0
votes
2answers
29 views

Prove that the order of $U(n)$ is even when $n>2$.

I'm trying to provide a solution to the following claim: "The order of $U(n)$ is even when $n>2$." Note: here, $U(n)$ is the set of all positive elements that are less than and relatively prime ...
2
votes
2answers
45 views

How to count the number of elements of given order?

I am trying to prove the following result. Let $G$ and $G'$ be two finite abelian groups. Besides, they have the same number of elements of any given order. Prove that $G\cong G'$. My attempt is ...
2
votes
4answers
150 views

there is no injective group homomorphism from $\mathbb Z\times\mathbb Z$ into $\mathbb Z$

there is no injective group homomorphism from $\mathbb Z\times\mathbb Z$ into $\mathbb Z$ But i don't know why it is true. should i investigate all group homomorphisms from $\mathbb Z\times\mathbb ...
3
votes
3answers
148 views

Basic Group Theory question

This is not so much a plea of ignorance, but rather me trying to see whether intuitively I actually understand what is going on in group theory. The question asks What group is ...
0
votes
1answer
47 views

generators of groups from exact sequence

Suppose I have a middle term exact sequence of finitely generated abelian groups $G \longrightarrow H \longrightarrow K$. How do I get the generators of $H$ if I know the same for other two groups?
0
votes
0answers
45 views

Finite generated abelian group $G$ and $H<G$. What is the rank of $(G/H)/(G/H)_t$?

I saw another question about this problem here. However there are quite different answers from my expectation. Anyway, here are my trials. Trial 1 : By structure theorem, $G\cong G_t\oplus F_1$ ...
5
votes
1answer
97 views

Additive non-abelian group?

Sometimes I see in books the term "additive abelian groups". In my opinion, when we use addition to represent the group operation, we already have in mind that the operation is commutative. So ...
3
votes
5answers
87 views

Proof that a group is abelian.

If $(G,*)$ is a group and $(a * b)^2 = a^2 * b^2$ then $(G, *)$ is abelian for all $a,b \in G$. I know that I have to show $G$ is commutative, ie $a * b = b * a$ I have done this by first using ...
1
vote
1answer
46 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 $[x,_0\, y]=x$; ...
4
votes
4answers
67 views

$n$th power map is an automorphism implies abelian group?

If $G$ is a finite group and $\phi(x) = x^n$ is an automorphism of $G$ does this imply $G$ is abelian? I've been reading this page. Def: A group $G$ is said to be $n$-abelian if $(ab)^n=a^nb^n$ ...
1
vote
2answers
37 views

A finite $\mathbb{Z}$-module whose submodules are totally ordered by inclusion.

I have the following problem: Let $M$ be a finite $\mathbb{Z}$-module such that set of the submodules is totally ordered by inclusion. Prove that there exist a prime $p$ such that $|M|=p^\alpha$ ...
2
votes
1answer
37 views

Show that, given an element $\bar{x}\in\mathbb{Q}/\mathbb{Z}$, there is an integer $n \ge 1$ such that $n\bar{x} = 0$. [duplicate]

Consider $\mathbb{Z}$ as a subgroup of the additive group $\mathbb{Q}$ of rational numbers. >Show that, given an element $\bar{x}\in\mathbb{Q}/\mathbb{Z}$, there is an integer $n \ge 1$ such that ...
2
votes
1answer
48 views

How to prove that $\mathbb Z/p$ is not a direct summand of any direct sum of copies of $\mathbb Z/n$?

How can I prove that $\mathbb Z/p$ ($p$ is a prime) cannot be a direct summand of any arbitrary direct sum of copies of $\mathbb Z/n$, where $p^2$ divides $n$?
5
votes
1answer
41 views

Let $G$ be a group and $a,b,c \in G$. Given that $abc$ and $cba$ are conjugated, prove that $G$ is abelian.

Let $G$ be a group and $a,b,c \in G$. Given that $abc$ and $cba$ are conjugated, prove that $G$ is abelian. In other words, if for any $a,b,c \in G$ there is a $g \in G$ so that $a b c = g c b a ...
2
votes
1answer
43 views

$\operatorname{Hom}(G,\Bbb C\setminus \{0\})$ non- isomorphic to $\operatorname{Hom}(G, \Bbb T)$?

Do you have an example of an abelian group $G$ for which $\operatorname{Hom}(G,\Bbb C\setminus \{0\})$ is not isomorphic to $\operatorname{Hom}(G, \Bbb T)$? $\Bbb C$ is the complex plane and $\Bbb ...
2
votes
2answers
61 views

The factor group $\mathbb{R}^{*}/\{-1,1\}$ is isomorphic to $\mathbb{Z}_{2}$. True or False.

I have been told that the answer for this question is False. And I'm trying to understand why. For what I understand so far, $\mathbb{Z} _2$ is an abelian group. I also understand that for a group ...
2
votes
1answer
74 views

What we know about $\mathbb{Q}/\mathbb{Z}$ as a group? [closed]

What we know about $\mathbb{Q}/\mathbb{Z}$ as a group? Are there any interesting properties? Best regards.
0
votes
0answers
58 views

Free groups vs. free abelian groups

I'm trying to solve this question in page 74 of Hungerford's book: A free abelian group is a free group (Section I.9) if and only if it is cyclic. I have no idea how to proceed, a solution or a ...
1
vote
2answers
529 views

What does it mean for a group to be Abelian?

I'm revising questions on groups for exams, and I still can't quite understand what an Abelian group is. Please help me understand, if anyone could give me a more simple explanation.
3
votes
1answer
61 views

Normal Abelian Subgroup does not imply Abelian Quotient Group

I'm a bit confused and just need some clarification about what I am missing in this: I have $S_4$ with normal subgroup $N=\lbrace(),(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\rbrace$. I know that $N$ is ...
3
votes
2answers
50 views

$mA = 0 = nC, \ \gcd(m,n) = 1 \Rightarrow $ every extension of $A$ by $C$ splits

This is Exercise 7.14(ii) from Rotman, Introduction to homological algebra, and I'm stuck on it. If $A$ and $C$ are abelian groups, with $mA = 0 = nC $ and $\gcd(m,n) = 1$ then every extension of ...
0
votes
0answers
51 views

Is my solution correct? Finite abelian groups are CLT groups.

I want to solve the following exercise from Dummit & Foote's Abstract Algebra text: Use Cauchy's Theorem and induction to show that a finite abelian group has a subgroup of order $n$ for each ...
0
votes
0answers
30 views

A proposition on exact sequence of inverse limit (Lang, Algebra, p. 165)

I am trying to understand this proof. My only question is that what are the vertical maps here?
2
votes
2answers
62 views

Determine the isomorphism class of M/T(M)

Let $M=\Bbb{Z}\oplus\Bbb{Z}\oplus\Bbb{Z}$ and $T: M\rightarrow M$ given by $T(x,y,z)=(4x+2z,2y,2x+10z)$. Show the cokernel $M/T(M)$ is an abelian group of order $72$, and determine its isomorphism ...
4
votes
1answer
53 views

$\biggl ( \prod_p G_p \biggr) /\biggl( \bigoplus_p G_p\biggr)$ is divisible

Let $G$ be an abelian group, $p$ a prime, then $G_p$ is the $p$-primary component of $G$, i.e. $$G_p = \lbrace g \in G \ | \exists \ n \in \mathbb{N} \ , p^ng = 0\rbrace$$ I have to prove that ...
2
votes
4answers
96 views

Homomorphism from $\mathbb{Z}\oplus \mathbb{Z}_2$ to $\mathbb{Z}$.

What sort of homomorphisms can I have from $\mathbb{Z}\oplus \mathbb{Z}_2$ to $\mathbb{Z}$? What about if I know the homomorphism sends the $\mathbb{Z}$ part to zero. In other words, if I know my ...
1
vote
1answer
44 views

automorphism group of a group of order $p^2$, where $p$ is prime

There is a corollary that states: "If $|P|=p^2$ for some prime $p$, then $P$ is abelian. More precisely, $P$ is isomorphic to either $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_p\times \mathbb{Z}_p$." I know ...
1
vote
1answer
43 views

showing that a group of order 45 is abelian

I'm trying to understand the following proof from Dummit & Foote (pg. 137) which shows why a group of order 45 is abelian. I understand everything but the last two sentences. Why is it that ...
5
votes
0answers
36 views

What are the invariant factors of $(\mathbb{Z}/(1000))^\times$?

I'm curious about the invariant factors of $(\mathbb{Z}/(1000))^\times$. I put down $$ (\mathbb{Z}/(1000))^\times\cong(\mathbb{Z}/(8))^\times\oplus(\mathbb{Z}/(125))^\times $$ It's easy to compute by ...
0
votes
0answers
14 views

Cyclic Factor group abelian proof [duplicate]

Show that if G is nonabelian, then the factor group G/Z(G) is not cyclic. I started to prove this via contrapositive. If G/Z(G) is cyclic, then G is abelian. I'm messing around with elements and ...
1
vote
2answers
29 views

Question about order of product of elements in a group

Let G be a finite abelian group. Prove that the product of all elements in G has order 2. I think i am supposed to use lagrange's theorem but how?
0
votes
1answer
31 views

If $G$ has more than one nontrivial elements with order 2, how to show that $\prod^n_{x=1}a_x=1$? [duplicate]

Let $G$ be an abelian group of order $n$, and $a_1,a_2,...a_n$ its elements. If $G$ has more than one nontrivial elements with order $2$, how to show that $\prod^n_{x=1}a_x=1$?
9
votes
1answer
118 views

Prove $G$ is abelian if $f(f(x)) = x$?

Let $G$ be a finite group and $f$ an automorphism such that $f(f(x)) = x$, and $f(x) = x$ if and only if $x=e$. Prove that $G$ is abelian and $f(x) = x^{-1}$. My attempt: ...
1
vote
1answer
36 views

How to find group homomorphisms from one group to another

I am trying to figure out all the homomorphisms from $\mathbb{Z}_2\times\mathbb{Z}_2$ to $\mathbb{Z}_2$. Is there a good process for doing such a think? I am getting lost...
4
votes
2answers
77 views

A non-abelian group such that $G/z(G)$ is abelian.

I'm looking for an example of a non-abelian group $G$ such that $G/z(G)$ is abelian, where $z(G)$ is the center of the $G$. In other words, I'm looking for a non-abelian group where $z(G)$ contains ...
3
votes
1answer
52 views

Classify abelian groups $A$ which are irreducible $End(A)$-modules

Classify abelian groups $A$ which are irreducible $End(A)$-modules. I think i did it for finite abelian group $A$ . A finite abelian group $A$ is irreducible iff order of $A$ a is power of prime. ...
6
votes
2answers
97 views

Rank of $(G/H)/(G/H)_t$ where $G$ is finitely generated abelian and $H$ is a subgroup.

Let $G$ be a finitely generated abelian group and $H$ be a subgroup. Let subscript $t$ denote the torsion subgroup. If $G/G_t$ is free of rank $n$ and $H/H_t$ is free of rank $m$, it is easy to embed ...
2
votes
2answers
75 views

$\langle x \rangle$ is a direct summand of a finite abelian group where $x$ is maximal order [duplicate]

Let $x$ be an element of a finite abelian group $G$ where $x$ has maximal order. Then I want to show that $\langle x\rangle$ is a direct summand of $G$. Note that I do not want to use finite abelian ...
0
votes
2answers
108 views

$\psi (m)\leq \phi (m)$ or $\psi (m) \geq \phi (m)$ when $\psi (m)\neq 0$?

(This is different than If $x^m=e$ has at most $m$ solutions for any $m\in \mathbb{N}$, then $G$ is cyclic) I was trying to solve this: Let $G$ be a finite abelian group of order $n$ for which the ...
2
votes
1answer
70 views

Let $H$ be a normal subgroup of $G$. If $H$ are $G/H$ are Abelian, should $G$ be abelian?

(a) Let $H$ be a normal subgroup of $G$. If $H$ are $G/H$ are Abelian, should $G$ be abelian? Attempt: : There's a counterexample to this claim, $G=D_3$ which is non abelian. But, what could be ...
0
votes
1answer
33 views

Why is $U(12) = U_{4} (12) ~ U_3(12)?$

Why is $U(12) = U_{4} (2)~ U_3(12)$ Attempt: Any subgroup $U_k(n) = \{x \in U(n)~~|~~x \mod k=1 , k ~|~n \}$ Hence : if $U(12) =\{1,5,7,11\}$ then : $U_4(12) =\{1,5\}$ and $U_3(12)=\{1,7\}$ We see ...
1
vote
1answer
39 views

Uniqueness FTOFAG

How do you prove uniqueness for the fundamental theorem of finite abelian groups? The book I'm using has this not very well written proof that I can't follow. So following this proof, I multiply by ...
0
votes
0answers
30 views

The Basis Theorem for Finte Abelian Groups

I am using Pinter's Abstract algebra book to prove the basis theorem for finite abelian groups (Every finite abelian group is a direct product of cyclic groups of prime power order.) $G$ is an abelian ...
0
votes
1answer
21 views

Abelian groups order help

Let $p$ be a prime number. Find the number of abelian groups of order $p^n$, up to isomorphism when n=2,3, and 5. I know the answer when $n=2$ and 3. And my professor said that there are 7 abelian ...
1
vote
1answer
34 views

direct limit of abelian groups

Let $I$ be a directed set and let $(A_i)_{i \in I}$ be a collection of abelian groups. Let $A = \varinjlim A_i$ be its direct limit. Suppose its maps are $\rho_{ij} : A_i \to A_j$ for $i \leq j$. I ...
0
votes
0answers
31 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
1answer
101 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 ...
3
votes
2answers
104 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 ...
2
votes
2answers
49 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 ...
1
vote
1answer
47 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 ...