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13

Let $G=\{z\in \mathbb{C}:z^n=1$ for some positive integer $n\}$. Then $G$ with complex multiplication is an infinite abelian group but every element has finite order.


11

This boils down to: Must $n$ be the identity in this diagram? $$\require{AMScd} \begin{CD} A @>f>> B @>g>> C @>h>> D @>j>> E \\ @| @| @VVnV @| @| \\ A @>f>> B @>g>> C @>h>> D @>j>> E \\ \end{CD} $$ At least $c-n(c)\in\ker h$ for all $c\in C$, hence $c-n(c)=g(b)$ for some $b\in B$. Since ...


8

Take an infinite number of generators, and specify that any element squared is the identity.


8

Consider $(\mathbb{Q},+)$ and the subgroup $(\mathbb{Z},+)$. Now take the quotient $\mathbb{Q}/\mathbb{Z}$.


6

For the non-free part: Take any two nonzero elements $x, y ∈ ℚ$ and show they satisfy $λx + μy = 0$ for some nonzero $λ, μ ∈ ℤ$, hence any two elements are linearly dependent. Thus, since $ℚ$ is not cyclic, it cannot have a basis. For the non-finitely-generated part: If $ℚ$ was finitely generated then without loss of generality (by finding the common ...


6

The only homomorphisms $\varphi:\prod_{n \in \mathbb{N}} \mathbb{Z} \to \mathbb{Z}$ are the obvious ones $$\varphi\left(a_1,a_2,\dots\right)=\sum_{n\in\mathbb{N}}a_nb_n,$$ where $\left(b_1,b_2,\dots\right)$ is a sequence of integers with $b_n=0$ for all but finitely many $n$. In particular, $\varphi$ is determined by its restriction to ...


5

I assume you are familiar with this lemma (or can prove it): If $n$ and $m$ are coprime, then $Z_{nm}=Z_n\times Z_m$ (where equality stands for isomorphism). Using this lemma you can decompose all of your groups to products of cyclic groups of prime power order. For example, \begin{eqnarray} Z_{33}\times Z_{15}\times Z_{15} &=& Z_{3\cdot11}\times ...


4

If $G$ is a group of order $p^2$, then $G$ is abelian. It follows immediately from the exercise. And then one knows $G \cong \mathbb{Z}/p^2 $ or $G \cong \mathbb{Z}/p \oplus\mathbb{Z}/p$ by the structure theorem. If $G$ is a group of order $p^3$, then $G$ doesn't have to be abelian. But it is abelian if the center of $G$ has order $\neq p$. In other words, ...


4

This already fails when $H$ is trivial group. Consider $G=\mathbb{Z}/p \oplus \mathbb{Z}/p$ for example.


4

Define the set $G$ to be the collection of all functions $f:\mathbb{N}\to \mathbb{Z}/2\mathbb{Z}$. We can then define $f+g$, for any two $f,g\in G$, in the natural way. Define $e$ to be the map such that $e(n) = 0$ for all $n\in \mathbb{N}$. It is not hard to see that $f+f = e$ for all $f\in G$.


4

The converse seems to be false. Let $G=\mathbb{Z}$, and let $g$ be the identity morphism. Let $H$ be the group $\prod_{n\in \mathbb{N}}\mathbb{Z}/n\mathbb{Z}$, and let $f:G\to H$ be the morphism sending each $m$ to the element of $H$ whose $n$-th coordinate is the image of $m$ in $\mathbb{Z}/n\mathbb{Z}$ by the natural projection. Then each $g_n$ factors ...


4

By correspondence theorem $N/M$ is normal in $G/M$. By third isomorphism theorem $$G/N \cong \frac{G/M}{N/M} $$But $G/M$ is abelian, and a quotient of an abelian group is abelian, so $G/N$ is abelian.


3

$$G/M\;\;\text{abelian}\;\iff\;G'=[G:G]\le M$$ But $$M\le N\implies\;\text{also}\;\;G'\le N\implies G/N\;\;\text{abelian}$$


3

You are almost done. Denote $A$ with an additive notation. If $\forall x \in A, x+x=0$, then it is well defined the $\mathbb{Z}_2$-vector space structure on $A$ $$\lambda x = \left\{ \begin{matrix} x & \mbox{ if } & \lambda = 1\\ 0 & \mbox{ if } & \lambda = 0 \end{matrix} \right. $$ for $\lambda \in \mathbb{Z}_2, x \in A$. Since $A \ncong ...


3

$(1,2,5)\in A_5$ and $(1,2,3)\in A_4$ but $$(1,2,5)^{-1}(1,2,3)(1,2,5)\notin A_4$$


3

If an abelian group $G\ne\{0\}$ is free, then it is isomorphic to $\mathbb{Z}^{(X)}$ (direct sum of copies of $\mathbb{Z}$), for some set non empty set $X$. Then $\mathbb{Z}$ is an epimorphic image of $G$. Since $\mathbb{Z}$ is not divisible, $G$ can't be divisible. But $\mathbb{Q}$ is divisible. The group $\mathbb{Q}$ is not finitely generated: if ...


3

I will write a very useful lemma, Lemma: If the index of $H$ in $G$ is the smallest prime dividing $G$, then $H$ is normal. This lemma proves $1$ directly. 2) Notice that whhen you show that it has a uniqe Sylow-p subgroup, you also shows that it is normal in G. $b)$ Let $H$ be the subgroup of order $7$ and $K$ be a subgroup of order $4$. It is clear ...


2

1) Consider the action of $G$ on the lateral classes of the subgroup ( call it H ) $$ g \cdot kH = gkH $$ This is an action. Consider the kernel K of the action, i.e $$K = \lbrace g \in H | g \cdot wH = w H \ \ \forall \ wH \rbrace$$Then $K \trianglelefteq G$; moreover $K \subseteq H $ ( simply consider $wH = H $ in the definition of $K$). $G/K$ is ...


2

Another approach, using the rather important property that finite $\;p$- groups have non-trivial center, which also gives for free the existence of such groups (and even of a normal subgroup of order $\;p^k\;$ , for any $\;0\le k\le n\;$. Take $\;1\neq z\in Z(G)\implies \langle z\rangle\lhd G\;$ , so $$\left|G/\langle z\rangle\right|=p^{n-1}$$ Apply ...


2

The conditions $\sigma\circ\psi_{G}=\varphi_{G}$ and $\sigma\circ\psi_{H}=\varphi_{H}$ are determining for $\sigma$. This because: ...


2

The statement is not true. You can verify that $A_4$ is a counterexample.


2

Show any finitely generated subgroup of $\mathbf Q$ is cyclic. Since $\mathbf Q$ is not cyclic, it cannot be finitely generated. $(1)$ It cannot be free: it is not cyclic, so any putative basis has at least two elements. But if $x,y$ are elements of the basis, we know $\mathbf Z^2\simeq \langle x,y\rangle=\langle x'\rangle\simeq\bf Z$ which is impossible. ...


2

There are many counter examples, some of the more natural ones are: $( \mathbb{N}, \oplus )$ where $\oplus$ is bitwise Exclusive OR of the binary representations between two natural numbers. Every non-zero numbers has order $2$. Consider any polynomial rings over any finite field. If we strip away its multiplication, we get an infinite abelian group with ...


2

It is in fact that easy. You can make it even simpler by noting you can conclude $\mathbb Q \subseteq \langle \frac{1}{c} \rangle$ from your assumption. Since $\frac{1}{2c}\in\mathbb Q$ we have $\frac{1}{2c}=\frac{k}{c}$ for some $k\in\mathbb Z$, but there is no $k\in\mathbb Z$ such that $2k=1$.


2

A very good example to serve its usefulness is Is it possible to have a group $G$ such that 0$(G/Z(G))=91$?solve this and get the beauty of the result


2

Here are my comments expanded into an answer which does not use Bezout's identity (I will in fact at the end use this to prove Bezout's identity purely using group theory). Let's take a more general setup and prove something a bit stronger: Let $G$ be any finite group and assume that $n$ is coprime to $|G|$. Then for any $x$ in $G$ there is a $y\in G$ such ...


2

Perhaps you don't know it by the name "Bezout Equality", but it is probably the easiest way to go here. It says that if $\;x,y,\in\Bbb Z\;$ and if $\;\;$ g.c.d. $(x,y)=d\;$ , then there exist $\;m,n\in\Bbb Z\;$ with $\;mx+ny=d$ . In our case, there exist $\;a,b\in\Bbb Z\;$ s.t. $\;a\mathcal O(G)+bn=1\;$ , and then for any $\;g\in G\;$ we get ...


2

Hint: If $n$ is composite, say $n=jk$, then if $o(a)=n$ then $o(a^j)=k$, so composite $n$ will not satisfy the requirement. Now it's your turn: See what happens if $n$ is prime.


2

Hint: It is easy to show that $e^{i\theta_1}\cdot e^{i\theta_2}=e^{i(\theta_1+\theta_2)\pmod{2\pi}}$ (Since, $e^{2i\pi}=1$ ). This would show that the group operation of $S_2$ is binary. If $e^{i\theta}$ is identity, then $e^{i\theta_1}\cdot e^{i\theta}=e^{i\theta_1}=e^{i(\theta_1+\theta)\pmod{2\pi}}$, can you find what $\theta$ is? Similarly, you can ...


2

Let $M$ denote the number of cyclic subgroups of $\mathbb{Z}_{p^2} \times \mathbb{Z}_{p^2}$ of order $p^2$. If I'm understanding correctly, it seems that you're really only asking why $M = p^2 + p$. First, you need to be aware that: 1) If $G_1, G_2$ are groups then $(a,b) \in G_1 \times G_2$ has order lcm$(|a|, |b|)$. 2) A cyclic group of order $p^2$ will ...



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