Choose the correct statements;a)Every non-abelian group has a non-trivial abelian sub-groupb)Every non-trivial abelian group has a cyclic sub-groupc)The smallest order for a group to have a non-abelian proper sub-group is 12d)There exist a group containing elements a and b such that o(a) = 0(b) = 2 and o(ab) = 3Correct answer is option 'A,B,C,D'. Can you explain this answer?

Question

Answer

Explanation of Group Theory Statements
The statements provided are related to fundamental concepts in group theory. Let's analyze each statement individually:
a) Every non-abelian group has a non-trivial abelian sub-group
- This statement is true. Every non-abelian group contains a non-trivial abelian subgroup due to the existence of normal subgroups. For instance, in any group, the center is always a non-trivial abelian subgroup if the group is non-abelian.
b) Every non-trivial abelian group has a cyclic sub-group
- This statement is also true. By definition, any abelian group must contain at least one element. The subgroup generated by any non-identity element is cyclic, thus every non-trivial abelian group has a non-trivial cyclic subgroup.
c) The smallest order for a group to have a non-abelian proper sub-group is 12
- This statement is true. The smallest non-abelian group is the symmetric group S3, which has an order of 6. However, S3 does not have a proper non-abelian subgroup. The smallest group that meets this criterion is the alternating group A4, which has an order of 12 and has non-abelian proper subgroups.
d) There exist a group containing elements a and b such that o(a) = o(b) = 2 and o(ab) = 3
- This statement is true. An example of such a group is the symmetric group S3, where both transpositions can be of order 2, and their product can yield an element of order 3. Thus, these elements exist in a group structure.
Conclusion
All statements A, B, C, and D are correct based on the principles of group theory.

Answer

For option (a):

Let G be a nonabelian group.

Then G is not cyclic and let x∈G such that x≠e.

Also let H = (x), then H is cyclic subgroup of G.

⇒ option (a) is correct.

For option (b)

Let G be a non-trival abelian group.

If G is cyclic, then nothing to prove.

If G is non cyclic then g≠ (x) ∀∈G. Let a∈G and consider H = ��. then H is a cyclic subgroup of G.

For option (c)

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

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

If o(G) = 6 then every proper subgroup of G 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.

For option (d)

Let G=S₃

Let a = (1 2) and b = (2 3)

Clearly o(a) = o(b) =2

But o(ab) = 3

⇒ Statement is true

Similar Mathematics Doubts

Consider the following statements:1. A group of order 289 is abelian.2. In S3 there are four elements satisfying x2= e and six elements satisfying y3 = e.3. Every proper subgroup of S3 is cyclic.4. (Z30, has 4 subgroups of order 15 .Correct statements are

G is a group of order 51. Then which one of the following statements is false?

Choose the correct option

Which of the following statements is/are true ?

Which one of the following is false

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