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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

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!


2

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 $8$. ...


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.


1

See https://oeis.org/A091317 That deals with the case of $n$ prime (and in the text the case $n=p^2$. Thanks to @GerryMyerson for this (and the reference to AO14662). As he remarks, there is unlikely to be any useful characterisation known. So this is essentially an open problem. But the table does give you several more examples.


1

the best way to do this is to think that the first factor in the decomposition of $G$ under cyclic groups is of maximum order, more precisely its order is the exponent of $G$, to joust at Cauchy result who assure that for all prime $p$ dividing the order of $G$, it also divides the exponent of this group. As noted in the comments, if $p$ and $q$ are distinct ...


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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