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In order to study the subgroup structure of the p-group $G\simeq (Z/pZ)^{m}\rtimes(Z/pZ)$, I need to solve the following exercice from the Book (Dummit & Foote p101):

Exercice:

Let $H$ be a normal subgroup of $G$ of prime index $p$ then for all $K\leq G$ either

  • $K\leq H$ or
  • $G=HK$ and $|K:K\cap H|=p$.

Please refer to any other idea to define the subgroups of $G$. Any help would be appreciated so much. Thank you all.

Community
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Nourddine Snanou
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1 Answers1

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Suppose that $H$ is a normal subgroup of $G$ of prime index. Moreover, assume that $ K$ is not a subgroup of $H$. Then, recall the following lemma:

If $H \le K \le G$ is a tower of groups. Then, $|G:H| = |G:K||K:H|$.

See Subsubgroups are subgroups of subgroups / Multiplicative Property of the Index for a discussion of this result.

Now, since $H$ is a normal subgroup, $HK$ is a subgroup of $G$. Hence, by the statement above we have that $|G:H| = |G:HK||HK:H|$ which implies that $p = |G:HK||HK:H|$. Since $p$ is prime $p \mid |G:HK|$ or $p | |HK:H|$, but not both. If $|HK:K| = 1$, then $K = HK$ which would mean $K \leq H$, contrary to our assumption. Hence, $|HK:K| = p$ which means that $|G:HK| = 1$, or $G = HK$. Now, apply the second isomorphism theorem to conclude that $|K:K\cap H| = p$.

Mike
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  • Thank you very much. But how can I applied this result on the p-group $G\simeq (Z/pZ)^{m}\rtimes(Z/pZ)$. There is a nonablian subgroup of G of the form $(Z/pZ)^{k}\rtimes(Z/pZ)$ with $k<m$ ?. – Nourddine Snanou Jul 08 '19 at 12:12
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    @NourddineSnanou You need to provide more context. What do you want to know about these groups? Why do you suspect this result is useful? – Mike Jul 08 '19 at 13:07
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    I see that if the first assertion holds, then every subgroup of $G$ is elementary abelian. I want to know in what condition we have this case. Else, is there a nonablian subgroup of $G$ of the form $(Z/pZ)^{k}\rtimes(Z/pZ)$ with $k<m$. – Nourddine Snanou Jul 08 '19 at 20:47