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10

I think the claim is a bit more complicated than that. For simplicity, let's look at a quadratic polynomial. If the roots are $r$ and $s$, and we impose the condition that the leading coefficient be $1$, then the polynomial is $$(x-r)(x-s)=x^2-(r+s)x+rs=x^2+bx+c,$$ with $b=-(r+s)$ and $c=rs$. So we can find $b$ and $c$ if we know $r$ and $s$. The ...

10

If $\phi:G \rightarrow G'$ is an isomorphism, then an element $g \in G$ has order $n \iff \phi(g)$ has order $n$. Notice that $\mathbb{C}^\times \times \mathbb{R}^\times$ contains elements of order $4$, e.g. $(i, 1)$. On the other hand, $\mathbb{R}^\times \times \mathbb{R}^\times$ does not contain elements of order $4$.

6

Question $1$ is true, if $P$ and $Q$ are $p$-sylow subgroups then they are conjugate. In an abelian group every subgroup is normal. Question $2$ is true, if you conjugate a $P$-sylow subgroup you get another (because they have the same order), so if there is only one then the subgroup is invariant under conjugation, hence normal. Question $3$: any group in ...

5

It's possible to identify the isomorphism types of both groups in a fairly straightforward manner. Namely, thanks to the exponential map, we have an isomorphism $$\mathbb{R}^{\times} \cong \mathbb{R} \times C_2$$ where $C_2$ denotes the cyclic group $\{ \pm 1 \}$ of order $2$, and similarly $$\mathbb{C}^{\times} \cong \mathbb{R} \times S^1.$$ Abstractly, ...

4

An abelian group is simple iff it is finite of prime order $\;p\;$ , and in that case its automorphism group is cyclic of order $\;p-1\;$ , so all its automorphisms commute with each other. If $\;f\;$ is an endomorphisms of such a cyclic group of order a prime then it is either the trivial homomorphism or the identity one (as it has no non-trivial subgroups)...

4

The Question: "Is the category of free abelian groups accessible?" There's an ambiguity: what should we take to be the morphisms of the category of free abelian groups? The following discussion will work if we consider any of the following categories: $\mathsf{Free}$: free abelian groups and all homomorphisms $\mathsf{Free}_\mathrm{inj}$: ...

4

I don't know if there's a nice way to do this. Here's a not nice way. Start with the observations 1) a locally cyclic group is either torsion or torsion free, and 2) every automorphism of a group lifts to an automorphism of its injective hull. The point of 2) is that it means we can get our result by proving automorphism groups of injective hulls of ...

3

Let $G = \bigoplus_{n \geq 0} \mathbb{Z}$ and consider the subgroups $K = \mathbb{Z} \oplus \mathbb{Z} \oplus \bigoplus_{n \geq 2} 0$ and $H = \mathbb{Z} \oplus \bigoplus_{n \geq 1} 0$. Then $H \subseteq K \subseteq G$ and $G/H \cong G \cong G/K$, but $H \cong \mathbb{Z} \ncong \mathbb{Z}^2 \cong K$.

3

Let $D$ be a divisible abelian group, and let $D'$ be a divisible torsion abelian group. The inclusion $j:D\to D\oplus D'$ satisfies your requierements, because Since $D$ is divisible we know $\mathbb F_p\otimes D=D/pD=0$ for every prime $p$, and the same holds for $D'$. Since $D'$ is torsion, $j\otimes \mathbb Q$ is the map $1_D\otimes \mathbb Q$, which ...

3

The most important theorem (maybe) for Abelian groups is Fundamental Theorem of Finitely Generated Abelian group : For any finitely generated abelian group $G$, it is isomorphic to \begin{align} \mathbb{Z}^{r}\times \mathbb{Z}/p_{1}^{e_{1}}\times\cdots\times\mathbb{Z}/p_{r}^{e_{r}} \end{align} for some $r, e_{1}, \dots, e_{r}\geq 0$ and prime ...

3

Let $G$ be an abelian group satisfying the above two properties. Then there is a nonzero homomorphism $\phi: G \rightarrow \mathbb{Z}$. The image of $G$ is a nonzero subgroup of $\mathbb{Z}$, say $n\mathbb{Z}$ for $n \geq 1$. So as you say, we can choose to be $\phi$ surjective. So there is an exact sequence of abelian groups $$0 \rightarrow \textrm{Ker }... 2 No. Consider the group G = \mathbb{Z} \left[ \frac{1}{2} \right]. Any proper subgroup of it must have elements with bounded denominators, and hence must be the infinite cyclic subgroup generated by \frac{1}{2^n} for some n. But G itself is not cyclic; in fact it can only be generated by infinitely many elements, since finitely many elements would ... 2 Your guess is correct. To show that pA is smaller than A, consider the map f:A\to pA given by f(a)=pa. Is this map injective? Is it surjective? 2 Your definition for pA is correct (at least it makes a lot of sense, considering what we want to prove) Here is a short proof for all the problem. Notice f:A\rightarrow A defined by f(a)=pa is a homomorphism ( we need commutativity for this), the image is pA, so it is automatically an abelian subgroup. Now notice that by Cauchy's theorem there is ... 2 The element c is constructed naturally: we have$$p^rb = na_1 = (p^kt)a_1=p^{k-r}p^rta_1.$$If you put c = p^{k-r}ta_1, then it is clear that c \in <a_1>. Indeed we prove that r \le k to guarantee that p^{k-r} is an integer. For your third question, observe that b + A_1 = a+ A_1. If the period of a is  u < p^r then (ub + A_1) = (ua+ ... 2 The only subgroups of \mathbb Z are those of form n\mathbb Z. In this case we need to find n so that 12,42\in n\mathbb Z. So n is a common divisor of 12 and 42, hence n\in \{\pm1,\pm2,\pm3,\pm6\} Clearly, the smallest of all of these options for n\mathbb Z is when n=\pm6. So the subgroup you want is the subgroup of multiple of 6. ... 2 Yes: in an abelian group every subgroup is normal. Correct. The group A_5 is simple and it has 60 elements. Then \langle(1,2,3,4)\rangle is a Sylow 2-group and it is not normal. 1 [I assume all topological groups are Hausdorff.] This follows easily from Pontryagin duality. Suppose G is a torsion compact abelian group. By Pontryagin duality, continuous homomorphisms G\to \mathbb{R}/\mathbb{Z} separate points of G. But since G is torsion, any homomorphism G\to\mathbb{R}/\mathbb{Z} lands in the torsion subgroup of \mathbb{... 1 Hint: \Bbb Z is cyclic, so all of its subgroups are cyclic, i.e., each subgroup is generated by only one element. What is the generator of your subgroup? 1 The First Prüfer Theorem: An abelian group of bounded exponent is isomorphic to a direct sum of cyclic groups. If G is an abelian group of exponent m\in\mathbb{N}, then G is a direct sum of cyclic groups according to the theorem above, from which we may assume that$$G\cong\bigoplus_{p}\,\bigoplus_{k=1}^{l(p)}\,\left(\mathbb{Z}/p^k\mathbb{Z}\right)^{...

1

You cannot say much other than $\ker f\cap \ker g\leq\ker(f\cdot g)$. Indeed, if $f$ is any homomorphism $f:A\rightarrow B$ with $B$ abelian then there exists a homomorphism $g:A\rightarrow B$ such that: $\ker(f\cdot g)=A$. Simply take $g: x\mapsto f(x)^{-1}$, as inversion is a homomorphism in abelian groups. $\ker(f\cdot g)=\ker f$ . Just take $g: x\... 1 Yes, they are isomorphic. This follows from the following: The groups$(\mathbb{R},+)$and$(\mathbb{R}^2,+)$are isomorphic (which you already know); For any two nonzero elements$a,b$of the group$(\mathbb{R},+)$are "equivalent" in the sense that a group isomorphism of$(\mathbb{R},+)$sends one to the other. (What's the isomorphism?) Now$\mathbb{Z}$... 1 Hint: If$\phi:\mathbb{R}\to\mathbb{R}^2$is a group isomorphism, what is$\phi(\mathbb{Z})$? In addition, what is the$\mathbb{R}$-dimension of the$\mathbb{R}$-span of$\phi(\mathbb{Z})$? 1 The answer is$n = \sum_{k} p_k^{n_k}$. That follows from the more general result that, for finite nilpotent groups$G$and$H$,${\rm mindeg}(G \times H) = {\rm mindeg}(G) + {\rm mindeg}(H)\$. I haven't checked the reference myself, but it is proved in D. Wright. Degrees of minimal embeddings of some direct products. Amer. J. Math., 97:897–903, 1975. See ...

1

A (adjA)= |A| I here adj A = [a -a a -a ] is not in G. so A^-1 = adjA/|A| doesnot mean in G

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