Re: G' cyclic

On Apr 29, 5:50 pm, "marcelo" <@alpha@@lll.it> wrote:
Let G' be the commutator subgroup [G,G], G finite. If G' is cyclic of order
p, where p is the smallest prime dividing |G|, then G is nilpotent.
The (one-line) proof says: "if p is the smallest prime dividing |G| then G'
<= Z(G) (why? *) and G is nilpotent (why? **)".

My opinion for (*) is that G' is in Z(G) because it is cyclic, then it is
abelian (and minimal because of its order). Since Z(G) is abelian too, then
either G'=Z(G) or G' < Z(G).
About (**), I believe that 1 nrm G' nrm G is a central series.
Indeed it is G/G' <= Z(G/G') because G/G' is abelian, and we also have G'=
G'/1 <= Z(G/1) = Z(G) (as proved in (*)).

Am I correct?
Thanks in advance.

For (*): Let G'=<x> and g in G, so that g acts by conjugation as an
automorphism of a cyclic group of order p. The automorphism group of
a cyclic group of order p is cyclic of order p-1, so the automorphism
has order dividing p-1. However the order of the automorphism is also
a divisor of the order of g, and p-1 is smaller than any prime factor
of the order of g (or of G). Therefore the automorphism has order 1
and is the identity. In other words g^-1*x*g = x and x*g= g*x, so g
centralizes x. Since this is true for all g in G, x and <x>=G' are in
the center of G.

Your reasoning for (**) is correct.

.

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