Let G be finite with |G| > 1. If Aut(G) acts transitively on G − {e} then G ∼= (Z/(p))n for some prime p.

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Expert Solution

Proof:

Elements that are linked by an automorphism have the same order, so the hypothesis in the theorem implies all non-identity elements of G have the same order.

Let p be a prime factor of |G|. Cauchy’s theorem gives us an element with order p, so all non-identity elements have order p. Thus |G| is a power of p. Since G is a non-trivial p-group, it has a non-trivial center. Elements that are linked by an automorphism are both in or both not in the center, so by the hypothesis of the theorem every non-identity element of G is in the center.

Thus G is abelian. Since each non-zero element has order p, we can view G as a vector space over Z/(p), necessarily finite-dimensional since G is finite. Picking a basis shows G ∼= (Z/(p))n for some n.

Elements that are linked by an automorphism have the same order, so the hypothesis in the theorem implies all non-identity elements of G have the same order.

Let p be a prime factor of |G|. Cauchy’s theorem gives us an element with order p, so all non-identity elements have order p. Thus |G| is a power of p.

Since G is a non-trivial p-group, it has a non-trivial center. Elements that are linked by an automorphism are both in or both not in the center, so by the hypothesis of the theorem every non-identity element of G is in the center.

Nipun Maity answered 2 years ago

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