Is every group of order $136$ nilpotent?

Asked Aug 27 '21 at 21:17

Active Aug 27 '21 at 21:39

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Is every group of order $136$ nilpotent?

The answer is no. Consider the group $G=\mathbb{Z}_{17}\rtimes_\phi \mathbb{Z}_8$ where $\phi: \mathbb{Z}_8 \to {\rm Aut}(\mathbb{Z}_{17})\simeq \mathbb{Z}_{16}$ is a non-trivial homomorphism (such homomorphism exists because $8$ divides $16$). If $G$ was nilpotent it would be the direct product of its Sylow subgroups, namely $G = \mathbb{Z}_{17} \times \mathbb{Z}_8$. But since $\phi$ is not trivial this can't be the case (as $G$ would be abelian).

Is my solution correct?

edited Aug 27 '21 at 21:39

Shaun

44,997

asked Aug 27 '21 at 21:17

Giorgos Giapitzakis

6,416

2

Yes, yours is fine. Alternatively, $D_{34}\times C_4$ would work. – David A. Craven Aug 27 '21 at 21:24
@DavidA.Craven Thanks. $D_n$ is nilpotent iff $n=2^k$ right? – Giorgos Giapitzakis Aug 27 '21 at 21:25
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Yes. Otherwise the reflection acts on an odd order rotation, and thus the Sylow $2$-subgroup cannot be a separate normal subgroup from the odd Sylows. – David A. Craven Aug 27 '21 at 21:26
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@DavidA.Craven Or $D_{68} \times C_2$ or $D_{136}$. – Geoffrey Trang Aug 27 '21 at 21:36
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There are fifteen groups of order 136, five of which are nilpotent. – David A. Craven Aug 27 '21 at 21:56
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Related: For which $n$ is every group of order $n$ nilpotent? and Pete Clark's answer to this question. – Arturo Magidin Aug 28 '21 at 02:12
@David A. Craven About your last remark: could you say where such information can be found ? – Jean Marie Aug 28 '21 at 12:22
@JeanMarie Search Google for the Magma online calculator. Then input: n:=136; {* IsNilpotent(SmallGroup(n,i)):i in [1..NumberOfSmallGroups(n)] *}; – David A. Craven Aug 28 '21 at 13:16
@David A. Craven A big thank. – Jean Marie Aug 28 '21 at 13:30

Is every group of order $136$ nilpotent?

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