36

Let $(\mathbb{Q},+)$ be the group of the rational numbers under addition. For any set $A$ of primes, let $G_A$ be the set of all rationals $a/b$ (in lowest terms) such that every prime factor of the denominator $b$ is in $A$. It is clear that $G_A$ is a subgroup of $\mathbb{Q}$, and that $G_A = G_{A'}$ iff $A = A'$. Since there are uncountably many sets ...


33

One example is the group consisting of all finite subsets of $\mathbb N$, with the group operation being symmetric difference. The group is countably infinite, but for each finite or infinite $A\subseteq \mathbb N$ there's a subgroup consisting of the finite subsets of $A$.


33

You've rediscovered the $10$-adic integers. Such "infinite decimal expansions of integers" form a ring using the usual rules for adding and multiplying decimals. However, they do not form a field. A quick and easy way to see this is that $10$ has no inverse (since if you multiply $10$ by anything, then the last digit of the product will be $0$). Even ...


27

You first have to define a probability measure over the sample space $\Omega = \mathbb{N}$. In your question, the sigma algebra contains all singleton sets, i.e., $\{i\}$ for $i \geq 0$. Suppose there is a probability measure on $\Omega$ such that $\epsilon =\Pr(\{0\})= \Pr(\{1\}) = \Pr(\{2\}) = \cdots$ By definition of probability, we have $$ 1 = \Pr(\...


27

For a very simple example, let $G=\mathbb{Q}$ and $H=\mathbb{Z}$. Since $\mathbb{Q}$ is divisible (for any $x\in\mathbb{Q}$ and any nonzero integer $n$, there exists $y\in\mathbb{Q}$ such that $x=ny$), any quotient of $\mathbb{Q}$ is also divisible, so $\mathbb{Z}$ is not isomorphic to any quotient of $\mathbb{Q}$.


26

When you write Probability = $\dfrac{\text{n(E)}}{\text{n(S)}}$, you're assuming that you're drawing a number uniformly at random, which means that every number has the same probability to be drawn. This formula is valid if $\text{E}$ is a finite set, but not if $\text{E}$ is infinite. In fact, we can show that there is no way to draw uniformly at random ...


26

No. Suppose $G$ is an uncountable group. Every element $g$ of $G$ belongs to a countable subgroup of $G$, namely the cyclic subgroup $\langle g\rangle$. Thus $G$ is the union of all of its countable subgroups. Since a countable union of countable sets is countable, $G$ must have uncountably many countable subgroups.


21

Yes, it's possible. For example, let $G=\langle x,y: y^{-1}xy=x^2\rangle$ and $H=\langle x^2\rangle$. Then $\langle x\rangle = H\cup Hx$, so $H=y^{-1}Hy\cup y^{-1}Hxy$, and so $yH=Hy\cup Hxy$. You'll get a similar example from any group $G$ with a subgroup $H$ that is properly contained in some conjugate of $H$.


20

Take the decomposition in base $2$ (or your favourite number), any $n \geq 0$ can be written in an unique way as $\sum_{k=0}^{m} a_k 2^k$. Now separate between odd and even powers, $$n = \sum_{k=0}^{\lfloor \frac{m}{2} \rfloor} a_{2k} 2^{2k} + \sum_{k=0}^{\lfloor \frac{m-1}{2} \rfloor} a_{2k+1} 2^{2k+1}$$ Take $A$ to be the integers that are the sum of even ...


19

The existence of such groups is apparently proved in the paper Shelah, Saharon "On a problem of Kurosh, Jónsson groups, and applications". Word problems, II (Conf. on Decision Problems in Algebra, Oxford, 1976), pp. 373–394, Stud. Logic Foundations Math., 95, North-Holland, Amsterdam-New York, 1980. I haven't seen the paper itself, but here is the ...


18

The reason why this may seem wierd is because we tend to think of $\mathbb{R}$ as a one dimensional, single object, so the idea that quotienting out by some non-trivial subobject doesn't change it is naturally somewhat odd. However, let me give a more intuitive example. Suppose you have $A = \mathbb{Z}$ (or $\mathbb{Q}$, ... or really any other abelian group)...


16

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 $4$.


16

What you're missing here is that when you say $$\text{Probability}=\frac{n(E)}{n(S)}$$ You're forgetting that $N(E)$ and $N(S)$ are both infinite, so you're claiming: $$\text{Probability}=\frac\infty\infty$$ You can't make the assumption that this equals one.


15

Take free product $A*B$ of the cyclic groups $A = \langle a\rangle, B = \langle b\rangle$ of order $m$ and $n$ respectively and thoughtfully look at the element $ab \in A * B$.


15

Consider an equilateral triangle in the plane and let $r$, $s$ and $t$ be the motions of the plane given by reflection with respect to each of the sides of the triangle. Then $a=rs$ is a rotation of angle $2\pi/3$ around the vertex of the triangle which is the intersection of the sides with respect to which $r$ and $s$ reflect. Similarly, $b=st$ is a ...


14

No, that’s clearly not what it means: a group of size $2$ is not an infinite group. You’re to find an infinite group $G$ in which every element except the identity has order $2$, meaning that if $g\in G$, and $g$ is not the identity element $1_G$ of $G$, then $g^2=1_G$. Of course $1_G^2=1_G$ as well, so your problem is really to find an infinite group $G$ in ...


14

$F_2$ acts on a certain tiling of the hyperbolic plane. It looks sort of like this: The above tiling is acted on by the modular group $\Gamma \cong \text{PSL}_2(\mathbb{Z})$, which naturally sits as a subgroup inside of the full group $\text{PSL}_2(\mathbb{R})$ of isometries of the hyperbolic plane. Abstractly, this group is the free product $C_2 \ast C_3$. ...


13

If you have an example for fields, you have an example for rings and abelian groups at the same time: Take the field of rational polynomials $F(x)$ where $F$ is a field. The map $x\mapsto x^2$ defines a ring homomorphism of $F(x)\to F(x)$ which is necessarily injective (since $F(x)$ is a field) but not onto (its image is $F(x^2)$. The image is an isomorphic ...


12

Yes they are. This is know as the Tarski Problem, and was recently solved by Sela and Kharlampovich-Myasnikov. The solution however is very difficult and spans more than 100 pages. Sela's approach at least uses sophisticated ideas from geometric group theory.


11

First note that for every $\alpha\in\mathbb R^\times$ we have that $x\mapsto\alpha\cdot x$ is an automorphism of $(\mathbb R,+)$. Also note that if $f(x+y)=f(x)+f(y)$ then $f(2)=f(1)+f(1)$, and by induction $f(n)=n\cdot f(1)$ for $n\in\mathbb N$, equally $f(k)=k\cdot f(1)$ for $k\in\mathbb Z$. This carries to rationals as well, so $f\left(\frac{p}{q}\right)=...


11

Let $S$ be the set of all subgroups of $G$. Consider the map $\phi: G \to S$ given $\phi(g) = \langle g \rangle$. If $\phi$ were injective, then we'd be done. Unfortunately, $\phi$ is not injective, but fortunately we can control how not injective it is. Indeed, every $H \in S$ has a finite number of pre-images since every cyclic group has finite number ...


11

Suppose that $f:\Bbb{R}^*\to\Bbb{Q}^*$ is a such homomorphism. Then we have some $x$ such that $f(x)=2$. Now take a cube root $\sqrt[3]{x}$ of $x$, which always exists in $\mathbb{R}^*$. Then $(f(\sqrt[3]{x}))^3 = f(x) = 2$, i.e. $f(\sqrt[3]{x})$ is a cube root of $2$. But $2$ has no cube root in $\mathbb{Q}^*$, so this is a contradiction.


11

Yes. What you've defined is called the 10-adic numbers. They don't form a field for two reasons, one of which is that $10$ is not invertible; the other reason is left as an exercise. Most books will only talk about the $p$-adic numbers for $p$ prime, and this is because the general case turns out to reduce to the prime case, essentially because of the ...


10

Rotman p. 324 problem 10.25: The following conditions on a group are equivalent: $G$ is divisible. Every nonzero quotient of $G$ is infinite; and $G$ has no maximal subgroups. It is easy to see above points are equivalent. If you need the details, I can add them here.


10

Hint. Let $G$ be a group in which $\mathbb{Z}$ is a subgroup of index $2$. $\hspace{10pt}\text{S}{\small \text{TEP}}\text{ I}.\hspace{8pt}$ Prove that subgroups of index $2$ are necessarily normal. $\hspace{10pt}\text{S}{\small \text{TEP}}\text{ II}.\hspace{5pt}$ This means that $G$ acts by automorphisms on $\mathbb{Z}$. What are the automorphisms of $\...


10

An example: $G=\langle a,b\mid a^2=b^2=1\rangle$. In it $|ab|=\infty$.


10

No. Just take $\mathbb{Z}_2\oplus \mathbb{Z}_2\oplus \mathbb{Z}_2 \oplus \ldots$. Every element has finite order, so cannot generate an infinite group.


10

Your first claim is false. In fact, there is not even a bijection between the sets $\Bbb{Z}$ and $\Bbb{R}^+$, by Cantor's diagonalization argument. However, the map $x \mapsto e^x$ does provide an isomorphism between the groups $(\Bbb{R}, +)$ and $(\Bbb{R}^+, \times)$.


10

Elements of order $2$ are $2$-divisible in the group $\mathbb{Z}_4^{\infty}$ but not in $\mathbb{Z}_2 \times \mathbb{Z}_4^{\infty}$ : If $x \in \mathbb{Z}_4^{\infty}$ is an element of order $2$, there exists $y \in \mathbb{Z}_4^{\infty}$ such that $x = 2y$ (because all the components of $x$ are either $0$ or $2$). There is no such $y$ if you take $x = (1, 0)...


10

The automorphism group, as an additive group, of the $p$-adic integers $\mathbb{Z}_p$ is the same as its group of units, which is isomorphic to $\mathbb{Z}_p\times\mathbb{Z}/(p-1)\mathbb{Z}$ for odd primes $p$. So $\mathbb{Z}_3\times\mathbb{Z}/2\mathbb{Z}$ is isomorphic to its own automorphism group.


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