Should be used with the (group-theory) tag. A group $(G,*)$ is said to be abelian if $a*b=b*a$ for all $a,b\in G.$

learn more… | top users | synonyms

1
vote
1answer
12 views

Lifting a decomposition of abelian $p$-groups.

Let $A$ be a finite abelian $p$-group and $x\in A$ an element of order $p$. Assume that have the following exact sequence : $$1\rightarrow \langle x\rangle \rightarrow A\rightarrow B_1\times B_2\...
0
votes
0answers
26 views

Can not see the use of Correspondence Theorem

In Herstein`s proof of Fundamental Theorem of Finite Abelian Groups, I don´t see the use of Correspondence Theorem. It says that exist some subgroup $Q$ of G such that $T=Q/B$. But I`m a bit confused ...
2
votes
1answer
109 views

Direct proof that infinite product of copies of $\mathbb{Z}$ is not projective

It is well-known that the abelian group $$A = \prod_{n=1}^\infty \mathbb{Z}$$ is not free (see, for example this MO question), and that over a PID being free is equivalent to being projective (see ...
0
votes
3answers
31 views

How to proceed in the proof of this statement.

I'm reading the proof of "Fundamental Theorem of Finite Abelian Groups" in Herstein Abstract Algebra, and I've found this statement in the proof that I don't see very clear. Let $A$ be a normal ...
0
votes
1answer
15 views

Commutativity of multiplication of cosets of the commutator subgroup

Take a group $H$ with a non-trivial commutator subgroup, and form the quotient group $H^{ab} = H/H'$. Now, take the cosets of the products of elements $a,b$ and $c,d$: $abH'$ and $cdH'$ in $H^{ab}$. ...
2
votes
1answer
110 views

$G$ contains a normal $p$-subgroup

Let $G$ be a non-abelian finite group with center $Z>1$. I want to show that if $G/Z$ is solvable then $G$ contains a normal $p$-subgroup for some prime $p$ with $p\mid |G:Z|$. $$$$ Since $G/...
2
votes
0answers
75 views

infinitely $p$-divisible elements in $A\otimes \mathbb{Z}_p$

Let $A$ be a (possibly non-finitely generated) torsion-free abelian group. Suppose that $A$ contains no infinitely $p$-divisible elements, then does the same hold for $A\otimes \mathbb Z_p$, where $\...
0
votes
2answers
64 views

Prove a group G is abelian if it satisfies x^2 = x for every x in G

I originally solved this problem by simply noting that x^2 = x implies x=e, so the only element in the group is the identity...but this is wrong. I am now stuck on this idea though and I have tried ...
1
vote
4answers
89 views

Show that the group is abelian

Let $M$ be a field and $G$ the multiplicative group of matrices of the form $\begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix}$ with $x,y,z\in M$. I have ...
1
vote
2answers
38 views

Let $L$ be a subgroup of $\mathbb{Z}^3$ of index $16$. What are the possibilities for $\mathbb{Z}^3 /L$?

Let $L$ be a subgroup of $\mathbb{Z}^3$ of index $16$. What are the possibilities for $\mathbb{Z}^3 /L$? Since $L$ has 16 elements, I think it might be $\mathbb{Z}_{16},\mathbb{Z}_2 \oplus \mathbb{Z}...
0
votes
1answer
36 views

$G \cong \mathbb{Z}_{p^{n_1}}\times \dots \mathbb{Z}_{p^{n_k}}$. If $p=2$ and $n_1>n_2$, prove that $L(G)\cong \mathbb{Z}_2$.

Let $G \cong \mathbb{Z}_{p^{n_1}}\times \dots \mathbb{Z}_{p^{n_k}}$ be a finite abelian $p$-group, in which $n_1\geq \dots \geq n_k$. Define $$L(G)=\{g\in G \;|\;\alpha(g)=g\; ,\forall \alpha\in Aut(G)...
1
vote
1answer
16 views

Proof that finite abelian groups are the direct sum of p-parts

A proof of this result was given, however I am having trouble understanding why the part in bold is true. Let $A$ be a non-trivial finite abelian group with $|A|$ having distinct prime divisors $...
0
votes
1answer
16 views

Are there any infinite (virtually) polycyclic groups with lattice orders that are not linear orders?

I am interested in noetherian group algebras, so I am learning about polycyclic groups. Specifically, I want to generalize some ideas that work well with $k[\mathbb{Z}^n]$ utilizing the lattice ...
1
vote
1answer
65 views

Generating sets and independent subsets in abelian groups.

Definition. A set $\{x_1,...,x_n\}$ of non zero elements in an Abelian group is independent if, whenever there are integers $m_1,...,m_r$ with $m_1 x_1 + \cdots + m_r x_r = 0$, then each $m_i x_i$ is ...
1
vote
1answer
62 views

$G$ a non-abelian group of order $p^3$. Show that $Inn(G)$ is abelian.

Let $G$ be a non-abelian group of order $p^3$ for prime $p$. Show that $Inn(G)$ is abelian. The center of $G$, $Z(G)$, is of order $p$ (can be seen in this question). I also know that $G / Z(G)=...
0
votes
1answer
34 views

Subgroups of a direct sum.

Let G be a finite abelian group and let $G = G_1 + G_2$ where the $G_i$ are cyclic. Add it is a $p$-group if you like. How do I prove that it isn't the case that, for some $H_i$, $G = H_1 \bigoplus ...
3
votes
1answer
32 views

rank of an abelian group and its embedment into vectorspace

I am confused about the rank of an abelian group. In this wiki page, https://en.wikipedia.org/wiki/Finitely_generated_abelian_group , we have that $\mathbb{Z}^n \oplus \mathbb{Z}_{q_1} \oplus \cdots ...
2
votes
1answer
38 views

Is there a strategy for expressing finitely generated abelian group as the direct sum of cyclic groups?

I know that every finitely generated abelian group can be expressed as a direct sum of cyclic groups. I wondering how easily we can find the cyclic groups given an abelian group. Specifically, one of ...
0
votes
0answers
74 views

Problem from Marcus' Number Fields

I have been stuck on the 27th problem of the 2nd chapter from Marcus' Number Fields. In it we're given $G$, a free abelian group of rank $n$ and its subgroup $H$, which is again of rank $n$. Now, ...
1
vote
1answer
74 views

Homomorphisms $\frac{\mathbb{Q}}{\mathbb{Z}} \longrightarrow \mathbb{Q}$

Can someone please show me a concrete example of a group homomorphism $$ \frac{\mathbb{Q}}{\mathbb{Z}} \longrightarrow \mathbb{Q}, $$ if it exists? I apparently cannot find any of it (except for the ...
1
vote
1answer
42 views

Defining Presheaves on Categories

I'm learning about sheaves and sheaf cohomology with the eventually goal of using these tools to study Riemann surfaces. My references are Forster's Lectures on Riemann Surfaces and Iverson's ...
0
votes
2answers
42 views

Pontryagin Dual of a Finite Abelian Group [closed]

Let $M$ be a finite abelian group. I want to show that the Pontryagin dual is a finite abelian group, and in particular I am interested in computing the elementary divisors/invariant factors of it. ...
3
votes
0answers
70 views

Show that $G$ is Abelian if and only if $f: G\times G \to G$ is a homomorphism.

Let $G$ be a group. Let $H$=$G\times G$ be the direct product of $G$ with itself. Define $f: H\to G$ to be $f((g,h))=gh$ for any $(g,h)\in H$. Show that $G$ is Abelian if and only if $f$ is a ...
0
votes
1answer
22 views

Let A be a finite group and P be a normal p sylow subgroup. What is the connection between P and $Tor_p(A)$

Let A be a finite group and P be a normal p sylow subgroup. can there be an element $g \in A$ where $order(g) = p^x$ where x>0 and $g \notin P$ ? what I really try to understand is the connection ...
2
votes
0answers
23 views

What triple “tensor product” is this? Is it just isomorphic to a double tensor product?

Consider the abelian groups $A = \Bbb{Q}^{\times}, B = \Bbb{Q}^{\times}, C = \Bbb{Z}^+$. What if we formed a product like: $A \star B \star C = \text{Free}_{\Bbb{Z}}(A \times B \times C)$ ...
1
vote
1answer
38 views

Finite abelian p-group with an element of maximal order

I want to know, following theorem comes from which book? Theorem . Suppose $G$ is a finite abelian $p$-group and $a \in G$ has maximum order, then there exists a subgroup $K⊆G$ such that: $ \langle ...
1
vote
2answers
84 views

Any group of order $15$ is abelian(without sylow theorem)

Prove that, any group of order $15$ is abelian (without help of Sylow's theorem or its application). What I have done so far is, by class equation we know that $|G|=|Z|+\sum\frac{|G|}{C(a_i)}$. Now ...
2
votes
1answer
33 views

Left adjoint to forgetful from modules to abelian groups

What is the left adjoint to the forgetful functor $U : R-\mathsf{Mod} \to \mathsf{Ab}$? Note here that $R$ is a general ring, not necessarily commutative. I've seen that they define it as $F A = R \...
0
votes
0answers
36 views

Prove that if $A,B,C$ are finite commutative groups and $A\times B\cong A\times C$ then $B\cong C$. [duplicate]

Prove that if $A,B,C$ are finite commutative groups and $A\times B\cong A\times C$ then $B\cong C$. Since $A,B,C$ are finite commutative groups hence we can write $A=\Bbb Z_{p_1^{\alpha_1}}\...
0
votes
2answers
29 views

Maximal subgroups of Z

In the ring of integers, the only maximal ideals are those generated by the prime elements. Is the same true for the group of integers? Are the only maximal subgroups of integers the ones generated by ...
0
votes
0answers
40 views

On conjugacy classes of involution

Let $G$ be a finite solvable group and $H$ be a normal subgroup of $G$. If all the involution of $H$ lie in a conjugacy class of $G$, then what can we say about the structure of $H$?
-1
votes
2answers
51 views

If K is a finite field, proof that $Gl_n(K)$ is not commutative [closed]

The following property was stated during a lecture in Algebra: If K is a finite field and $n \ge 2$ then $Gl_n(K)$ is a non-abelian finite group. I know how to proof that $Gl_n(K)$ is finite but,...
0
votes
2answers
48 views

Show that $o(g_1\cdot g_2)=o(g_1)\cdot o(g_2)$ [duplicate]

Let $G$ be an abelian group, $g_1,g_2\in G$ of finite order ($o(g_1)=m,o(g_2)=n)$ with $(o(g_1),o(g_2))=1$ (relatively prime). Show that $o(g_1\cdot g_2)=o(g_1)\cdot o(g_2)$. I have tried the ...
0
votes
3answers
169 views

Group Theory vs Graph Theory [closed]

I would like to know that, For a given graph can we find an associated finite group? If there are more than one such group, what are the differences and similarities between them? Here ...
2
votes
0answers
44 views

Isomorphic product of finite abelian groups

Suppose $X,Y,Z$ are finite abelian groups with $X \times Y \cong X \times Z$. How to show that $Y\cong Z$? If we assume that we can decompose $Y,Z$ into cyclic groups that are powers of primes, I ...
0
votes
1answer
61 views

Center of a group $G$, when the commutator subgroup has index 2 [closed]

Suppose that $G$ is a finite group, $M=G^{'}Z(G)$, $|\frac{G}{M}|=2$ and there is an element $x\in G$ such that $|C_G(x)|=4$. Is it true that $|Z(G)|>1$ ? ($G^{'}$ is the commutator subgroup and $Z(...
0
votes
3answers
38 views

A finite abelian group must contain an element which is the l.c.m. of the orders of its elements.

Let $G={g_1,...,g_n}$ be a finite abelian group of order $n$ and let $m =$ l.c.m.$(|g_1|,...,|g_n|)$. Since $G$ is finite (without loss of generality) suppose $g_1\cdots g_n = g_1$. We know $(g_1\...
1
vote
1answer
109 views

Abelian group in short exact sequence

If we have a short exact sequence of continuous group homomorphisms between abelian groups $$0 \rightarrow \mathbb{Z} \oplus \mathbb{Z} \rightarrow X \rightarrow \mathbb{Z} \rightarrow 0,$$ can we ...
1
vote
1answer
40 views

Unique subgroup of index 2 in a finite abelian group.

Suppose $G$ is a finite abelian group, all elements of which are their own inverse. If the order of $G$ is greater than $2$, then prove or disprove that the subgroups of index $2$ in $G$ are not ...
0
votes
0answers
25 views

Property of Abelian Group

I am reading the Handbook of the Mathematical Logic and in the page 8 say: An abelian group $G$ has every element of order $\leq$ n if $G$ is a model of $\forall x [ x = 0 \vee 2x = 0 \vee \...
1
vote
2answers
69 views

What is $\mathbb Z \oplus \mathbb Z / \langle (2,2) \rangle$ isomorphic to?

This question came up after I'd solved the following exercise: Determine the order of $\mathbb Z \oplus \mathbb Z / \langle (2,2) \rangle$. Is the group cyclic? I had no trouble solving the ...
2
votes
0answers
41 views

Index of a maximal subgroup among normal abelian subgroups

Let $P$ be a $p$-group and $A$ maximal among abelian normal subgroups of $P$. Show that: 1) $A=C_P(A)$. 2) $|P:A|\mid (|A|-1)!$. 1) If $A$ is an abelian normal subgrup of a certain group $...
1
vote
1answer
102 views

Doubt about a kernel

I was reading the proof of Lemma 1.25 in this thesis and I thought I understood it, but I think I don't. The thing that I don't see clearly is in page 26 where he is showing that $\textrm{ker}\ \eta\...
3
votes
1answer
40 views

What is $\mathbb Z^2/\text{Im}(\phi)$ isomorphic to in the following case?

Let $\phi:\mathbb Z^2\to\mathbb Z^2$ be the map $(x,y)\mapsto (x+y,2y)$. I need to find $\mathbb Z^2/\text{Im}(\phi)$. My guess is that this is isomorphic to $\mathbb Z_2$ but I am having ...
1
vote
2answers
50 views

The center of a group is an abelian subgroup

Let $(G,\circ)$ be a group and let $Z(G):=\{x \in G : ax=xa \ \forall \ a \in G\}$ be the center of $G$. How can I show that $Z(G)$ is an abelian subgroup of $G$? What I did so far: $Z(G)$ is a ...
-1
votes
3answers
48 views

Submodules of a module with a given property

I am curious about the submodules of a module with a given property. Let $M$ be an $R$-module. If $M$ is a finitely generated are the submodules of $M$ finitely generated? If $R=\mathbb Z$, $M$ ...
1
vote
1answer
44 views

Number of Nonisomorphic Subgroups of Finite Abelian Group

Lets say I have an abelian group $G$ with order $n$ and I am given the primary components of $G$ and their type. How can I determine how many nonisomorphic subgroups of $G$ there are? And as an ...
0
votes
2answers
29 views

Let G be a finite Abelian group of order $p^nm$, where p is a prime that does not divide m. Then $G=H\times K$ where H and K are the following sets.

I'm trying to follow this proof in my textbook, Contemporary Abstract Algebra by Gallian (p231) but I'm having trouble understanding what's going on. He writes Let G be a finite Abelian group ...
1
vote
0answers
70 views

Questions about completion chapter in Atiyah-Macdonald

I was reading the completion chapter of Atiyah-Macdonald. I have the following questions: (i) What is the topology in the completion group of the topological abelian group? I saw an answer here. But ...
1
vote
1answer
73 views

How do I prove this seemingly obvious property of subgroups

The statement is the following: Given an abelian group $G=\langle a_1,...,a_t\rangle$, and a subgroup $H$ of $G$, we need at most $t$ elements to generate $H$; i.e. $H=\langle b_1,...,b_t\rangle$ for ...