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I was reading a paper "ON THE EXISTENCE OF NONINNER AUTOMORPHISMS OF ORDER TWO IN FINITE 2-GROUPS" you can get it from here.
In Introduction page 2 He said "It is worth mentioning here that we need only treat the challenging case where p = 2 because it is well known that every finite p-group, p odd, with cyclic commutator subgroup is regular" and reference to two papers.
1. On finite p-groups with cyclic commutator subgroup, you can get it from here.
2. Finite p-groups with a cyclic commutator subgroup, you can get it from here.
But I can't get from this two papers that how p-group with odd p and cyclic commutator subgroup is regular

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  • $\begingroup$ This is a special case of the result that $p$-groups of nilpotency class less than $p$ are regular. See en.wikipedia.org/wiki/Regular_p-group $\endgroup$
    – Derek Holt
    Aug 18 '15 at 20:56
  • $\begingroup$ I know this but I want to get this result from those two papers $\endgroup$
    – user148528
    Aug 18 '15 at 21:05
  • $\begingroup$ Why you need to obtain the result from these two papers? $\endgroup$ Aug 21 '15 at 21:04
  • $\begingroup$ @Holt. Does cyclic commutator subgroup imply that the group has class less than $p$, for odd $p$? One can proceed directly and easily to prove that a group with cyclic commutator subgroup (or $\gamma_{p-1}$instead of $\gamma_2$) is regular. $\endgroup$ Aug 21 '15 at 21:11
  • $\begingroup$ @YassineGuerboussa Cyclic commutator subgroup does not imply that the group has class less than $p$ for odd $p$. For example in GAP SmallGroup(2187,194) is a $3$-group of nilpotency class $4$ but with cyclic derived subgroup. $\endgroup$ Jun 12 '19 at 8:47
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So that this question does not remain listed as unanswered, here is one way to see what is going on.

There is a result in the theory of regular groups, that says the following:

Lemma: Let $G$ be a finite $p$-group. If $\gamma_{p-1}(G)$ is cyclic, then $G$ is regular.

Suppose the prime $p$ is odd and that $\gamma_{2}(G)$ is cyclic. Then $\gamma_{p-1}(G)<\gamma_{2}(G)$ and as a subgroup of a cyclic group it must be cyclic. Then by the lemma above it follows that $G$ is regular.

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Proof of the Lemma. (Outline). - Reference page 25 of the lecture notes here (any errors in reproducing it below are my own).

If $p=2$ then $\gamma_{1}(G)$ is cyclic and the result is clear. Now consider odd $p$. Choose $x,y \in G$ and let $H= \langle x,y \rangle$. The goal now is to show that $\gamma_{p}(H) \leq \mho_{1}(H^{\prime})$, as this will then mean the group is regular (if this is unclear to you, take a look at the collection formula of P. Hall and the definition of regular).

If $\gamma_{p-1}(H) =1$ then it is clear. Otherwise if $\gamma_{p-1}(H) \neq 1$ then $\gamma_{p-1}(H)>\gamma_{p}(H)$. It follows that $\gamma_{p}(H) \leq \mho_{1}(\gamma_{p-1}(H)) \leq \mho_{1}(H^{\prime})$.

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Turning to your question about the paper and the two references given, the author is not saying that this result follows from those two references. The references are of papers where a particular type of group is studied. The following, seperate, sentence is then the author claiming the result which I have outlined above.

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