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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4
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1answer
67 views

On finite 2-groups that whose center is not cyclic

Let $G$ be a finite 2-group such that $\left|\dfrac{G}{Z(G)}\right|=4$, $Z(G)$ is not cyclic and $Z(G)$ has at least one element of order 4. Then prove that there exists an automorphism $\alpha$ of ...
1
vote
1answer
49 views

Commutators Calculus

I was trying to understand the above Corollary but I have a problem, namely why in the second to last line $A_0 \leq \zeta_p(G)$? Any ideas? Definitions By recurrence we define $[x,_0\, y]=x$; ...
5
votes
2answers
60 views

Show finite group is $p$-group given some structure of group

Let $G$ be a finite group. If there exists an $a\in G$ not equal to the identity such that for all $x\in G$,$\phi(x) = axa^{-1}=x^{p+1} $ is an automorphism of $G$ then $G$ is a $p$-group. This is ...
4
votes
1answer
78 views

Finite Group with $n$-automorphism map

If $G$ is a finite group and $\phi(x) = x^{p+1}$ is an automorphism of $G$ with $order(\phi) |p$ then $G$ is a $p$-group...? If the order of $\phi$ is $1$ then $\phi(x) = x = x^{p+1} = x^px ...
0
votes
0answers
28 views

Show that 2 representations are not equivalent and find all the irreducible representations of G.

Show that 2 representations are not equivalent and find all the irreducible representations of $G$. The group $G=T_{16}$ has order 16 and presentation given by $G=\langle a,b : a^8=b^2=1, ...
1
vote
1answer
60 views

Finite $p$-group in which all its maximal subgroups are cyclic

Let $G$ be a finite $p$-group, $|G|=p^n$. Let $M_1,\dots,M_r$ be all the maximal subgroups and suppose they are cyclic. Why is $\Phi(G)\le Z(G)$? $\Phi(G)$ is the Frattini subgroup. I have no idea ...
1
vote
1answer
41 views

On $p$-groups with a unique minimal subgroup

If $G$ is a finite group with a unique minimal subgroup, we know that $|G|=p^n$. I have to prove that if $p\neq2$ then $G$ is cyclic. This is the contest. What I don't understand is the following ...
3
votes
2answers
142 views

Cauchy's Theorem $\#$ of elements order $p$

In the stronger statement of Cauchy's Theorem it states the the number of elements of order $p$ is a multiple of $p$. http://en.wikipedia.org/wiki/Cauchy%27s_theorem_(group_theory)#Proof_2 I noticed ...
0
votes
1answer
36 views

Proving group is $p$-group by contradiction

http://www.proofwiki.org/wiki/Group_is_P-Group_iff_All_Elements_have_Order_Power_of_P Is $k$ a prime or a prime power? Sorry for this stupid question but I can't tell what $k$ is in this context ...
3
votes
2answers
38 views

$|G|=p^3$ non abelian $\Longrightarrow\gamma_2(G)\lneq Z(G)$

Let $G$ be a non abelian group of order $p^3$. Hence $c=2$ ($c$ is the nilpotence class of $G$). I'll write down some notation, in order to be clear. Let $Z_0(G):=1$, $Z_1(G)=Z(G)$, $Z_{k+1}(G)$ ...
6
votes
1answer
48 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 ...
0
votes
1answer
37 views

$\exists a\in G-H$ such that $aHa^{-1}=H$

Let $G$ be a $p$-group with proper subgroup $H$. Show that there exists an element $a\in G -H$ such that $a^{-1} Ha = H$ Can you check my proof? Since $G$ and $H$ are $p$-groups their centers ...
4
votes
1answer
138 views

Subgroups of abelian $p$-groups

Let $A$ be an Abelian group of prime power order. It can be expressed as a (unique) direct product of cyclic groups of prime power order: $A = \mathbb{Z}_{p^{n_1}} \times \cdots \times ...
3
votes
1answer
112 views

Order and element set of the group with presentation $\langle a,b \;|\; a^9=1, b^3=a^3, [a,b]=a^3\rangle$

If we have a group presentation $G=\langle a,b \;|\; a^9=1, b^3=a^3, [a,b]=a^3\rangle$, how we will get the following values: The order of the group. The elements of the group written in terms of ...
0
votes
1answer
60 views

Sylow subgroups of soluble groups

Suppose $G \leqslant S_p$ acts transitively on $\{1,...,p\}$ for prime $p$. Let $P \leqslant G$ be a Sylow p-subgroup. Is it true that $G$ is soluble <=> $P \triangleleft G$?
0
votes
1answer
103 views

Lower Exponent P Central Series

The lower exponent $p$-central series for a $p$-group $G$ is defined by $G=P_1(G) > P_2(G) > \ldots > P_c(G) = 1$, where $$P_i(G)=[P_{i-1}(G), G] P_{i-1}(G)^p.$$ If $G_i=G/P_i(G)$ and ...
2
votes
0answers
188 views

number of elements of each order in p-groups $Z_{p^n}\rtimes Z_p$ and $Z_{p^n}\times Z_p$ [closed]

Do $p$-groups $\mathbb{Z}_{p^{n}}\rtimes \mathbb{Z}_p$ and $\mathbb{Z}_{p^{n}}\times \mathbb{Z}_p$ have the same number of elements of each order? (The prime $p$ is odd.)
1
vote
1answer
79 views

Classification of automorphism groups of groups of order $p^4$

For the purpose of classifying another algebraic structure which is parametrised by the choice of a group and of an automorphism I need the classification up to isomorphism of automorphism groups of ...
1
vote
1answer
38 views

about z(G) by concept of symplectic spaces

if G be a finite p-group and G' be isomorphic to zp, what we can say about z(G) by concept of symplectic spaces? is [G:z(G)] a perfect square? ((i take the elementary abelian group G/Z(G) as a ...
1
vote
1answer
32 views

Writing $p$-groups using $p$-adics

Is it possible to write any finite abelian $p$-group as $\mathbb{Z}_p^n/\mbox{im }(A)$ for some $n\times n$ matrix $A$ over $\mathbb{Z}_p$? Here $\mathbb{Z}_p$ denotes the $p$-adic integers.
1
vote
2answers
91 views

No group of following property. Is this true?

Let $p$ be a prime greater than 3 and $G$ be group of order $p^5$. Is it true that there is no group $G$ of order $p^5$ such that the order of frattini subgroup is $p^3$ and the order of center is ...
2
votes
2answers
105 views

Let $G$ be a finite $p$-group with has more than one maximal subgroup. Prove that $G$ has at least $p+1$ maximal subgroups.

Let $G$ be a finite $p$-group with has more than one maximal subgroup. Prove that $G$ has at least $p+1$ maximal subgroups. I don't have idea. Help me. Thanks in advanced. EDIT: I found a result ...
1
vote
1answer
100 views

Nonabelian group of order $p^4$ [closed]

Let $P$ be a nonabelian group of order $p^4$, where $p$ is a prime, and let $A$ be a subgroup of $P$ maximal with the property of being normal and abelian. Prove that $A$ is of order $p^3$. Thanks a ...
8
votes
2answers
157 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 $i>0$, $Z_{i}$ ...
1
vote
0answers
25 views

characters in p-groups

I would like to know that some examples of p-groups with $|cd(G)| = dl(G)=3$, such that $cd(G)$ to denote the set of degrees of the irreducible characters of $G$ and $dl(G)$ to denote the derived ...
5
votes
0answers
74 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 ...
1
vote
2answers
63 views

$G$ a finite $p-$group, $H < G$ (i.e, $H$ a subgroup of $G$, but $H \neq G$), prove that $H \neq N_G(H)$ [duplicate]

I have one problem, that I think it's pretty hard to solve. The problem reads: Let $G$ be a finite $p-$group, and $H < G$ (i.e $H \le G$, and $H \neq G$). Prove that $H \neq N_G(H)$. Here are ...
1
vote
1answer
193 views

Any irreducible representation of a $p$-group over a field of characteristic $p$ is trivial.

In general, we know that if $G$ is a finite group and $K$ is a field, then $K[G]$ (the group algebra) is semisimple whenever $\operatorname{char}(K)$ does not divide the order of $G$. However, this ...
1
vote
0answers
87 views

Orders of Elements in Minimal Generating sets of Abelian p-Groups

I'm looking for as much information about the orders of elements in minimal generating sets of finite abelian $p$-groups as possible. What I really need is complete knowledge about the possible orders ...
3
votes
0answers
60 views

An example where $Z(Z_G(A))$ is not a subset of right Engel elements in a finite $p$-group

Find a counter example to the following statement: Let $G$ be a finite $p$-group such that $G/Z(G)$ has exponent $p$. Let $A$ be a normal abelian maximal subgroup of $G$, and $Z_G(A)$ be the ...
4
votes
0answers
156 views

Is there an analogue of outer Space to study outer automorphisms of free pro-$p$ groups?

I would like to know if there is an analogue of Culler & Vogtmann's outer space to study outer automorphisms of free pro-$p$ groups. Perhaps an initial guess of such a space would be a moduli ...
0
votes
2answers
330 views

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 ...
1
vote
2answers
77 views

Non-abelian group of order $p^3$ without semidirect products

I am trying to read a proof that there are at most two non-abelian groups of order $p^3$ if $p$ is an odd prime. The proof presents it as two cases: the first, where every non-identity element has ...
0
votes
1answer
90 views

Without using any Sylow theorem, if every element is a $p$-element then $G$ is a $p$-group

How can we prove the following theorem without using any Sylow theorem? Let $p$ be a prime. In a finite group $G$, if every element is a $p$-element then $G$ is a $p$-group. Or is it possible to ...
7
votes
1answer
930 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. ...
0
votes
1answer
82 views

Quotient groups of $p$-groups

Suppose I am trying to show that a group $G$ is solvable and I gotten to having $Z(G)$ be a p-group and $G/Z(G)$. Now if I can show that $G/Z(G)$ is also a $p$-group, then both are solvable implying ...
1
vote
0answers
76 views

Lift a group action from a quotient

Let $p$ be a rational prime and $H$ be a finite cyclic group of prime order $l$ prime to $p$, i.e. $(l,p) = 1$. Let $G$ be a finite abelian group of $p$-power order. If I can write an (abelian) group ...
0
votes
0answers
49 views

Definition of $p$-class of $p$-group?

I read this sentence somewhere: ... an epimorphism from the finitely presented group $A$ to the largest $p$-group of $p$-class $c$ which is a quotient of $A$ ... May I clarify how is "largest ...
2
votes
1answer
433 views

Subgroups and quotient groups of finite abelian $p$-groups

Motivation The fundamental theorem of finite abelian groups gives us a concise description of the isomorphism types of finite abelian $p$-groups $G$ (in the following, $p$ is a fixed prime). The ...
2
votes
0answers
105 views

Properties of Finite and Infinite $p$-Groups

By a $p$-group, we mean a group in which every element has order a power of $p$. It is well known that finite $p$-group has non-trivial center. But, an infinite $p$-group may have trivial center. ...
3
votes
1answer
49 views

Proving that a $p$-group operating on a finite set of order not divisible by $p$ has a fixed point.

Let $G$ be a finite group of order $p^e$ for some prime $p$. Let $S$ be a set of size not divisible by $p$. I know that $|G| = |Stab(s)|\cdot|O_s|$ = (stabilizer of $s$)(orbit of $s$) So if there ...
6
votes
1answer
145 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 ...
6
votes
1answer
68 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 ...
0
votes
0answers
69 views

Extensions of elementary abelian p-groups by themselves.

How many distinct extensions of $C^k_p$ by $C^l_p$ are there where $C^n_p$ is elementary abelian of order $p^n$, p a prime?
7
votes
3answers
2k 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 ...
1
vote
1answer
69 views

Understanding a proof about finite $p$-groups

I can't follow the reasoning of the author,in this proof: let $G$ be a finite $p-$group. If $H$ is a proper subgroup of G, then $H<N_G(H)$ (clearly $N_G(H)$ is the normalizer of $H$ and p is ...
5
votes
2answers
87 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 ...
1
vote
0answers
33 views

P-groups and homomorphisms… [duplicate]

If $f: G \rightarrow H$ is a surjective homomorphism, $G$ finite, prove the following: 1) If $P$ is a p-subgroup of $G$, then $f(P)$ is a p-subgroup of $H$. 2) If $S$ is a Sylow-p-subgroup of $G$, ...
4
votes
1answer
106 views

Nilpotency class of normal product of dihedral 2-groups

Pessimistically paraphrasing Polya: “if Jack cannot answer a question, there is an easier question Jack also cannot answer.” Hence I ask: Given positive integers $a,b$ describe the set $$N_{a,b} = ...
4
votes
1answer
189 views

Existence of a normal subgroup in a finite group.

Let $G$ be a finite group. If a Sylow $p$-subgroup $P$ of $G$ is contained in the centre, then does there exist a normal subgroup $N$ of $G$, such that $P \cap N = \{e\}$ and $PN=G$? Thanks in ...