All Questions
34 questions
8
votes
1
answer
353
views
Structure of a single automorphism of a finite abelian p-group
A finite abelian $p$-group $H$ is homogenous when it is the direct sum of cyclic groups of the same order $p^r$, i.e. $H \cong \big(\mathbb{Z}/p^{r}\mathbb{Z}\big)^{e}$. Every finite abelian $p$-group ...
4
votes
2
answers
227
views
Maximal subgroups of finite abelian $2$-groups
Suppose $G$ is a finite abelian $2$-group, and $S$ is a subset of $G$, $\langle S\rangle=G$,$S^{-1}=S$,$e\notin S$. How to determine whether there exists a maximal subgroup $M$ of $G$, such that $S$ ...
5
votes
1
answer
313
views
Projective representations of a finite abelian group
Projective representations of a group $G$ are classified by the second group cohomology $H^2(G,U(1))$. If $G$ is finite and abelian, it is isomorphic to the direct product of cyclic groups
$$
G\cong ...
3
votes
1
answer
203
views
normalizer quotient is $\operatorname{GL}_2(p)$
Let $p$ be a prime and let $w$ be a primitive $p$-th root of unity in $\mathbb{C}$. There is an element $e$ of order $p$ in $G=\operatorname{PGL}_n(\mathbb{C})$ where $n=pk$ and
$$e=\left[\left(\begin{...
2
votes
0
answers
193
views
Decomposition of finite abelian groups of even order if there is an involution
Let $G$ be a finite abelian group and $\sigma :G\rightarrow G$ an automorphism of order two ($\sigma\circ \sigma =id_G$). Denote by $F$ and $A$ the subgroups of fixed and anti-fixed points of $\sigma$ ...
15
votes
1
answer
753
views
Finite abelian groups with fewer automorphisms than a subgroup
It is not too hard to find examples of finite groups which have fewer automorphisms than one of their subgroups. For example $\mathcal D_4 \times \mathbb Z/2\mathbb Z$ (where $\mathcal D_4$ is the ...
2
votes
0
answers
408
views
Conceptual proof of fundamental theorem of finite abelian groups
I'm looking for a conceptual proof of the following statement:
Lemma: Let $G$ be a finite abelian $p$-group. Let $a$ be an element of maximal order. Then $G=\langle a \rangle \times H$ for some ...
1
vote
0
answers
125
views
Is the commutator of the holomorph of generalized quaternion group abelian?
Let $Q_{2^{n}} = \langle x, y \mathrel\vert x^{2^{n-1}}=y^4 = 1, x^{2^{n-2}}=y^2, y^{-1}xy = x^{-1} \rangle$ be the generalized quaternion group of order $2^{n}$.
Let $\operatorname{Hol}(Q_{2^{n+1}})$ ...
3
votes
0
answers
251
views
Commuting real elements in finite groups
Let $p$, $q$, $r$ be three distinct odd primes, and $G$ a finite group with $|G|$ divisible by $p$, $q$, $r$ to the first power only. Let $x,y,z \in G$ be of order $p,q,r$ respectively. Assume
(a.) $[...
5
votes
1
answer
204
views
Characteristic subgroups of a finite abelian $2$-group
I have recently stumbled across the problem of describing the characteristic subgroups of a finite abelian group. With some discussions with some mathematicians in my lab, I managed to obtain a "...
3
votes
1
answer
474
views
Structures of subgroups of a finite abelian p-group
$\newcommand\la{\langle}\newcommand\ra{\rangle}$Let $G=\mathbb{Z}/p^{i_1}\times\cdots\times\mathbb{Z}/p^{i_r}$ with $i_1\leq\ldots\leq i_r$ be a finite abelian $p$-group. Then there can be many ...
8
votes
2
answers
501
views
On $p$-groups with abelian automorphism group
Let $G$ be a $p$-group of order $p^{n}\geq p^{7}$ and its automorphism group is elementary abelian $p$-group. Then, it is clear that $G$ is nilpotent of class $2$. However, the converse is not true in ...
1
vote
1
answer
344
views
Bound for order of a group depending on number of elements of maximal order
This question has been partly answered in MSE, see here.
In a paper On the Number of Elements of maximal order in a Group, it is proven that an arbitrary group $G$ with a finite number of elements of ...
10
votes
2
answers
865
views
Is there a non-degenerate quadratic form on every finite abelian group?
Let $G$ be a finite abelian group. A quadratic form on $G$ is a map $q: G \to \mathbb{C}^*$ such that $q(g) = q(g^{-1})$ and the symmetric function $b(g,h):= \frac{q(gh)}{q(g)q(h)}$ is a bicharacter, ...
0
votes
1
answer
557
views
Abelian groups and their subgroups
It is well known that every finite abelian group is a direct product of cyclic groups. So for every $n$ every finite abelian group of exponent $n$ is a direct product of cyclic groups of order at most ...
3
votes
1
answer
153
views
On decomposition of finite Abelian groups
It is easy to see that for any finite Abelian group $G$ and any numbers $a,b$ with $|G|=ab$ there exist a subgroup $A\subset G$ and a subset $B\subset G$ such that $|A|=a$, $|B|=b$ and $G=A+B$, where $...
13
votes
1
answer
791
views
How nearly abelian are nilpotent groups?
It is not uncommon to read that "nilpotent groups are 'close to abelian'."1,2
Can this sentiment be made precise
in the sense of the
Turán and Erdős definition of "the probability that two elements of ...
1
vote
0
answers
109
views
Symmetric analogue of "alternating bihomomorphism is skew of 2-cocycle" theorem
Let $G$ be a finite abelian group. It is well-known that every alternating bihomomorphism $\Omega:G\times G \to \mathbb{C}^\times$ arises as the skew $\kappa/\kappa^T$ of a 2-cocycle $\kappa \in Z^2(G,...
5
votes
0
answers
444
views
Subgroups and quotients of an abelian pro-finite group
It is well known that every subgroup $H$ of a finite abelian group $G$ is isomorphic to a quotient of $G$.
I'm wondering whether there is a counterpart for profinite groups.
For example is it true ...
8
votes
2
answers
501
views
Exact sequence of $n$th powers of abelian groups
Let $A,B,C$ be finitely generated abelian groups. Assume that there is an exact sequence $$0 \to C \to A^n \to B^n \to 0,$$where $A^n = A \oplus \dotsc \oplus A$ as usual. It is not assumed that $A^n \...
2
votes
0
answers
100
views
Alternating bihomomorphism is skew of 2-cocycle - relative situation
Let $G$ be a finite abelian group. It is well-known that every alternating bihomomorphism $\Omega:G\times G \to \mathbb{C}^\times$ (i.e. $\Omega(g,g)=1$) arises as the skew $\kappa/\kappa^T$ of a 2-...
1
vote
1
answer
145
views
Non-degeneracy of product of group pairings
For $G$ finite abelian group, let $\eta,\omega:G \times G \to \mathbb{C}^\times$ be group pairings. What can I say about the (non-)degeneneracy of the product pairing $\eta \cdot \omega$ in terms of ...
32
votes
3
answers
3k
views
Is there a nice explanation for this curious fact about cyclic subgroups?
Here's something that I noticed that quite surprised me.
Let $G$ be a finite abelian group. Consider the following expression.
$$
\nu(G) = \sum_{\substack{H \leq G \\ H \text{ is cyclic}}} |H|
$$
It ...
6
votes
2
answers
232
views
Finding an "optimal" quotient in a free group
Consider the abelian free group $G = \mathbf{Z}^n$ of rank $n$ and a finite subset $A \subset G \setminus \{0\}$. Since $G$ is residually finite, there is a subgroup $H \subset G$ such that $A \cap H =...
3
votes
2
answers
318
views
Character kernels in the lattice of subgroups of a finite abelian group
I am looking for any efforts that have been made to characterize the character kernels (equivalently, the subgroups yielding cyclic quotients) inside the lattice of subgroups of a finite abelian group....
6
votes
1
answer
332
views
Zero-sum sets in union-closed families
The Davenport constant $D(G)$ of a finite abelian group $G$ is the minimum integer $n$ such that whenever $a_1, \ldots, a_n \in G$ (not necessarily distinct), there is a non-empty $I \subseteq [n]$ ...
2
votes
1
answer
159
views
Counting elements with certain word length in abelian groups
Given a (finite) abelian group $G = \langle S \mid R \rangle$, has the problem of counting the number of elements which can be expressed as a word (in $S$) of length $\leq k$ been studied? If so, ...
2
votes
1
answer
198
views
The special subgroups of a finite abelian group of rank two
Let $G=\langle a_{1}\rangle\times\langle a_{2}\rangle$ such that $|a_{i}|=2^{k_{i}}$ and $k_{1}>k_{2}$ and $H$ be a subgroup of $G$ that there exists an automorphism of $G$ such that fix only ...
9
votes
1
answer
3k
views
Automorphism group of a finite group
I would like to ask if there exists an explicit description of $\mathrm{Aut}(G)$, the group of automorphisms of a finite group $G$, in particular, when $G$ is abelian. E.g., if $G = \mathbb{Z}/m\...
1
vote
0
answers
242
views
How many subgroups of order $\prod_{1}^{n} p_{i}^{n_{i}}$ are there in the finite product of cyclic groups?
All of the following ${p_{i},q_{i}}$are prime numbers, ${n,m,k}$ are pre-assigned integers.
Consider the product of cyclic groups $\prod_{1}^{n} \mathbb Z_{p_{i}^{n_{i}}}$ then we asked the question:
...
2
votes
3
answers
1k
views
Finite / uniquely divisible abelian groups
Is there any counter example for the following statement?
STATEMENT:
Let $0 \to F \to A \to Q \to 0$ be a short exact sequence of abelian groups.
Assume that $F$ is a finite group, and $Q$ is a ...
9
votes
2
answers
1k
views
On the existence of a direct summand containing a fixed subgroup
Let $G$ be a finite abelian group, and $g_1, \ldots, g_n \in G$ such that the cyclic groups that they generate are in direct sum $\langle g_1 \rangle \oplus \cdots \oplus \langle g_n \rangle$. Is it ...
1
vote
2
answers
1k
views
Maximal subgroups of a finite p-group
I want to prove the following:
Let $G$ be a finite abelian $p$-group that is not cyclic.
Let $L \ne {1}$ be a subgroup of $G$ and $U$ be a maximal subgroup of L then there exists a maximal subgroup $...
4
votes
2
answers
1k
views
Cyclic subgroups of finite abelian groups
I learned from MO Subgroups of a finite abelian group that the problem of enumerating subgroups (not up to isomorphism) of finite abelian groups is a difficult one.
Are there simple formulas if one ...