(n) is cyclic if and only if n is 1, 2, 4, or 2p, where p is an odd prime and k ≥ l.

Hence the correct option is n!/2

we have  2⁵ = 32(mod 29)
implies       2⁵ = 3(mod 29)
implies       (25)³ = (3)³ (mod 29)
implies       2¹⁵ = -2(mod 29)
implies       (2¹⁵)² = (-2)²(mod 29) implies       2³⁰ = 4(mod 29)
implies       2²⁸ = 4/4 (mod 29)
implies       2²⁸= 1 (mod 29)
Hence, The order of 2 in the field z₂₉ is 28.

Let A - { 1 , 2 , 3 }
Then, the non-empty even subsets
ⁿC₂ + ⁿC₄ + ... = 2^n-1 - 1
(even set is the set having even number of elements) of a set having three elements r { 1 , 2 } , {2 , 3 }, {1, 3}
Number of subsets = 2^3-1 - 1 = 3 Hence, the number of non-empty even subsets of a set having n elements is 2^n-1 - 1.

Given A_{4 =} is an even permutation}
A₄ has a subgroup of order 6 is false.
Because in this example converse of Lagrange’s theorem fails.
Hence, option (c) is correct.

Let order of <0, 2> + (3,1) is k, then k <0, 2> + (3, 1) + k(3, 1) = <0, 2>

=> Thus, k will be 4.
For this, k 4(3,1) = (12,4) = (0,4). Thus, 4 <0, 2> + (0, 4) ε <(0, 2)>

We know that

Given 
So,
 

Let G be a group of order 8 generated by 'a' and ‘b’ s.t. a⁴ = b² = 1 and ba = a³b. This group is dihedral group. (i.e. D⁴)
Therefore the order of the centre of G is 2.