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Let $S$ be the collection of (isomorphism classes of) groups $G$ which have the property that every element of $G$ commutes only with the identity element and itself. Then

  1. $|S| = 1$
  2. $|S| = 2$
  3. $|S| \geq 3$ and is finite.
  4. $|S| = \infty$
in Set Theory & Algebra by Boss (29.8k points) | 78 views

1 Answer

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Answer C 

Since an element always commutes with its powers, we have that every element has order 2. In particular, this implies a=a−1a=a−1 for every a∈Ga∈G. Also, a group with this property must be abelian, since [a,b]=aba−1b−1=abab=(ab)2=e[a,b]=aba−1b−1=abab=(ab)2=e.

Now, if the group has more than two elements, all these should commute, which contradicts the initial conditions. So, the group must have one or two elements and is isomorphic either to trivial group or Z/2ZZ/2Z

by Active (1.8k points)

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