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43

Any infinite group $G$ must have infinitely many abelian subgroups. Note that for each $x \in G$, there is a cyclic subgroup $\langle x \rangle$, which is abelian. If there is an $x$ such that $\langle x \rangle$ is infinite, then $\langle x \rangle$ has infinitely many abelian subgroups. If no such $x$ exists, there must be infinitely many distinct finite ...


30

Take the product $G = S_3 \times \Bbb Z$, which is non abelian since it has a non-abelian subgroup, namely $S_3$. However, $\{1\} \times n\Bbb Z$ are abelian subgroups of $G$ for every $n \geq 0$.


12

Consider the subgroups of $\mathrm{SO}(3)$ (visualized as the rotational symmetries of the $2$-sphere) representing rotations about a fixed axis through the center of this sphere. There are infinitely many choices of this axis, each of which specifies an (abelian) subgroup isomorphic to $\mathrm{U}(1)$.


6

No -- you can't say "$k\otimes_{\mathbb Z} m\bar a$" unless $k$ is itself a multiple of $m$. Otherwise it is not a member of the left factor of the tensor product. Instead, we can note that $\mathbb Z$ and $m\mathbb Z$ are isomorphic as groups, so the tensor product ought to be the same (up to isomorphism) as $\mathbb Z\otimes_{\mathbb Z}\mathbb Z/m\mathbb ...


5

Your argument is not conclusive, because the elements you list do not generate $\mathbb{Z}^{\mathbb{N}}$, but only $\mathbb{Z}^{(\mathbb{N})}$ (the subgroup of sequences with only a finite number of nonzero terms). A much simpler example is $G=\mathbb{Z}$ and $A=2\mathbb{Z}$. There's no $B$, because two nonzero subgroups of $\mathbb{Z}$ always have nonzero ...


5

The multiplicative group of matrices $$G = \left\{ \begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix} \middle\vert\, a,b,c \in \mathbb{F}_3\right\} \subset \operatorname{Mat}(3 \times 3, \mathbb{F}_3)$$ is a counterexample (it is isomorphic to the group in Joanpemo's answer). For all those who are not willing to check this example on ...


4

Nope. The (non-trivial) semidirect product $\;C_3\ltimes(C_3\times C_3)\;$ has exponent $\;3\;$ and it is certainly non-abelian.


4

It's not that the free abelian group $F$ is constructed "with respect to" $\varphi$, even though the wording in your quote could lead one to think that. If you construct a free group, the input the construction is $S$ alone, and (in the formalism assumed here) the output of the construction is $F$ together with the map $\varphi$. The definition of "free ...


3

A free abelian group on a set $S$ is determined by a pair $(F,\varphi)$, where $F$ is an abelian group and $\varphi\colon S\to F$ such that the following property is satisfied: for every map $f\colon S\to G$, where $G$ is an abelian group, there exists a unique group homomorphism $g\colon F\to G$ with $g\circ\varphi=f$. Note that such $\varphi$ must be ...


2

Here is yet another, more abstract point of view: if $S$ is a set, then the free group $F_S$ by itself is not quite yet a representative of the functor $\mathsf{Ab} \to \mathsf{Set}$, $G \mapsto \operatorname{Map}_{\mathsf{Set}}(S, U(G))$ (recall that we're looking for a left adjoint of the forgetful functor). Indeed, the existence of a free group is ...


2

The subgroup $G = \langle \alpha,\beta \rangle$ of $S_6$ is clearly transitive and, since $\langle \alpha \rangle$ is transitive on $\{1,3,4,5,6\}$, $G$ is $2$-transitive. Then $\beta$ of order $3$ stabilizes two points, so $|G|$ is divisible by $6 \times 5 \times 3=90$. Also $G \le A_6$ and $A_6$ has no subgroup of index $2$ or $4$, so we must have $G=A_6$....


2

The set of $2\times 2$ matrices with real entries is non-Abelian when the operator is multiplication, but it has an infinite number of Abelian subgroups. For example consider any subgroup of the form $$\{A | A = \begin{bmatrix} p^n & 0\\ 0 & 1\end{bmatrix} \mbox{ where } n\in \mathbb{Z}\}$$ where $p$ is a constant and can be any prime.


2

Yes, this is true. Theorem: Let $G$ be a finite abelian group. The following two statements hold. (A) Each subgroup of $G$ is isomorphic to a quotient group of $G$, (B) Each quotient group of $G$ is isomorphic to a subgroup of $G$. For a proof see [this MSE question]( Is every quotient of a finite abelian group $G$ isomorphic to some subgroup of $G$?, ...


2

Yes, you are correct. The order of the Cartesian product of two groups is the product of the order of the groups. Good job!


1

The most usual way would be to look at the order of the elements in your given group. Using your example where $|G| = 8$, suppose you find that $G$ contains an element of order $8$. Then $G$ must be $\newcommand{\Ints}{\mathbb{Z}} \Ints_8$ because neither $\Ints_4 \times \Ints_2$ and $\Ints_2 \times \Ints_2 \times \Ints_2$ have an element of order $8$. ...


1

We have an exact sequence of abelian groups (with $\Sigma$ another index set) $$ 0\to B\xrightarrow{\phi}\mathbb{Z}^{\oplus \Lambda}\xrightarrow{\kappa}\mathbb{Z}\to 0 $$ We want to construct a map $\psi: \mathbb{Z}\to \mathbb{Z}^{\oplus \Lambda}$ such that $\kappa\circ \psi=\mathrm{id}_\mathbb{Z}$. Since $\kappa$ is surjective there exists some $F\in \...


1

Whether or not you include 0 as a positive integer is largely a matter of convention - I think most people would say "no". If you don't include 0, then you are correct that the absence of an additive identity means it's not a group. There are also no inverses, as you note, whether or not you include 0. The answer given in the book should have read "There ...



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