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4

We have $$x^3 = x \Longrightarrow x^2 = 1 \Longrightarrow x = x^{-1}$$ for all $x$. Thus $$ab = (ab)^{-1} = b^{-1}a^{-1} = ba$$ for all $b , a \in G $.


4

Using the map $$\begin{cases} \mathbb{R}\to S^1=\{z\in\Bbb C : |z|=1\} \\x\mapsto e^{2\pi i x}\end{cases}$$ we see this is a surjective group homomorphism--the group operation on $S^1$ is multiplication and $e^{z+w}=e^ze^w$--with kernel $\Bbb Z$, hence, by the first isomorphism theorem, we get $S^1$ is isomorphic to $\Bbb R/\Bbb Z$


4

Look at the definition of $\mathbb R/\mathbb Z$: Two elements $a,b\in\mathbb R$ are in the same coset if and only if $a-b\in\mathbb Z$. Now look at the definition of $a\equiv b$ mod $1$, it says that $a=b+k\cdot 1$ for some $k\in\mathbb Z$, which is equivalent to $a-b\in\mathbb Z$. So yes, you can think of $\mathbb R/\mathbb Z$ as "$\mathbb R$ modulo 1".


3

In additive abelian group the word ‘additive’ refers to the symbol used for the operation $({+})$ and, in principle, it has nothing to do with the group being abelian. It's true that in most cases the additive notation is used for abelian groups (or, more generally, for commutative operations), but this is not universal. For instance, the two operations on ...


3

it is obvious that every subgroup of every abelian group is abelian. for the inverse $S_3$(group of permutations of 3 objects) is not an abelian group but it's subgroups(except $S_3$) are abelian(beacuse subgroups are of order $1$ or $2$ or $3$ that are abelian).


3

$[A_0,_p,x] = [[A_0,_{p-1},x],x]=1$, so $x$ centralizes $[[A_0,_{p-1},x]$, and $N$ does also, so $C_G([A_0,_{p-1},x])$ contains $\langle N,x \rangle = G$. i.e. $[[A_0,_{p-1},x] \in \zeta_1(G)$. Then we get $x$ centralizes $[[A_0,_{p-2},x]$ modulo $\zeta_1(G)$, so $[[A_0,_{p-2},x] \le \zeta_2(G)$, etc.


3

Every abelian group of bounded exponent satisfies (1) and (2). As I pointed out in my comment, every torsion group satisfies (1), because every nontrivial subgroup contains a cylic group of prime order, which is clearly minimal. So let $G$ be an abelian group of exponent $n$. Let $H$ be a proper subgroup. Then there exists $g \in G \setminus H$, and by a ...


3

Note that the assumption is equivalent to $x^2=e$ for all $x\in G$. Let $x,y\in G$. Then $$xy=y^2 x y x^2=y(yx)(yx)x=yx.$$


2

The quotient group has order $24/6=4$. An abelian group of order $4$ is either $C_4$ or $C_2\times C_2$.


2

let,G be group.h is subgroup of G. given that G is abelian i.e. ab=ba for all element of G. let,x,y be any element of H. H is subgroup of G means H is also subset of G. therefoer x,y are also elt. of G therefore xy=yx for all elt. of H so H is abelian. For converse look at S3.


2

$abab=a^2b^2\implies a^{-1}abab=a^{-1}a^2b^2\implies bab=ab^2\implies bab b^{-1}= ab^2b^{-1}\implies ba=ab$


1

For the alternative approach that @user133281 suggested, see Isaacs' 'Finite Group Theory' exercise 1A.8. Using group actions, it is easier, I think.


1

Let $A=\left(\begin{array}{cc}1&0\\0&2\end{array}\right)$, $B=\left(\begin{array}{cc}0&1\\2&0\end{array}\right)$, $C=\left(\begin{array}{cc}2&0\\0&1\end{array}\right)$, $D=\left(\begin{array}{cc}0&2\\1&0\end{array}\right)$. Then $ABCD=\left(\begin{array}{cc}1&0\\0&16\end{array}\right)$ and ...


1

The exact sequence $G \xrightarrow{f} H \xrightarrow{g} K$ gives a short exact sequence $\DeclareMathOperator{\im}{im}$ $$0 \to \im(f) \to H \to H/\im(f) \cong \im g \to 0$$ Then $G,K$ finitely generated (f.g.) abelian groups $\implies H$ f.g. (since $\im g$ is a subgroup of $K$, hence is f.g., using only that $\mathbb{Z}$ is a Noetherian ring, and also ...


1

As a divisible group, $G:=\mathbb{R} / \mathbb{Z}$ may be written as a direct sum of Prüfer groups and copies of $\mathbb{Q}$. First we have $$G \simeq \mathrm{Tor}(G) \oplus G/ \mathrm{Tor}(G),$$ where $\mathrm{Tor}(G)$ is the torsion subgroup of $G$. According to one of my previous answer, $$\mathrm{Tor}(G) \simeq \mathbb{Q} / \mathbb{Z} \simeq ...


1

$G$ is a group of order $24$, and $H$ is a normal subgroup of order $6$. As such, $G/H$ is of order $4$. The fundamental theorem of finitely generated abelian groups tells you that this group is isomorphic to $\mathbf{Z}_4$ or $\mathbf{Z}_2 \times \mathbf{Z}_2$. (Note that each individual part is of prime power order, if you had an abelian group of order $12 ...


1

There is only one operation defined for the group, namely $*$, so if you want to be pedantic/exact, $a*b$ is a valid statement, while $ab$ is not defined. However, in practice we shorten the notation, so $a*b$ can be written as $ab$. So your final expression is equivalently $ab=ba$ or $a*b=b*a$. They are the same with slightly different notation.


1

$$x^2 = e ,\forall x \in X \Rightarrow (xy)^2 = x^2y^2 \Rightarrow (xy)(xy) = x^2y^2 \Rightarrow xyx = x^2y \Rightarrow yx=xy.$$ For $x,y \in G$ of course.



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