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6

Sure. Fix a Hamel basis $B$ for $\Bbb R$ over $\Bbb Q$, fix some $r\in B$ and map $B\setminus\{r\}$ to $0$, and $r$ to $1$. Then you're done. If you want to avoid the axiom of choice, you can't. It is consistent that every homomorphism from $\Bbb R$ to $\Bbb Q$ is continuous, and therefore its image is connected. But this means that every homomorphism is ...


5

There isn't a unique isomorphism; if you're given a short exact sequence $0 \to A \to E \to B \to 0$ of vector spaces, to exhibit an isomorphism $E \cong A \oplus B$ from here you need to pick a splitting, which means a section $s : B \to E$ of the map $E \to B$. Given a splitting, the isomorphism $A \oplus B \to E$ has components the given map $A \to E$ and ...


4

$\newcommand{\Span}[1]{\langle #1 \rangle}$$\newcommand{\Order}[1]{\lvert #1 \rvert}$There is no such example. Suppose $$G = \Span{x_{1}} \times \dots \times \Span{x_{n}},$$ where, setting $\Order{x_{i}} = 2^{e_{i}}$, we have $e_{1} \ge e_{2} \ge \dots \ge e_{n} > 0$. Setting $y_{i} = x_{i}^{2^{e_{i} - 1}}$, we have that the involutions of $G$ are the ...


4

Use the fact that any $h_1,h_2\in H$ can be written $h_1 = \phi(g_1),h_2 = \phi(g_2)$ for some $g_1,g_2\in G$.


4

Big Hint using the fact that $\mathbb R$ is abelian, we can very easily prove that $$(a+ib)(c+id)=...=(c+id)(a+ib)$$ for $a,b,c,d\in\mathbb R$.


4

Hint Write $\mathbb R$ as a vector space over $\mathbb Q$. Pick a basis $B$ and define a vector space homomorphism by sending $B$ to $\{ 1\}$. Show it has the desired property. Second solution Use Zorn's Lemma to define a maximal homomorphism from a subgroup of $\mathbb R$ to $\mathbb Q$ which is identity on $\mathbb Q$.


3

I want to thank Tobias Kildetoft for the useful suggestion of using the Chinese Remainder Theorem. So we have that $120 = 8 \; 3 \; 5$ $\mathbb{Z}_{120} = \mathbb{Z}_{3} \times \mathbb{Z}_{8} \times \mathbb{Z}_{5}$ And from this $\mathbb{Z}_{120}^* = \mathbb{Z}_{3}^* \times \mathbb{Z}_{8}^* \times \mathbb{Z}_{5}^*$ Now, from a general result: ...


3

Let $G$ be such a maximal group and assume $q=\frac nm\notin G$ with $m\in\Bbb N$. Let $\tilde G=\frac 1mG$. Then $G\subseteq\tilde G$ (just add $m$ copies of $\frac 1m x$ to obtain $x$) and $\frac nm\in \tilde G$ (because $\Bbb Z\subseteq G$). By maximality, $\tilde G=\Bbb Q$. Hence $\frac1mq\in\tilde G$ and $q\in G$, contradiction.


3

kaiten has already provided a good answer to your question, so I just want to make some remarks about the general theory and show why computation of the Smith normal form gives the desired answer. Given a free module $M$ with rank $n < \infty$ over a PID $R$, every submodule $N \leq M$ is also free of finite rank. Moreover, there is a basis $y_1, ...


3

No. Hint: Find an quotient with torsion.


3

Hint: see if you can show that one of these groups is cyclic, whereas the other is not.


3

Hint: Given $a$ and $b$, find an element $x$ such that $axb=bxa$. (I can think of two $x$s that would work.)


3

No, most non-finitely generated abelian groups are not products of cyclic groups. For instance, a product of infinitely many nontrivial cyclic groups is uncountable (since any product of infinitely many sets with more than one element is uncountable). So any countable abelian group that is not finitely generated cannot be a product of cyclic groups. A ...


2

We will prove that every abelian divisible group $(G, +)$ is the homomorphic image of $\mathbb{Q}^{(X)}$ for some set $X$. The first thing you will want to do is select a set $X$. After all, not every $X$ will work. We define $X = G$, and we will show that there exists a surjective homomorphism $\mathbb{Q}^{(G)} \to G$. We will begin by defining a map ...


2

$4 \cdot tor(A)=0$ means, that the torsion part of $A$ consists of factors of the form $C_2$ and $C_4$ only. Hence you have to write down all combinations of $5$ factors of $C_2, C_4, \mathbb Z$, namely the group is of the form $$\mathbb Z^a \times C_2^b \times C_4^c$$ with $a+b+c=5$.


2

Note that $2^8\equiv 1\pmod{51}$, so the given argument breaks down. In fact $2$ as an element of $B$ has order $8$. Each of our groups has order $32$. So by the Fundamental Theorem, each group is a direct sum of groups of order a power of $2$. The order of any element of $A$ divides $4$. We will show that the group $B=\left(\mathbb{Z}/51\right)^\times$ ...


2

You will have to think a little more. Show that $g a g^{-1} \ne 1$, and then compute $(g a g^{-1})^{2}$. In $S_{3}$, take $a = (12)$ and $g = (123)$ and see what happens. $a g$ of order $2$ implies $a g a g = 1$. Now multiply on the left by $a$.


2

If a finite abelian group $G$ is non-cyclic, it is the direct sum of more than one cyclic $p$-groups, by the Fundamental theorem on finite abelian groups. Each of these factors has an element of order $p$, which generates a subgroup of order $p$ in $G$, and these subgroups are distinct since their generators are.


2

Given any subgroup of $\Bbb Q$ containing $\Bbb Z$, consider it as a subgroup of $\Bbb Q$ without considering that it contains $\Bbb Z$. You have shown that it is not maximal, so there is a proper subgroup containing it. That proper subgroup must contain $\Bbb Z$ as well, so the answer is no.


2

Every subgroup where there is an element with a nonzero right coordinate will have finite index, since $\langle (0,n) \rangle$ and $\langle (0,2n) \rangle \subseteq \langle (1,n) \rangle$ both would have finite index in the group, so we have there are infinitely many subgroups of finite index. There are finitely many subgroups of a given index $n\in ...


2

Here is a proof that appears in Weil's gem Number theory for beginners and in several other books. Let $m$ be the order of $A$ and let $\psi(n)$ be the number of elements of order $n$ in $A$. Then, by Lagrange's theorem, $m=\sum_{d\mid m} \psi(d)$ because every element has an order that is a divisor of $m$. If $\psi(d)>0$, then there is an element of ...


2

If $X$ has at least $3$ elements, then the group of all invertible maps on $X$ contains $S_3$, which is not abelian. Concretely, let $x_1, x_2, x_3$ be three distinct elements of $X$. Then the map that swaps $x_1$ and $x_2$ does not commute with the map that swaps $x_1$ and $x_3$.


1

1. When $\varphi$ is the identity, the semi-direct product coincides with the direct product. 2. It generalizes the notion of direct product. 3. \begin{eqnarray*} (a,b)+(\varphi(-b)(-a),-b)&=&(a+\varphi(b)\varphi(-b)(-a),0)\\ &=&(a+\varphi(b-b)(-a),0)\\ &=&(a+\varphi(0)(-a),0)\\ &=&(a-a,0)\\ &=&(0,0) ...


1

Note: Not at all a complete answer. For prime numbers, $\phi(p)=p-1$. So all prime numbers are cyclic. The cyclicity of any prime number will be 1. The only even cyclic number is 2; this implies that all numbers divisible by 4 will have a cyclicity of only 1 (Excluding its prime factorization representation).


1

Yours is one possible way to show this. One can also directly use the universal property of the free abelian group: Given an abelian group $G$ let $A(G)$ be the free abelian group on the underlying set of $G$. Then by the universal property of the free abelian group there exists a unique group homomorphism $\varphi \colon A(G) \to G$ with $\varphi(g) = g$ ...


1

The last step of your proof follows from the fact that for groups $G_1$ and $G_2$ and normal subgroups $N_1 \subseteq G_1$ and $N_2 \subseteq G_2$ we have that $N_1 \times N_2 \subseteq G_1 \times G_2$ is a normal subgroup with $(G_1 \times G_2)/(N_1 \times N_2) \cong (G_1 / N_1) \times (G_2 / N_2)$. To see the isomorphism notice that the group epimorphism ...


1

I believe that Propositions 1.10 and 1.16 of this paper of mine give (a generalization of) what you are looking for. The notation may take some getting used to. Indeed all the key ingredients of the discussion above appear in this somewhat abstracted context...and they comprise the proof.


1

In fact, $(P)$ is wrong: there is only one normal subgroup $H$ of order $4$ in $S_4$, and $S_4/H\simeq S_3$. See Show that $S_4/K \cong S_3$.


1

You are absolutely correct about (Q). As to (P), you should know there is a unique normal subgroup $V$ of order $4$ in $S_{4}$. This is a transitive subgroup, so $S_{4} = V T$, where $T$ is the stabilizer of $4$, say. But then $T \cong S_{3}$.


1

Your calculations are correct. $G_1$ is already expressed in primary decomposition and so to find the invariant factors you just list the powers of each prime right justified and multiply them vertically: \begin{array}{rrr} & 2 & 4 \\ 3 & 3 & 9 \\ \hline 3 & 6 & 36 \\ \end{array} The primary decomposition of $G_2$ is $C_3 \times ...



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