Skip to content
Part CCSIR NET December 2024no-finite-group-is-divisible

No finite group is divisible

A group G is said to be divisible if for every y ∈ G and for every positive integer n, there exists x ∈ G such that xn=yx^{n} = y. Which of the following groups are divisible?

  1. A.Q\mathbb{Q} with ordinary addition✓ Correct
  2. B.C\mathbb{C} \ {0} with ordinary multiplication✓ Correct
  3. C.The cyclic group of order 5
  4. D.The symmetric group S5S_{5}

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests standard counterexample.

See pricing

50 are analysed free — try those first.

The trap it tests

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

Drill statements like this

Related counterexample: (ℤ/2^kℤ) is cyclic for every k

More on this topic

The chapter behind this: Structure of finite abelian groups — free to read

From Groups › Finite abelian groups

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram