0
votes
0answers
23 views

Prove that $U(m) = U_{m/n_1} (m) \times U_{m/n_1} (m) \times \cdots\times U_{m/n_k}$

If $m = n_1 n_2 \cdots n_k $ where $\gcd(n_i~,n_j)=1 ~~ \forall i \neq j$, then prove that: $$U(m) = U_{m/n_1} (m) \times U_{m/n_1} (m) \times \cdots\times U_{m/n_k}$$ where $\times$ refers to the ...
0
votes
0answers
40 views

Let G be a finite abelian group. Let p be a prime number.

i) Let $G$ be a finite abelian group. Prove that the product of all elements in $G$ has order $2$. ii) Let $p$ be a prime number. Use (i) to prove: $((Z/pZ)\setminus\{0\}, \cdot )$ only has one ...
0
votes
1answer
18 views

Evidence about the group $\left ( (\mathbb{Z}/p\mathbb{Z})^*,\underset{p}{ \odot} \right )$

Be $p$ an odd prime number. Show that the group $\left ( (\mathbb{Z}/p\mathbb{Z})^*,\underset{p}{ \odot} \right )$ has a unique element of order $2$, namely $\overline{p-1}$, and show that ...
1
vote
1answer
18 views

Be $G=\{ e,g_1,g_2,\ldots, g_n \}$, $|G|=n+1$. Suppose $G$ has a unique element of order $2$, say $g_1$. Show that $eg_1g_2\ldots g_n=g_1$.

Be $G=\{ e,g_1,g_2,\ldots, g_n \}$ an abelian group of order $n+1$. Suppose $G$ has a unique element of order $2$, say $g_1$. Show that $eg_1g_2\ldots g_n=g_1$. I have serious difficulties with ...
0
votes
0answers
37 views

Let $G$ be a finite abelian group and let $p$ be a prime that divides order of $G$. then $G$ has an element of order $p$

Let $G$ be a finite abelian group and let $p$ be a prime that divides order of $G$. then $G$ has an element of order $p$ Proof When $G$ is abelian. First note that if $|G|$ is prime, then $G \approx ...
5
votes
2answers
95 views

Equivalence relation to make a group commutative

A while ago I was wondering if there is a "natural" way to make a commutative group out of an arbitrary one. I played with the idea a bit and here is what I came up with. Define a binary relation ...
3
votes
2answers
87 views

If the intersection of a normal subgroup and the derived group is {e}, show that N is a subset of Z(G).

I think my reasoning is wrong, but if the intersection only contains the identity, doesn't that imply that the only commutator in N is {e}, so doesn't that mean N is automatically commutative? Why was ...
0
votes
0answers
46 views

Let G be a group and let diag G denote the subset of GxG defined by diag G:={(g,g) |g∈G}(The subset diag G is known as the diagonal of G in GxG.) [on hold]

Let G be a group and let diag G denote the subset of GxG defined by diag G:={(g,g) |g∈G}(The subset diag G is known as the diagonal of G in GxG.) (a) Show that diag G is a subgroup of GxG. (b) Show ...
2
votes
2answers
25 views

Supose that G is a finite abelian group that does not contain a subgroup isomorphic to $Z_p \oplus Z_p$ for any prime $p$. Prove that G is cyclic.

Supose that G is a finite abelian group that does not contain a subgroup isomorphic to $Z_p \oplus Z_p$ for any prime $p$. Prove that G is cyclic. Attempt: If $G$ is a finite abelian group, then let ...
2
votes
1answer
19 views

Determining the structure of the abelian group, integral matrix

I am revising for my upcoming university exams and I have a past exam question that I am finding particularly challenging... a) Consider the integral matrix $$R=\begin{bmatrix} 2 & 2 & ...
1
vote
1answer
38 views

Can we give of the fact that a group of order $9$ is abelian without using an argument involving the product of two cyclic groups of order $3$?

A group of order $9$ is always abelian. I've seen proofs of this result, but I would like to prove it the following way: Let $G$ be a group of order $9$. If $G$ has an element $a$ of order $9$, then ...
-1
votes
1answer
40 views

conjugacy classes and order of group

Suppost that $k_G(A)$ denotes the number of conjugacy classes of $G$ that intersects $A$ non-trivially ($A$ is an arbitrary subset of $G$) and $M=G^{'}Z(G)$. Also suppose that $G$ is non-solvable, ...
2
votes
1answer
51 views

Isomorphisms between finite abelian groups and cyclic groups

If G is abelian of order 175 and H is cyclic of order 25 and there is a homomorphism from G onto H then what is G isomorphic to? I can see how G is isomorphic to either $C_{25} * C_7$ or to $C_5 * ...
0
votes
2answers
45 views

If $G$ is non-abelian, then $Inn(G)$ is not a normal subgroup of the group of all bijective mappings $G \to G$

Let $(G,\cdot)$ be a group and let $\mathfrak{S}(G)$ be the set of all bijective mappings from $G$ to $G$. Show that: If $G$ is non-abelian, then $Inn(G):=\{\kappa_a \vert a\in G\}$ is not a ...
1
vote
1answer
33 views

Number of elements in a group

The group $G$ consists of the binary strings of length $5$ under addition $\mod 2$ in each component. (It is isomorphic to $(\mathbb Z_2)^5$, the direct product of $5$ copies of $\mathbb Z_2$.) Let ...
2
votes
2answers
79 views

How many subgroups of $\Bbb{Z}_4 \times \Bbb{Z}_6$?

I have been trying to calculate the number of subgroups of the direct cross product $\Bbb{Z}_4 \times \Bbb{Z}_6.$ Using Goursat's Theorem, I can calculate 16. Here's the info: Goursat's Theorem: Let ...
4
votes
1answer
121 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 ...
1
vote
2answers
74 views

$g\in G$ maximal order in $G$ abelian then $G=\left<g\right>\oplus H$

If $g\in G$ is an element of maximal order in a finite abelian group $G$ then exists $H\leq G$ such that $G=\left<g\right>\oplus H$ Attempt: Using fundamental theorem I know that ...
0
votes
1answer
50 views

Subgroups of finite Abelian groups

I am interested in finding all of the subgroups (up to isomorphism) of a finite Abelian group $A$. I know the following: -- A finite Abelian group $A$ can be represented as a direct product of ...
2
votes
1answer
36 views

Let $G$ be an Abelian group with odd order. Show that $\varphi : G \to G$ such that $\varphi(x)=x^2$ is an automorphism

Let $G$ be an Abelian group with odd order. Show that $\varphi : G \to G$ such that $\varphi(x)=x^2$ is an automorphism. I was able to show that the $\varphi$ function is a homomorphism and ...
2
votes
1answer
39 views

Let $G$ be a group, where $(ab)^3=a^3b^3$ and $(ab)^5=a^5b^5$. Prove that $G $ is an abelian group.

Let G be a group, where $(ab)^3=(a^3)(b^3)$ and $(ab)^5=(a^5)(b^5)$. Prove that $G$ is abelian group. Thank you in advance. Any help is appreciated.
1
vote
1answer
23 views

Does this condition gaurantee the cyclicity of a finite abelian group?

Let $G$ be a finite abelian group in which there are at most $n$ solutions of the equation $x^n = e$ for each posivite integer $n$. How to determine if $G$ is cyclic or not?
0
votes
1answer
44 views

Which elements of this cyclic group would generate it?

Let $n$ be a given arbitrary positive integer, and let $U_n$ denote the group of all the positive integers less than $n$ and relatively prime to $n$ under multiplication mod $n$. Then for which values ...
0
votes
2answers
68 views

Possible difference between $\mathbb{Z}$-modules and vector spaces

Suppose $G$ is a free abelian groups, i.e. a free $\mathbb{Z}$-module; we have a set $S \subset G $ such that $S$ spans $G$. Can we conclude that the rank of $G$ as a $\mathbb{Z}$-module is $ \leq ...
0
votes
2answers
38 views

Abelian and residually finite groups

If $G$ is a finitely generated abelian group then $G$ is residually finite. I don't know if the result holds or not. I tried to follow the definition but could not go far. Any hint will be highly ...
2
votes
1answer
49 views

Maximal $\mathbb{Z}$-submodules in $\mathbb{Q}$

Is it true that $\mathbb{Q}$ viewed as $\mathbb{Z}$-module ( i.e. abelian group ) has not maximal $\mathbb{Z}$-submodules ? Why ?
0
votes
1answer
39 views

Commutative group

Let $A$ be an nonempty set, and $f: A^3 \rightarrow A$ mapping which satisfies: $f(x,y,y)=f(y,y,x)$ for each $x,y \in A$ ...
0
votes
1answer
23 views

Order of elements in a disjoint cycle

What's the difference between a subgroup and a cyclic subgroup? $A_4 = \{e,(123),(132),(124),(142),(134),(143),(234),(243),(12)(34),(13)(24),(14)(23)\}$ And if I was looking for a subgroup of order ...
2
votes
4answers
83 views

Order of elements in a disjoint cycle

Just wondering how to find the order of each element in this group: $A_4 = \{e,(123),(132),(124),(142),(134),(143),(234),(243),(12)(34),(13)(24),(14)(23)\}$ I tried writing each elements not in ...
1
vote
1answer
29 views

Let $A$ finitely generated abelian group, and $A_1 \le A$. I have to prove that $rk(A)=rk(A_1)+ rk(A/A_1)$

Let be $A$ a finitely generated (f.g.) abelian group, and let be $T$ its torsion, then by the structure theorem of f.g. abelian gruops we have that $A/T \simeq \mathbb{Z}^d$, so we can define $b$ the ...
4
votes
1answer
39 views

uniqueness of groups in an exact sequence

I was wondering how unique are the groups making up to an exact sequence. Suppose we have three groups $A, B, C$ such that the sequence $$ A \rightarrow B \rightarrow C $$ is exact. I wanted to know ...
6
votes
2answers
58 views

If $G$ is a finite union of some of its abelian subgroups, then the index of the center of the group is finite

If $G$ is a finite union of some of its abelian subgroups, then the index of the center of the group is finite Would I not simply state that by Lagrange's theorem, $Z(G)$ can divide into the ...
3
votes
1answer
38 views

Shorter proof for some equvalences

Let $(G,\cdot)$ be a group show that A) $G$ is abelian B) For all $x,y\in G: (xy)^{-1}=x^{-1}y^{-1}$ C) For all $x,y\in G: (xy)^{2}=x^{2}y^{2}$ D) There existst an $n\in \mathbb{Z}$ such that for ...
3
votes
1answer
108 views

Is the group $(G,*)$ abelian?

Let $(G,*)$ be a finite group in which the sets $C_a$={$x\epsilon G$|$ax=xa$} have the same cardinality, for all $a \epsilon G$ \{e}. My question is: is the group abelian?
4
votes
2answers
103 views

Explicitly computing the isomorphism class of the tensor product of two finite abelian groups

How do I compute the isomorphism class of $A\otimes_\mathbb{Z} B$, where $A$ and $B$ are abelian of finite order? I can do this for a few examples, but I am unsure of how to proceed in the ...
3
votes
1answer
51 views

isomorphism between divisible, totally ordered, abelian groups

Let $G$, $H$ be divisible, abelian, linearly ordered groups, whose cardinalities are equal and satisfy $\mu := |G|=|H|>\aleph_{0}$. These are supposed to be (order!) isomorphic. And just about ...
0
votes
1answer
61 views

Using the conjugacy class equation [duplicate]

Let $G$ be a group of order $p^2$. Use the class equation to prove that $G$ is abelian. The conjugacy class equation, at least how I remember it, is $$ |G| = |Z(G)|+\sum_{x\in I \backslash Z(g)} ...
0
votes
1answer
32 views

Subgroups of a finite abelian group

Let $G$ be a finite abelian group, and let $K$ be a subgroup of $G$. Does $G$ necessarily have a subgroup $H$ such that $H\cong G/K$ and $H\cap K=\langle 0\rangle$? I'm not sure where to start.
0
votes
2answers
58 views

Abstract Algebra. Let $\mathit{G} $ be an abelian group. Show that the elements of finite order in $\mathit{G}$ form a subgroup of $\mathit{G}$.

Let $\mathit{G} $ be an abelian group. Show that the elements of finite order in $\mathit{G}$ form a subgroup of $\mathit{G}$, called the torsion subgroup of $\mathit{G}$. let $g \in G$ I know that ...
2
votes
1answer
100 views

Classifying abelian groups up to isomorphism

List all abelian groups (up to isomorphism) of the given orders: a) $144$, b) $600$ a) For order $144$, I feel confident with this one so far: $\mathbb{Z}_4 \oplus \mathbb{Z}_{36}$ Elementary ...
3
votes
1answer
77 views

Finite abelian p-group and an element of maximal order

I'm studying for an exam and I'm having trouble understanding the proof given for the following statement: Suppose $G$ is a finite abelian $p$-group and $a \in G$ has maximum order, then there ...
-2
votes
1answer
36 views

Prove that the following are equivalen for abelian group

Let $(G, *)$ be a group. Prove that the following are equivalent: a. $G$ is abelian. b. $aba'b' = e$ for all $a,b \in G$. c. $(ab)^{2}$ = $a^2b^2$ for all $a, b \in G$.
1
vote
1answer
49 views

subgroup proof.

Prove that if $G$ is an abelian group, then $H =\{ x \in G\mid x^{2} = e \}$ is a subgroup of $G$. I did show that $H$ is close, associative, have identity and inverse element. Then my prof said I ...
4
votes
1answer
54 views

Splitting of exact sequence of groups when middle group has split subgroup.

I am trying to show that a short exact sequence of abelian groups splits. I have a short exact sequence, $$0\rightarrow \mathbb{Z} \rightarrow G \rightarrow \mathbb{Z}_2 \rightarrow 0$$ and I know ...
0
votes
1answer
191 views

Question about Finite Abelian Groups [duplicate]

Let $(G, .)$ be a finite abelian group, $G=\{x_1, ..., x_n\}$ and let $x=x_1. \cdots. x_n$. Show that $x^2=e$ Suppose $G$ has no element of order 2 or that $G$ has more than one element of order 2. ...
2
votes
1answer
125 views

Classify Artinian $\Bbb Z$-modules

How can I classify Artinian $\mathbb{Z}$-modules as Noetherian $\mathbb{Z}$-module? (A $\mathbb{Z}$-module is Noetherian iff it is finitely generated). Any hint will be helpful. I have seen the ...
0
votes
3answers
89 views

Commutator Subgroup is Normal Subgroup of Kernel of Homomorphism

Please help to understand this problem. Let $G$ be a group, $H$ an abelian group, $\phi : G \rightarrow H$ a homomorphism. Show that $C(G) \lhd \mathrm{Ker}(\phi)$ I must be misunderstanding ...
1
vote
1answer
66 views

Prove that a group is abelian if every element commutes with exactly K other elements

Let $(G,*)$ be a finite group with the property that every element, aside the neutral element $e$, commutes with exactly K other elements. Alternatively speaking, the centralizer of every element is a ...
0
votes
2answers
56 views

“Classify $\mathbb{Z}_5 \times \mathbb{Z}_4 \times \mathbb{Z}_8 / \langle(1,1,1)\rangle$”

I have a question that says this: Classify $\mathbb{Z}_5 \times \mathbb{Z}_4 \times \mathbb{Z}_8 / \langle(1,1,1)\rangle$ according to the fundamental theorem of finitely generated abelian groups. ...
0
votes
1answer
52 views

If a group has only one commutator, why does that mean it is abelian?

I understand that if $aba^{-1}b^{-1} = e$ then $ab$ is commutative, but I don't see how having multiple commutators will prevent the group from being abelian