2010
DOI: 10.1016/j.jalgebra.2009.10.013
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Powerful p-groups have non-inner automorphisms of order p and some cohomology

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Cited by 32 publications
(39 citation statements)
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(21 reference statements)
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“…So in particular if G is a nonabelian regular p-group and is the Frattini subgroup of G, then H n (G/ , Z( )) = 0 for all n; Schmid then asks whether this result holds more generally. Abdollahi [1] has given some cases where the result holds, and in the final section we prove the following result.…”
Section: Theorem 2 Let G Be a Finite P K -Central P-group Of Coclassmentioning
confidence: 55%
See 1 more Smart Citation
“…So in particular if G is a nonabelian regular p-group and is the Frattini subgroup of G, then H n (G/ , Z( )) = 0 for all n; Schmid then asks whether this result holds more generally. Abdollahi [1] has given some cases where the result holds, and in the final section we prove the following result.…”
Section: Theorem 2 Let G Be a Finite P K -Central P-group Of Coclassmentioning
confidence: 55%
“…By the result of Uchida [24] we know that A is cohomologically trivial if H r (Q, A) = 0 for just one integer r. In [20] Schmid investigates when the cohomology is non-trivial, he proves that if G is a regular p-group and Q = G/N is not cyclic then H n (Q, Z(N)) = 0 for all n. So, in particular, if G is a non-abelian regular p-group and is the Frattini subgroup of G then H n (G/ , Z( )) = 0 for all n, Schmid then asks whether this holds more generally. Abdollahi addresses this question in [1] (and uses the alternative definition of p-central as mentioned in our Introduction) and poses the following more general question: By Uchida's result we will be able to restrict our attention to H 0 (Q, Z(N)). Recall, H 0 (Q, A) = A Q /A τ , where A Q denotes the fixed points of A under the action of Q, and A τ denotes the image of A under the trace map τ = τ Q .…”
Section: Theorem 1 Let G Be a Finite Group Of Exponent P And Nilpotementioning
confidence: 99%
“…Indeed, Deaconescu [12] proved it for all finite p-groups G which are not strongly Frattinian. Moreover, Abdollahi [2] proved it for finite p-groups G such that G/Z(G) is a powerful p-group, and Jamali and Visesh [14] did the same for finite p-groups with cyclic commutator subgroup. In the realm of finite groups, quite recently, the result has been confirmed for semi-abelian p-groups by Benmoussa and Guerboussa [6], and for p-groups of nilpotency class 3, by Abdollahi, Ghoraishi and Wilkens [3].…”
Section: Introductionmentioning
confidence: 94%
“…Furthermore, as a result due to Abdollahi in [2], we know that if G is a finite p-group such that G has no non-inner automorphisms of order p leaving Φ(G) elementwise fixed, then d(Z 2 (G)/Z(G)) = d(G)d(Z(G)). Thus, in view of this previous matter, we may assume that the condition d(Z 2 (G)/Z(G)) = d(G)d(Z(G)) holds.…”
Section: Berkovich Conjecture For Two Generator Normally Constrained mentioning
confidence: 99%
“…The conjecture has been established for p-groups of class 2 and 3 [2,3,11], for regular p-groups [13], for p-groups G in which G/Z(G) is powerful [1], for p-groups G in which (G, Z(G)) is a Camina pair and p 2 [9], for 2-groups with a cyclic commutator subgroup [10], and for p-groups of order p m and exponent p m−2 [14]. It is worth noting that most of the noninner automorphisms given in these results leave either Φ(G) or Z(G) elementwise fixed.…”
Section: Introductionmentioning
confidence: 99%