# Questions tagged [p-groups]

Use with the (group-theory) tag. Refers to questions concerning finite groups of prime power order or infinite p-groups such as Prüfer groups, pro-p-groups, and Tarski monsters. This tag is not for p-adic number systems.

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### Meta cyclic p-group

While studying meta cyclic p groups, I came across an interesting class of meta cyclic groups which can be written as semi-direct product of two cyclic p-groups of order $p^m$ and $p^n$ respectively. ...
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### On classification of groups of order $p^5$

Can someone suggest me some source where the author has classified all non-isomorphic groups of order $p^5$ ? Edit 1 : I need complete classification (not upto isoclinism), and also in finitely ...
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### Structure of non abelian finite p-groups

I am familiar with the concepts of direct products, semi-direct products, wreath products and central products of groups. After seeing the classification of finite $p$-groups upto order $p^4$,(Theory ...
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### Help to understand a step in the fix point theorem

Let $G$ be a $p$-group and $X$ be a finite set on which G operates. We define $X_G:=\{x \in X: g \circ x=x \; \forall g \in G\}$ set the of all fixed points. For the proof of $|X|\equiv |X_G|$ mod $p$ ...
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### Let $G$ be a p-group. Let $H$ be a proper subgroup of $G$. Show that there exists $g$ $\in$ $G \setminus H$ such that $gHg^{-1}=H$.

Let $G$ be a p-group. Let $H$ be a proper subgroup of $G$. Show that there exists $g$ $\in$ $G \setminus H$ such that $gHg^{-1}=H$. I tried to use a counting argument. Let's assume by contradiction ...
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### Group of order 81 and exponent 3

I know that there is one group only of order 81 which is non abelian and of exponent 3. But I have no idea how to prove it. This question is personal. I looked at Wims, which suggested the answer. ...
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### How many subgroups does a Prüfer group have?

Let $Z_{2^\infty}:=\{z\in \mathbb C:z^{2^n}=1,$for some $n\in \mathbb N\}$. This is a countable group. But I am not sure about the cardinality of the set of its subgroups. Does it have uncountably ...
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### Let $p$ be a prime number and let $G$ be a finite $p\text{-group}$. Let $M$ be a maximal subgroup of $G$.

QUESTION: Let $p$ be a prime number and let $G$ be a finite $p\text{-group}$. Let $M$ be a maximal subgroup of $G$. Show that $M$ is a normal subgroup of $G$ and that $| G: M | = p$. THE HINT GIVEN ...
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### Is the inverse image of a $p$-group under the canonical homomorphism also a $p$-group?

Let $G$ be a finite group and $H$ a normal subgroup of $G$. Let $f:G\rightarrow G/H$ be the canonical homomorphism. Let $Q\leq G/H$ be a $p$-subgroup of $G/H$. I have to show that $f^{-1}(Q)\leq G$ ...
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### Quotient of quotient groups and Sylow $p$-subgroups

Let $G$ be a finite group and $N$ a normal subgroup of $G$. Let $\pi:G\rightarrow G/N$ be the canonical homomorphism. Suppose $P$ is a Sylow $p$-subgroup of $G$. Then $\pi[P]\leq G/N$. Note that $G$ ...
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### Unique subgroup of index p in an inifite p-group

A question that recently came up in my homework: Let $p$ be a primer number and let $G$ be a $p$-group. Prove that if $G$ has a unique subgroup of index $p$ it must be that $G$ is cyclic. I know how ...
Let $G$ be a finite group. Let $\mathfrak{P}$ denote the set of prime numbers and $n$ the order of $G$. Since $n>0$, there exists a unique family $(\nu_q(n))_{q\in\mathfrak{P}}$ of elements of $\... 0answers 38 views ### Every finite group contains a Sylow$p$-subgroup Let$G$be a finite group of order$n$and$p$a prime number. Write$n=p^rm$for some$r\in\mathbb{N}$and$m\in\mathbb{N}_{\geq1}$such that$m\not\in p\mathbb{Z}$. Let $$E=\{X\subset G\ :\ |X|=p^r\}... 2answers 46 views ### A cyclic subgroup of a p-group Let G\ne\{e\} be a p-group. Then there exists x\in Z(G) such that x\ne e. Let k>0 such that x^{p^k}=e: i.e. let p^k be the order of the element x. I would like to show that the ... 1answer 99 views ### G has a unique normal subgroup of order p iff G is cyclic center. Let G be a p-group. Proof that Z(G) is cyclic if and only if G has a unique normal subgroup H of order p. I am supposed to prove it without using Sylow theorems. I already prove the first ... 1answer 42 views ### A condition implying powerfulness of finite p-groups, where p is odd For an odd prime p, a finite p-group is called powerful if [G,G] \subseteq G^p, where G^p = \langle g^p ~:~ g \in G \rangle. Prove that : [G,G] \subseteq [G,G,G] G^p implies G is powerful. ... 0answers 85 views ### Pattern in Number of Conjugacy Classes of p-groups I was playing around with the number of conjugacy classes of p-groups in GAP and made the following conjecture: If there is a group of order p^{2n} with k conjugacy classes then there is a ... 1answer 38 views ### Classifcation of p-groups with cyclic subgroup of index p I am trying to read the Classification of p-Groups with cyclic subgroup of index p done in Cohomology of Groups by Brown. He starts his proof by construction the exact sequence, 0→\mathbb{Z}_q→G→H→0 ... 1answer 57 views ### A p-group that is divisible. Suppose G is an infinite p-group and H is a minimal infinite subgroup of G (where p is a given prime). If H=pH then H is divisible. How can this be? An element of H will only be ... 0answers 49 views ### Are there any references on extensions G of a cyclic group C_2 by 2-groups P? Are there any references on extensions G of a cyclic group C_2 by 2-groups P such that 1\neq a\in C_2 is a square element in G? In other words, if G/{C_2}\cong P, where P is a 2-group,... 0answers 110 views ### Finite intersection property for sets containing generating elements of derived subgroups of quotients What I need to prove is a consequence of the following theorem. Theorem A. Let G be a finite p-group and suppose that its derived subgroup G' is generated by 2 elements. Then there exists x\... 1answer 59 views ### Are there examples of subgroups of \Bbb Z[\frac1n]/q\Bbb Z not totally ordered by inclusion? I have the theorem that the Prufer P-groups (of which \Bbb Z[\frac12]/\Bbb Z is one example) are the only infinite groups whose subgroups are ordered by inclusion. That this property holds for \... 1answer 51 views ### On generalized fitting subgroup I can't understand so much the second paragraph (page 160) of the proof of the lemma 31.17(1) (pages 160) in M. Aschbacher, Finite Group Theory about generalized fitting subgroup. Here I post the ... 1answer 41 views ### Order of conjugacy class of a p-group I' m trying to solve the following exercise: Prove that, if G is finite of order p^n, p prime, n\geq3 and |Z(G)|=p, then G contains a conjugacy class of order p. I know that every class of ... 3answers 100 views ### If P\le G is a sylow-p and Q is any p subgroup, then Q\cap P = Q\cap N(P). If P\le G is a sylow-p and Q is any p subgroup, then Q\cap P = Q\cap N(P). I'd appreciate any help. I have a proof from some old notes but it says that it is sufficient to prove that if ... 2answers 71 views ### Equivalent definitions of a p-group In Dummit and Foote, p.139, a p-group is defined as a group of order p^\alpha for \alpha \geq 1. I also found online a definition of a p-group as a group in which every element has a power of ... 1answer 65 views ### Let G be a group with p^3 elements, p prime. Find the cardinal of the set \{\operatorname{C}(x)\mid x\in G\}. Let p be a prime number and (G,\cdot) be a group with p^3 elements. We denote by \operatorname{C}(x) the centraliser of x\in G. If |\operatorname{Z}(G)|=p, then find the cardinal of the ... 2answers 78 views ### Abelian group, invariant and p-group For an abelian group G I have shown that$$ G=Z(p^{l_1})\times...\times Z(p^{l_r})\Rightarrow G^p\cong Z(p^{l_1-1})\times...\times Z(p^{l_r-1}) \, \text{and ord} (G/G^p)=p^r.$$A$p$-group$G$is ... 0answers 70 views ### A question about the Frattini argument and normalizers$G$is a finite group and$N$is a normal subgroup of$G$. Let$H/N$be any nontrivial subgroup of$G/N$of prime power order. Then we have$|H/N|=p^n$, for some prime$p$and$n\geq 1$. Let$P$... 5answers 85 views ### About an order of a p-group [duplicate] I'm trying to show that if G is a Group, then$|G| = p^2 \Rightarrow G$is abelian. The path I'm taking relies on supposing that$|Z(G)| = p$and forming the quotient group$\overline{G} = G/Z(G)$. ... 2answers 119 views ### Q: how to describe these results by a descendants tree in gap I wrote an implement to find the "fullyInvariantGroups" in GAP and the results appeared as below: ... 1answer 62 views ### Finite groups with elements of order a prime power Consider a finite group$G$where the order of each element is a power of a certain prime number$p$, then$G$is a$p$-group. My question: are there groups that are not$p$-groups, but for which the ... 1answer 50 views ### Multiplicative group of order$2^k$has proper subgroup containing set of all squares Specifically, I'm trying to solve the following problem: Let$G$be a multiplicative group of order$2^k$where$k\geq1$. Show that$G$has a proper subgroup$H<G$containing the subset$S=\{g^2:...
Let $P$ be a finite $p$-group and $Q$ a proper subgroup of $P$. Define the normalizer tower of $Q$ in $P$ as follow: \begin{equation} N^0(Q) = Q \mbox{ and } N^i(Q) = N_P(N^{i-1}(Q)) \end{equation} ...