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7

Any element of order $p^2$ lies in a $p$-Sylow subgroup. Since all $p$-Sylow subgroups are conjugated, it is enough to consider the canonical $p$-Sylow subgroup given by upper triangular matrices with diagonal $1,\dotsc,1$. But if $A$ is such a matrix, then $(A-1)^n=0$ and hence $(A-1)^p=0$, i.e. $A^p=1$.


7

Try $$(x_1,x_2,\ldots)\mapsto \sum_{k\in\mathbb N}\frac 1kx_k$$ Note that the sum is in fact finite.


5

Answer in the form of a keyword which will answer your question and its generalization to $\mathbb Z^n$: Smith normal form.


5

Yes, the prime numbers generate the positive rationals under multiplication: every element can be written as a product of powers of primes (corollary to existence of prime factorizations), and they satisfy no multiplicative relations (corollary to uniqueness of prime factorizations) so it is freely generated by them. The free abelian group generated by a set ...


5

Unless I'm missing something obvious, this should be true. I'm not sure what you mean by $2 \times 3 \times 5 \times 7 \times 11 \times 13...$ not being a positive rational. In the group theory setting we only consider finite words in the generators. To put it another way, the free abelian group over some set $\Lambda$ is isomorphic to $$\bigoplus_{\lambda ...


4

Given a ring $R$, the notation $R^{\times}$ is used to denote the multiplicative group of units in $R$. Recall that $r \in R$ is a unit if there is $s \in R$ such that $rs = sr = 1$. If $R$ is a field, then every non-zero element has a multiplicative inverse, so $R^{\times} = R\setminus\{0\}$. As $\mathbb{C}$ is a field, $\mathbb{C}^{\times} = ...


4

This is Theorem 17.2 in Fuchs' "Infinite Abelian Groups" (vol.1), the result is originally due to Prüfer and Baer: Theorem. A bounded Abelian group is a direct sum of cyclic groups. As a corollary you obtain that any indecomposable bounded Abelian group is cyclic.


3

I use the answer given by Rory Daulton to your previous question. Take for $n\geq 1$ $H_n=\{2^{n-1}\}\cup 2^n \mathbb{Z}$. If $a,b\in H_{n+1}$, then $a,b\in 2^n\mathbb{Z}$, hence $a-b\in 2^n\mathbb{Z}\subset H_n$. We also have by the above that $2^{n-1}$ is not in $\{a-b, a,b\in H_{n+1}\}$. Hence your hypothesis are satisfied. But for any $n\geq 1$, ...


3

If $x\to x^{-1}$ is homomorphism so $(ab)^{-1}=a^{-1}b^{-1}$ so $$b^{-1}a^{-1}=a^{-1}b^{-1}$$ which you wanted to omit it. Now $$b=aba^{-1},~~~ab=a^2ba^{-1}$$ and so $a^2b^2=a^2b(aba^{-1})=a^2(bab)a^{-1}=(a^2ba^{-1})(a^2ba^{-1})=(ab)^2$. Therefore $x\to x^{2}$ is group homomorphism. If you do the other side iii to ii, you'll see that we cannot escape from i. ...


2

$f(ab) = abab = f(a) f(b) = aa bb$. Hence $a^{-1} f(ab) b^{-1} = ba = a^{-1} f(a) f(b) b^{-1} = ab$.


2

Hint: Why does the group have an element of order $p_i$ for each $i$? Once you've figured this out, think about how to "combine" these elements to get a generator for the group.


2

$$ \bigoplus_{n\in\Bbb N}\Bbb Z\simeq\bigoplus_{\text{$p$ prime}}\bigoplus_{t\in\Bbb N}\frac1{p^t}\Bbb Z\stackrel{\Sigma}\longrightarrow\Bbb Q. $$ Here $\Sigma$ denotes the addition map. Obviously this map has a big kernel.


2

A variant on user148212' s argument . Let $g$ be a $p$-element in $G$, that is, $g^{p^{k}} = 1$ for some $p$. Since we are in characteristic $p$, we have $(g - 1)^{p^{k}} = 0$, that is, $g-1$ is nilpotent. Then proceed as in Martin Brandenburg's answer, $(g-1)^{p} = 0$, so that reversing the argument $g^{p} = 1$.


2

You are nearly there. Let $x \in G$. If $x \notin H$, you have that $\langle H, x \rangle > H$, so that $\langle H, x \rangle \cap S \ne \{ 0 \}$. That is, there are an integer $n$, and $h \in H$ and $s \in S$ such that $0 \ne nx + h = s$, that is $n x = -h + s \in H + S$. Clearly $n \ne 0$, as $H \cap S = \{ 0 \}$.


2

The reason is because $\mathbb Z$ is cyclic and infinite - it is generated by $1$. Hence, the image of a homomorphism $\theta:\mathbb Z \to \mathbb Z$ will be generated by $\theta(1)$. If $\theta(1) = 0$ this gives the trivial homomorphism. Otherwise, $\theta(1) = a \in \mathbb Z$ and therefore $\theta(n) = an \quad \forall n \in \mathbb Z$. In general, ...


2

One issue is torsion. If $f:\mathbb{Z}\to A$ is a homomorphism of abelian groups, we have $\ker f = n\mathbb{Z}$ for some $n$. If $n>1$, then $f(\mathbb{Z})$ has $n$-torsion, namely $n\cdot f(1) = f(n) = f(0)$. So if $A$ is a group with no torsion (say, a free abelian group), the only morphisms $f:\mathbb{Z}\to A$ are either trivial ($n=1$) or ...


1

One possible answer is that any such endomorphism becomes multiplication by an element in the ring $Z$ (the ring of endomorphisms of Z), and so lifts to an endomorphism on the field of fractions (the rationals), where it is seen to be injective - in general field homomorphisms are injective because fields have no nontrivial ideals. P.S. The natural numbers ...


1

The definite source is Maximal Orders by I. Reiner. My hopefully mostly accurate recollections below. If $G$ is cyclic of order $n$, then $\Bbb{Q}[G]\cong \Bbb{Q}[x]/\langle x^n-1\rangle$. The polynomial $x^n-1=\prod_{d\mid n}\Phi_n(x)$ is a product of pairwise distinct irreducible cyclotomic polynomials $\Phi_n(x)$. By the Chinese remainder theorem we thus ...


1

It is enough to describe $\mathbb{R} \otimes_{\mathbb{Z}} \mathbb{R}$. If $c=|\mathbb{R}|$, there is an isomorphism of $\mathbb{Q}$-vector spaces, hence of abelian groups $\mathbb{R} \cong \mathbb{Q}^{(c)}$. It follows $\mathbb{R} \otimes_{\mathbb{Z}} \mathbb{R} \cong \mathbb{Q}^{(c \times c)} \cong \mathbb{Q}^{(c)} \cong \mathbb{R}$. Notice that this is ...


1

No-Sylow proof: Just conjugate over $\bar{\mathbb{F}}_q$ to reach the Jordan normal form of this element, then the remaining is the same as Martin Brandenburg's answer.


1

Hint: index the copies of $\mathbb Z$ with the primes and send $(0,0,\dots,0,1,0, \dots)$ with $1$ on the copy indexed by $p$ to $p\in\mathbb Q$. (Note: this is for $(\mathbb Q_{>0}, \cdot)$. The same idea works more generally for countable abelian groups, the only work is finding a bunch of generators.) Edit: Here's the general idea: let $A = \{a_1, ...


1

Any short exact sequence whose first map is $f$ will be isomorphic to your original short exact sequence. Thus, your conjecture can only be true if the projection $K \to L$ is an isomorphism. For general computation, you can use the ker-coker sequence. e.g. one way to arrange everything is $$ \begin{matrix}0 &\to& G &\to& H &\to& K ...


1

The answer would seem to be no. A finitely generated Abelian group is a direct product of finitely many cyclic groups (some of which may be infinite). In your situation, the cyclic factors would be of bounded order, so if your group were finitely generated, it would be finite (and even cyclic by the indecomposability condition you impose). Later edit: In ...


1

Yes, this is a direct consequence of the classification theorem for finitely generated abelian groups: For every finite abelian group, there holds $$ G \cong \mathbb{Z}_{n_1} \oplus \mathbb{Z}_{n_2} \oplus \cdots \oplus \mathbb{Z}_{n_k}$$ where $n_i$ divides $n_{i+1}$ Therefore, in your case, as all your multipliers are distinct primes, the only ...


1

Yes, this is a particular case of the Chinese remainder theorem (where you consider the ring structure of $G$ as a $\mathbb{Z}$-algebra).


1

In every group $(G,\cdot)$, for all $p,q\in G$ the equation $x \cdot p = q$ has a unique solution for $x$, and similarly for $p \cdot x = q$ (although it need not be the same $x$). This means that the group table must be a "magic square", i.e. each element must occur precisely once in each row and each column.


1

No, since there's a row with repeated elements. In particular, the given operation is not associative, since: $$ b * (c * c) = b * a = b \neq c = a * c = (b * c) * c $$


1

I like your idea that if $U(n)$ has an element of even order, then the order of $U(n)$ is even by Lagrange's Theorem. On the other hand, for $n>2$, the order of $n-1$ in $U(n)$ is 2. Another approach to this problem is to work with properties of the Euler phi function since $o(U(n))=\varphi(n)$.


1

There is no counterexample with $A,B,C$ finitely generated abelian groups. There is, more generally, no counterexample with $A,B,C$ finitely generated modules over any noetherian ring $R$. To see this, consider the exact sequence $$0\rightarrow Hom(C,A)\rightarrow Hom(B,A)\rightarrow Hom(A,A)$$ The original sequence splits if and only if this sequence ...



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