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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.

This tag is for questions specifically pretaining to finite p-groups and their properties, such as isoclinism, schur covers and projective representations, and cohomology. The tag can also be used for questions about infinite p-groups such as Prüfer groups, pro-$p$-groups, or Tarski monsters. This tag is not for use with p-adic number systems, though it may be used for the $p$-adic integers $\displaystyle\varprojlim \mathbb{Z}/(p^n\mathbb{Z})$.

687 questions
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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? Is there a nice way or do we just check all…
Hui Yu
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123
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1 answer

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 this property holds? If this question is too broad,…
Alexander Gruber
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48
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1 answer

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. This is what I have so far: Suppose $|G| = p^n$. For…
user104986
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37
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4 answers

A normal subgroup intersects the center of the $p$-group nontrivially

If $G$ is a finite $p$-group with a nontrivial normal subgroup $H$, then the intersection of $H$ and the center of $G$ is not trivial.
7171717171
32
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12 answers

Showing group with $p^2$ elements is Abelian

I have a group $G$ with $p^2$ elements, where $p$ is a prime number. Some (potentially) useful preliminary information I have is that there are exactly $p+1$ subgroups with $p$ elements, and with that I was able to show $G$ has a normal subgroup $N$…
Jimmy Valmer
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26
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1 answer

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 full blown analogs proved with vastly…
Alexander Gruber
  • 28,037
22
votes
5 answers

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 $G=\Bbb Z/p\Bbb Z\oplus \Bbb Z/p\Bbb Z$. This is…
Pedro
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22
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2 answers

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's what I've got so far: If $H$ is normal,…
Benjamin Lu
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18
votes
3 answers

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 $\langle g \rangle$ that has order $p$?
user85362
17
votes
3 answers

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 values of $n$ too (for squarefree n, there is a…
Peter
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17
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2 answers

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. Now I'm trying to take care of that, but there…
Bruno Stonek
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17
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2 answers

Group with order $p^2$ must be abelian . How to prove that?

Possible Duplicate: Showing non-cyclic group with $p^2$ elements is Abelian I must show that a group with order $p^2$ with $p$ prime must be a abelian. I know that $|Z(G)| > 1$ and so $|Z(G)| \in \{p,p^2\}$. If I assume that the order is $p$ i get…
user42761
16
votes
1 answer

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 $G$. What I have tried: each element of the center…
WLOG
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16
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3 answers

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 $(\bigoplus _\alpha \mathbb Z/4) \oplus (\bigoplus_\beta \mathbb Z/2)$?
user46652
  • 529
15
votes
1 answer

Group of order $p^{n}$ has normal subgroups of order $p^{k}$

Q: Prove that a subgroup of order $p^n$ has a normal subgroup of order $p^{k}$ for all $0\leq k \leq n$. Attempt at a proof: We proceed by Induction. This is obviously true for $n=1, 2$. Suppose it is true for $m \leq n-1$. Now, take the group $G$…
TheManWhoNeverSleeps
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