The smallest order for a group to have a non-abelian proper sub-group is (Answer should be integer) _________.Correct answer is '12'. Can you explain this answer?

Question

Answer

Let the smallest order for a group to have a non abelian proper subgroup isn.

Clearly n≠1,2,3,4,5 because if o(G)<5 then G is abelian.

If o(G)= 6 then every proper subgroup ofG is cyclic. So, o(G) ≠ 6.

If o(G) = 7, 11 then G is cyclic. So. o(G) ≠ 7.11.

If o(G)= 8, 9, 10, then every proper subgroup of G is cyclic.

If o(G) = 12 then let G=S₃*ℤ₂

⇒ G has a proper subgroup which is non abelian.

Answer

Explanation:

Definition:
A group is said to be abelian if the group operation is commutative, i.e., for all elements a, b in the group, a * b = b * a.

Smallest Order for a Non-abelian Proper Sub-group:

Claim:
The smallest order for a group to have a non-abelian proper sub-group is 12.

Proof:
- Let's consider the smallest non-abelian group, which is the symmetric group of degree 3, denoted by S3. The order of S3 is 6.
- The sub-groups of S3 are {e}, {1, (12), (13), (23)}, and S3 itself.
- All these sub-groups are abelian, and there are no non-abelian proper sub-groups in S3.
- Now, let's consider the dihedral group D6, which is the symmetry group of a regular hexagon. The order of D6 is 12.
- The sub-groups of D6 are {e}, {r, r^5}, {r^2, r^4}, {s, sr^3}, {sr, sr^5}, {1, r^3, s, sr^3}, and D6 itself.
- The sub-group {s, sr^3} is a non-abelian proper sub-group of D6.
- Therefore, the smallest order for a group to have a non-abelian proper sub-group is 12.
Therefore, the correct answer is 12.

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