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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 ...

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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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