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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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Is there a characterization of groups with the property $\forall N\unlhd G,\:\exists H\leq G\text{ s.t. }H\cong G/N$?

A common mistake for beginning group theory students is the belief that a quotient of a group $G$ is necessarily isomorphic to a subgroup of $G$. Is there a characterization of the groups in which ...
95
votes
2answers
8k views

More than 99% of groups of order less than 2000 are of order 1024?

In Algebra: Chapter 0, the author made a remark (footnote on page 82), saying that more than 99% of groups of order less than 2000 are of order 1024. Is this for real? How can one deduce this result? ...
30
votes
1answer
9k views

A $p$-group of order $p^n$ has a normal subgroup of order $p^k$ for each $0\le k \le n$

This is problem 3 from Hungerford's section about the Sylow theorems. I have already read hints saying to use induction and that $p$-groups always have non-trivial centres, but I'm still confused. ...
25
votes
1answer
306 views

References on the theory of $2$-groups.

Many theorems about odd order $p$-groups fail miserably for $2$-groups. These can range from simple $2$-group exceptions (e.g. Frobenius complements can be either cyclic or generalized quaternion) to ...
18
votes
4answers
10k views

There exists only two groups of order $p^2$ up to isomorphism.

I just proved that any finite group of order $p^2$ for $p$ a prime is abelian. The author now asks to show that there are only two such groups up to isomorphism. The first group I can think of is $G=\...
16
votes
3answers
678 views

infinite abelian group where all elements have order 1, 2, or 4

Let $A$ be a (not necessarily finitely generated) abelian group where all elements have order 1, 2, or 4. Does it follow that $A$ can be written as a direct sum $(\bigoplus _\alpha \mathbb Z/4) \oplus ...
14
votes
3answers
8k views

Does every group whose order is a power of a prime p contain an element of order p?

I need to know if every group whose order is a power of a prime $p$ contains an element of order $p$? Should I proceed by picking an element $g$ of the group and proving that there is an element in $\...
14
votes
3answers
882 views

Known bounds for the number of groups of a given order.

The number of nonisomorphic groups of order $n$ is usually called $\nu(n)$. I found a very good survey about the values. $\nu(n)$ is completely known absolutely up to $n=2047$, and for many other ...
14
votes
1answer
175 views

The equivalence classes of $N\sim M\Leftrightarrow G/N\cong G/M$.

Let $G$ be a finite group. Given some $N\unlhd G$, define $$\mathfrak{C}_N:=\{M\unlhd G : G/M \cong G/N\}.$$ How are the subgroups in $\mathfrak{C}_N$ related? Is there some other description ...
10
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1answer
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Power of commutator formula

A few people remember a commutator formula of the form $[a,b]^n = (a^{-1} b^{-1})^n (ab)^n c$ where $c$ is a product of only a few commutators (say $n-1$) of them. Here $a,b$ are in a (free) group ...
10
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2answers
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Which $p$-groups can be Sylow p-subgroups with trivial intersection?

Every cyclic p-group is a Sylow p-subgroup of a finite group whose distinct Sylow p-subgroups intersect trivially in pairs (and there is at least one pair). For instance, let q be a prime congruent ...
10
votes
1answer
1k views

Non-abelian $p$-group; abelian subgroups of index $p$

I'm trying to prove the following problem: (a) Let $G$ be a non-abelian $p$-group with an abelian subgroup of index $p$. Then the number of abelian subgroups of $G$ of index $p$ is either $1$ or $p+...
10
votes
1answer
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If $H$ is a proper subgroup of a $p$-group $G$, then $H$ is proper in $N_G(H)$.

Let $H$ be a proper subgroup of $p$-group $G$. Show that the normalizer of $H$ in $G$, denoted $N_G(H)$, is strictly larger than $H$, and that $H$ is contained in a normal subgroup of index $p$. Here'...
9
votes
2answers
3k views

Classify groups of order 27

Let $|G|=27$. Prove that all subgroups of index $3$ are normal. Classify all groups of order $27$. I can do the first one, but the classification is overwhelming. I don't even know where to start. ...
9
votes
1answer
2k views

Conjugacy classes of a $p$-group

This is a problem from Preliminary Exam - Spring 1984, UC Berkeley For a $p$-group of order $p^4 $, assume the center of $G$ has order $p^2 $. Determine the number of conjugacy classes of $G$. ...
9
votes
2answers
194 views

The $i$-th center $Z_{i}(G)$

Let $H$ be a normal subgroup of a $p$-group $G$, $H$ is of order $p^i$. Prove that $H$ is contained in the $i$-th center $Z_{i}(G)$. Recall that we define $Z_{0}(G)=1$, and for $i>0$, $Z_{i}$ is ...
9
votes
2answers
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Characterizations of the $p$-Prüfer group

I'm an undergrad student fairly keen on algebra. Over the different algebra courses I've taken, I've often encountered the so-called $p$-Prüfer group on exercises but somehow never got around to them. ...
9
votes
2answers
808 views

A question on $p$-groups, and order of its commutator subgroup.

$\textbf{QUESTION-}$ Let $P$ be a p-group with $|P:Z(P)|\leq p^n$. Show that $|P'| \leq p^{n(n-1)/2}$. If $P=Z(P)$ it is true. Now let $n > 1$, then If I see $P$ as a nilpotent group and ...
8
votes
2answers
810 views

What are central automorphisms used for?

A central automorphism is an automorphism $\theta$ for which $x^{-1}\theta(x)\in Z(G)$ for each $x\in G$. It's not difficult to prove that the set of central automorphisms forms a subgroup of $\...
8
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1answer
449 views

Generic properties of $p$-groups

I have the impression that most people in algebra believe the following statement to be true, but I have no reference for it. Fix a natural number $n$. Consider for each prime $p$ the set of all ...
8
votes
2answers
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If $|G| = p^n$ then $G$ has a subgroup of order $p^m$ for all $0\le m <n.$

Prove that if $|G| = p^n$ then $G$ has a subgroup of order $p^m$ for all $0\le m <n.$ Since $G$ is of prime-power order I know $|Z(G)| \ne e$ so there is an $a\in Z(G)$ with order $p$ such that $p ...
8
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2answers
552 views

Nonabelian $p$-groups all of whose proper subgroups are abelian.

Theorem. Let $G$ be a finite, non-abelian $p$-group all of whose proper subgroups are abelian. Then $|G'|=p$. Take a counterexample of minimal order. Assume that exist a $H$ such that $1<H<G'...
8
votes
1answer
274 views

Unipotent action of pro-$p$-group

Say $p$ and $\ell$ are distinct prime numbers. Let $G$ be a pro-$p$-group which acts continuously on a finite-dimensional $\mathbb{Q}_\ell$-vector space $V$. Assume that the action of $G$ on $V$ is ...
8
votes
1answer
202 views

Solvable group determined by the action on its Fitting subgroup and the isoclass of some of its Sylows

Let Q ≤ P be finite p-groups, H ≤ Aut(Q). Is it really true that there is at most one p-solvable group G such that $Q \unlhd G$, $C_G(Q) \leq Q$, P is a Sylow p-subgroup of G, and the map from G to ...
8
votes
1answer
158 views

Power automorphisms which is not inner

I want to know an example of a small order non-abelian $p$-group $G$ with a power automorphism which is not inner, i.e. an automorphism of the form $g\mapsto g^k$ for all $g\in G$, but non-inner. In ...
8
votes
0answers
108 views

Reference request: groups of order $p^4$.

I am looking for a textbook or a paper which include the classification of groups of order $p^4$ ($p$ is prime) using generators and relations. In particular I like to understand which group $G$ "...
7
votes
3answers
1k views

Abelian $p$-group with unique subgroup of index $p$

Let $G$ be a finite abelian $p$-group with a unique subgroup $H$ of index $p$. It is a fact that $G$ is cyclic. This can be deduced from the classification theorem for finite abelian groups by writing ...
7
votes
2answers
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Subgroups of order $p$ and $p^{n-1}$ in a group of order $p^n$.

I have a group $G$ of order $p^n$ for $n \ge 1$ and $p$ a prime. I am looking for two specific subgroups within $G$: one of order $p$ and one of order $p^{n-1}$. I don't think I would use the Sylow ...
7
votes
2answers
272 views

Very generic question about Commutator and Center

Let $G$ be a finite group and $G'$ the commutator group of $G$. What can I say about $G' \cap Z(G)$? Could you be as specific as possible about p-Groups?
7
votes
2answers
2k views

If every element has prime power order and $Z(G) \neq 1$ then $G$ is a $p$-group.

In a finite group $G$ if every element is of some prime power order (prime may vary with element) and if $G$ has non trivial center then prove that $G$ is actually of prime power order. Deduce that ...
7
votes
3answers
230 views

$p$-groups in which the centralizers are normal

Let $|G|=p^n$ and $p$ a prime and let $|G:C_G(x)|\leq p$ for all $x\in G$. Prove: $(a)~~~~~C_G(x)\trianglelefteq G$ for all $x\in G;$ $(b)~~~~~G’\leq Z(G);$ ${\color{red}{{(c)~~~~~|G’|\...
7
votes
3answers
369 views

Deduce that if $G$ is a finite $p$-group, the number of subgroups of $G$ that are not normal is divisible by $p$

Given: Let $G$ be a group, and let $\mathcal{S}$ be the set of subgroups of $G$. For $g\in G$ and $H\in S$, let $g\cdot H=gHg^{-1}$ Question: Deduce that if $G$ is a finite $p$-group, for some prime $...
7
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1answer
165 views

Bound on the number of p-groups for fixed exponent

It's well-known that for each prime number $p$ there are exactly two groups of order $p^2$, five of order $p^3$, and fifteen of order $p^4$ (at least when $p>3$). I know that the classification of ...
7
votes
1answer
155 views

Groups of order $64$ with abelian group of automorphism

G. A. Miller in 1913 constructed the first example of a non-abelian group of order $64$ with abelian group of automorphisms. It is the group $$G=(C_8\rtimes C_4)\rtimes C_2=\langle x,y,z\colon x^8, y^...
7
votes
1answer
187 views

An example of a simple infinite $2$-group

Is there an example of a simple infinite $2$-group? Informations If a $2$-group is Artinian I know that it also locally finite, so the simple $2$-group cannot be Artinian. Take the subgroup ...
7
votes
1answer
109 views

Classify $p$-groups in which all groups of the same order are isomorphic

The answer to “Are two subgroups of a finite $p$-group $G$, of the same order, isomorphic?” is definitely no. Such groups are very rare. How rare? Can you classify all finite $p$-groups $G$ such that ...
7
votes
1answer
251 views

Frobenius kernel is regular normal elementary abelian p-subgroup?

I'm attempting Exercise 3.4.6 in Dixon & Mortimer's book on Permutation Groups: Let $G$ be a finite primitive permutation group with abelian point stabilisers. Show that $G$ has a regular ...
6
votes
2answers
145 views

There is no core free subgroup of order $p^2$ in a group of order $p^4$

By the classification of group of order $p^4$ ($p$ odd prime) from Burnside's book it seems to me that there is no core free subgroup of order $p^2$ in a group of order $p^4$. If I am not wrong there ...
6
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1answer
420 views

An infinite $p$-group may not be nilpotent

It is well-known fact that every finite $p$-group $G$ is nilpotent. I am asking to have a counter example when $G$ is infinite $p$-group. Thanks.
6
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2answers
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Is this Galois theory proof of Fundamental Theorem of Algebra correct?

I am studying Galois theory through Lang's Algebra and Dummit-Foote's Abstract Algebra. While studying the Fundamental Theorem of Algebra's proofs from both books I spent a lot of time to understand ...
6
votes
1answer
103 views

Definition of $\Omega$-group and $\Omega$-composition series

What are the definitions of $\Omega$-group and $\Omega$-composition series? No luck searching on the internet..
6
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1answer
275 views

Does every $p$-group of odd order admit fixed point free automorphisms?

Does every $p$-group of odd order admit fixed point free automorphisms? equivalently, Given an odd order $p$-group $P$, is there a group $C$ such that we can form a Frobenius group $P\rtimes C$? ...
6
votes
1answer
83 views

Are there asymptotically more nonabelian groups of order $p^k$ than there are abelian groups of order $\leq p^k$?

Let $\alpha(n)$ denote the number of isomorphism classes of abelian groups of order $n$ and $\alpha^\prime(n)=\sum_{m=1}^n\alpha(m)$. Similarly, define $f(p^k)$ to be the number of isomorphism ...
6
votes
1answer
347 views

Why are there so many groups (up to isomorphism) of order $p^n$ for $n>2$, especially when compared to groups of similar sized order?

While bounds on the number of isomorphism classes of groups of order $p^n$ where $p$ is prime have been known for quite a while (such as the work of Higman$^{[1]}$ and Sims$^{[2]}$) which give us the ...
6
votes
0answers
117 views

Abelian groups whose automorphism group is a $p$ group

$\def\Aut{\operatorname{Aut}}$ Let $G$ be a finite abelian group such that $\Aut(G)$ is an $p$ group ,that is, $|\Aut(G)|=p^n$ . Then can we determine the cyclic decomposition of $G$ or at least the ...
6
votes
0answers
301 views

Groups of order $p^5$

I am reading a paper "A Determination of order $p^5$" by H A Bender ($p$ is an odd prime). He divides the classification in two classes, one which contains an abelian subgroup of order $p^4$ and other ...
5
votes
5answers
2k views

Any group of order $2p$ has a subgroup of order $p~$($p$ being a prime)

I want to show without using Sylow theorem that Any group of order $2p$ has a subgroup of order $p~$($p$ being a prime) My attempt: Since $|G|=2p,$ even $\exists~a\neq e\in G$ such that $a^{-1}=a.$...
5
votes
1answer
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Difference between definitions of $p$-subgroup and Sylow $p$-subgroup

I'm reading Abstract algebra by Dummit and Foote and the following definitions are made: $1$. A group of order $p^{\alpha}$ for some $\alpha\geq1$ is called a $p$-group. Subgroups of $G$ which are $...
5
votes
3answers
552 views

A simple question on characteristic subgroups

Suppose $P$ is a finite abelian $2$-group and $U$,$V$ are characteristic subgroups of $P$ such that $|V:U|=2$. Does it follow that $P$ has a characteristic subgroup of order $2$?
5
votes
3answers
183 views

at least one element fixed by all the group

$G$ is a p-group and $S$ is a set that $G$ acts on. p does not divide $|S|$. Why is there at least one element $a\in S$ such that $|O(a)|=1$, or in other words, $G_a=G$? I tried to ask this question ...