Question:

Consider the following statements :
I. Every infinite group has infinitely many subgroups.
II. There are only finitely many non-isomorphic groups of a given finite order.
Then

Updated On: Nov 17, 2025
  • both I and II are TRUE
  • I is TRUE but II is FALSE
  • I is FALSE but II is TRUE
  • both I and II are FALSE
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The Correct Option is A

Solution and Explanation

To evaluate the correctness of the given statements, we need to analyze each statement individually.

Statement I: Every infinite group has infinitely many subgroups.

  • An infinite group is a group with an infinite number of elements.
  • Consider the simplest example of an infinite group, the group of integers under addition, denoted by \(\mathbb{Z}\).
  • The subgroups of \(\mathbb{Z}\) are of the form n\mathbb{Z}, where n is a positive integer. There are infinitely many such n, hence infinitely many subgroups.
  • In general, infinite groups tend to have many such subgroups (technically an infinite number of them).
  • Therefore, Statement I is true.

Statement II: There are only finitely many non-isomorphic groups of a given finite order.

  • A group has a finite order if it has a finite number of elements.
  • The statement postulates that for any given finite number of elements, there are only a finite number of distinct (non-isomorphic) group structures possible.
  • This is a well-known result in group theory: for any given finite order (say, n), there are only finitely many groups up to isomorphism.
  • For example, for order 4, the known groups up to isomorphism are the cyclic group \(\mathbb{Z}_4\) and the Klein four-group V_4.
  • Thus, Statement II is true.

Based on the above reasoning, both statements I and II are true. Therefore, the correct answer is that both I and II are TRUE.

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