All questions with tag [math: abelian-groups]

0

Free abelian group $F$ has a subgroup of index $n$?

Suppose that we have a free abelian group $F$. How can it be proved that $F$ has a subgroup of index $n$ which $n≥1$? Honestly, according to the Theorems, I just know that if we take $X$ as a base for $F$, then $$ F= \bigoplus_{\alpha \in X} \mathbb Z_\alpha \ $$ in which for all $ \alpha \in X$; $\mathbb Z_\alpha \ $ is a copy of $ \mathbb Z $....
2022-07-25 20:47:13
1

Determining the Smith Normal Form

Consider the indispensable matrix $$R = \left(\begin{matrix} 2 & 4 & 6 & -8 \\ 1 & 3 & 2 & -1 \\ 1 & 1 & 4 & -1 \\ 1 & 1 & 2 & 5 \end{matrix}\right).$$ Determine the framework of the abelian team offered by generators and also relationships. $$...
2022-07-25 20:43:11
0

Is it true that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as abelian groups?

I think the answer is yes. Sketch of the proof Consider $\mathbb{R}$ as a vector space over $\mathbb{Q}$. Let $\{e_\lambda:\lambda\in\Lambda\}\subset\mathbb{R}$ be its Hamel basis. Then $\{(e_{\lambda_1},e_{\lambda_2}):\lambda_1,\lambda_2\in\Lambda\}\subset\mathbb{R}^2$ is a Hamel basis of $\mathbb{R}^2$. Consider set theoretic bijection $i:\La...
2022-07-25 20:38:59
1

An abelian group of order 100

The first part of the problem asks you to prove that an abelian group $G$ with order $100$ must contain an element of order $10$. For this part, I use Sylow theorem to list possiblities for $H$ and $K$ where $|H|$=$2^{2}$ and $|K|$=$5^{2}$. $K$ must be normal since there is not subgroup of order $25$ while $H$ might not be normal since $1$$\eq...
2022-07-25 15:50:34
1

Converse of Lagrange's theorem for abelian groups

I'm trying to prove that the converse of Lagrange's theorem is true for finite abelian groups (i.e. "given an abelian group $G$ of order $m$, for all positive divisors $n$ of $m$, $G$ has a subgroup of order $n$"). This is an exercise from a book, and it is in the section on finite abelian groups, so I know I have to use the fundamenta...
2022-07-25 15:24:50
1

exact sequence of finite abelian groups which are squares

Let $0 \to A \to B \to C \to 0$ be an exact sequence of finite abelian groups. Assume that $B$ and $C$ is a square (i.e. there are groups $D,E$ such that $B \cong D^2$, $C \cong E^2$). Does this imply that also $A$ is a square? Of course we may assume that $A,B,C$ are finite abelian $p$-groups for some prime $p$.
2022-07-24 05:54:25
5

How to prove a group has a basis with exactly one element?

I am having problem with the adhering to inquiry. Intend I have a team $H$ which is a subgroup of $\mathbb{Z}\oplus\mathbb{Z}$, such that any kind of component $\begin{bmatrix} a \\[0.3em] b \end{bmatrix}$ is specified as: if $b=0$, after that $a=0$. Just how can I confirm that $H$ has a basis with specifically o...
2022-07-24 05:53:46
2

Classify finitely generated modules over the ring $\mathbb{C}[\epsilon]$ where $\epsilon^2=0$

Classify finitely generated modules over the ring $\mathbb{C}[\epsilon]$ where $\epsilon^2=0$ Since $\mathbb{C}[x]$ s noetherian we have that $\mathbb{C}[x]/(x^2)$ is too. And thus finitely generated $\mathbb{C}[\epsilon]$ modules are also finitely presented. I'm not sure where to go from here. I imagine I need to use the fact that $\epsilon$ is...
2022-07-24 05:30:12
1

Finding subgroups of index 2

Let $G = \prod_{i=1}^\infty \mathbb{Z}_2$ with enhancement mod 2. I am searching for subgroups of index 2. I see that taking the whole room and also getting rid of all series which have a 1 in a particular placement offers a subgroup of index 2. As an example the set of all series $\{(0,\cdot,\cdot,\ldots)\}$ which have a 0 in the first placemen...
2022-07-22 18:46:19
1

Trouble with decomposition of groups of order 2009

An inquiry claims: confirm or confirm or refute that there are just 2 non - isomorphic abelian groups of order 2009. I assume that it holds true due to the fact that I broke up $2009$ right into $7 \times 7 \times 41$ therefore this offers the teams $$\mathbb{Z}_{2009}$$ $$\mathbb{Z}_{7} \oplus \mathbb{Z}_{7\times 41}$$ where the first order s...
2022-07-22 18:46:01
2

Proving that a subgroup of a finitely generated abelian group is finitely generated

An inquiry claims: Using the isomorphism theories or otherwise, confirm that a subgroup of a finitely created abelian team is finitely created. I would certainly claim that for a finitely created abelian team $G$, there exists components $g_1,\dots, g_n$ such that a straight mix of them creates the entire team. Consequently as every component o...
2022-07-22 18:44:25
2

Torsion subgroup quotient

Let G be an abelian team, $T$ the torsion subgroup of $G$. If $G/T$ is torsion - free, after that $T$ and also $G/T$ have to be disjoint. $G=T \bigoplus G/T$ indicates this too. I do not recognize why they are disjoint to begin with.
2022-07-22 18:30:10
1

Finding subgroups of index 2 of $G = \prod\limits_{i=1}^\infty \mathbb{Z}_n$

I considered this question and also its solution. The solution makes use of the reality that every vector room has a basis, so there are vast subgroups of index 2 if $n=p$ where $p$ is prime. Exist vast subgroups of index 2 if $n$ is not prime? The trouble looks the very same (with marginal adjustment), yet the means we located the subgroups i...
2022-07-22 18:01:23
1

What are the Subgroups of the abelian group $\mathbb{Z}_5$?

How can I find the subgroups of abelian team $\mathbb{Z}_5$? From Lagrange is theorem, the dimension of the subgroup need to separate 5 in this instance. So the dimension of the subgroup need to be 1 or 5 ($\mathbb{Z}_5$ itself). Can there be any kind of dimension 1 subgroups of $\mathbb{Z}_5$?
2022-07-22 15:49:00
2

How can I prove an abelian group is not free?

How can I confirm an offered abelian team ; such as $\mathbb{Z}_4$ with enhancement mod 4, is not a free team? Should I take into consideration all the parts of the offered team and also confirm any one of them can not be a basis? Yet this strategy will certainly offer me a great deal of below teams to take into consideration. Exists any kind o...
2022-07-22 15:45:54
0

Normal and Abelian groups?

Possible Duplicate: A, B subgroups of G, B/A abelian. Show that BN/AN is abelian. Let A, B, N be subgroups of a group G such that A $\triangleleft$ B and B/A is Abelian. Also suppose N $\triangleleft$ G. Prove that AN $\triangleleft$ BN and that BN/AN is abelian. I'm awful at these kinds of questions and never know what I need to do! I know...
2022-07-22 15:45:54
1

A nonsplit short exact sequence of abelian groups with $B \cong A \oplus C$

A homework problem asked to find a short exact sequence of abelian groups $$0 \rightarrow A \longrightarrow B \longrightarrow C \rightarrow 0$$ such that $B \cong A \oplus C$ although the sequence does not split. My solution to this is the sequence $$0 \rightarrow \mathbb{Z} \overset{i}{\longrightarrow} \mathbb{Z} \oplus (\mathbb{Z}/2\mathbb{Z}...
2022-07-22 15:39:56
1

Simplify the category of finite abelian groups

Consider the category $\mathsf{FinAb}$ of finite abelian groups. The structure theorem tells us that we can write down a skeleton for this category (a set of representatives for the isomorphism classes), which consists of the groups $\mathbb{Z}/n_1 \oplus \cdots \oplus \mathbb{Z}/n_s$ with positive integers satisfying $n_1 | \cdots | n_s$. But ...
2022-07-22 15:32:23
1

how to show that a group is elementarily equivalent to the additive group of integers

Is there any kind of rather very easy means of revealing a team is elementarily equal to the additive team of the integers? I've located a straightforward characterization below: A ‘natural’ theory without a prime model , yet the evidence in Szmielew is paper is fairly lengthy and also far more basic, while I'm seeking something extra primary. ...
2022-07-22 15:25:13
1

Show that fiber products exist in the category of abelian groups.

Show that fiber products exist in the category of abelian groups. In fact, If $X, Y$ are abelian groups with homomorphisms $f: X \to Z$ and $g: Y \to Z$ show that $X \times_z Y$ is the set of all pairs $(x, y)$ with $x \in X$ and $y \in Y$ such that $f(x) = g(y)$.
2022-07-22 14:56:23