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6

Use the classification of finite abelian groups. Write $G=\bigoplus_{p\mid n}G_p$ where $G_p$ is a group of order $p^\nu$ where $p^\nu$ is the maximum power of $p$ dividing $n$. Show that $k(G)=\prod k(G_p)$. So this leaves us with the case that $n=p^\nu$ is a prime power. Show that in this case the minimum $(p^\nu-1)\cdot p+1$ is attained by $(\mathbb ... 5 It's sufficient to show that iii) implies i). Let$\Phi(G) $be the Frattini subgroup of G, i.e. the intersection of all maximal subgroups of G. There is a lemma due to Dlab that states $$\Phi(G) = \bigcap_{p \in \mathbb{P}} \ pG$$ where$\mathbb{P}$is the set of prime integers. So$\Phi(G) = G \Leftrightarrow G$has no maximal subgroups, and in this ... 5 Theorem If$N \unlhd G_1 \times G_2$, then either$N \subseteq Z(G_1 \times G_2)$or$N$intersects one of the factors$G_1$or$G_2$non-trivially. Proof. Assume that$N \cap (G_1 \times${$1$}$)$= {$(1,1)$} =$N \cap $({$1$}$\timesG_2)$. We will show that N is contained in the center of$G_1 \times G_2$. Fix an arbitrary$(n_1,n_2) \in N$. Let$g_1 ...

3

It's easier to see the contrapositive: If a group is abelian, then $e$ is its only commutator: $$aba^{-1}b^{-1} = aa^{-1}bb^{-1} = ee = e$$ Therefore, if the group has more than one commutator, at least one of them will be different from $e$, and so the group cannot be abelian.

3

My picture for $D_4$ shows $fr = r^{-1}f$ and not $rf$. See Zev Chonoles's exquisite pictures at http://math.stackexchange.com/a/686175/53934 too.

3

Apparently it's called a (commutative) inverse monoid. For further details, see Wikipedia1, 2, 3 or Lawson's Inverse Semigroups4. (I haven't proven that the sets of axioms are equivalent. You may want to reserve the bounty for someone who does so.)

3

A more elementary approach to showing $(iii) \Rightarrow (i)$ is to show the contrapositive: Suppose $G$ is not divisible, then for some minimal positive integer $n$, $nG < G$ is a proper subgroup. Since $G$ is abelian, $nG$ is normal. Now consider the quotient $G/nG$; this is again an abelian group where every element of $G/nG$ has order dividing $n$. ...

3

If you have a sequence $$0 \to F \to G \to H \to 0$$ then $H$ is a quotient of $G$. In general there will be no map $H \to G$ giving a section of $G\to H$. However, if such a section exists then the sequence is indeed split on the right (by definition), and therefore split. The important thing is not only the existence of an injection $H\to G$, but the ...

2

In [1], you can find that $k(\mathbb Z_n)\ge k(G)$ for any group group of order $n$. So $\max k(G)$ is found. For $\min k(G)$, Hagen give you a clue. [1] H. Amiri, S. M. Jafarian Amiri and I. M. Isaacs, Sums of element orders in finite groups, Comm. Algebra 37 (2009), no. 9, 2978-2980.

2

It seems as if you are misunderstanding the definition of the commutator. It is not the center of the group. Many times denoted $[G,G]$, it is defined to be the subgroup generated by the set $\{xyx^{-1}y^{-1} : x, y \in G\}$, where I am using multiplicative notation. In your example $\mathbb{Z}$, an element in the commutator will be, for example with $x = ... 2 The group structure of the torsion subgroup may be the same, but the group law may look very different! I find the following example to be interesting and related to your question. Let$E:y^2=x^3+1$with zero at$[0,1,0]$, and consider$E':y^2=x^3+1$where we now declare zero to be$[2,3,1]$. Then,$E$and$E'$are clearly birationally equivalent via the ... 2 To make this a more or less computational manner, consider $$\mathbb Z_5\times\mathbb Z_4\times\mathbb Z_8\big/\left\langle(1,1,1)\right\rangle \cong \mathbb Z^3\big/U,$$ where $$U= \left\langle \pmatrix{5\\0\\0}, \pmatrix{0\\4\\0}, \pmatrix{0\\0\\8}, \pmatrix{1\\1\\1} \right\rangle.$$ Now our goal is to find bases of$U$and$\mathbb Z^3$(both as ... 1 I'm assuming that$\langle(1,1,1)\rangle$means the subgroup generated by$(1,1,1)$, or in other words$\{(k\bmod 5,k\bmod 4,k\bmod 8)\mid k\in\mathbb Z\}$. In that case we can see that each of the cosets that make up the quotient must contain an element of the form$(0,x,0)$. Namely, assume that$(a,b,c)$is some element of the cosets; then by the Chinese ... 1 Let$|G|=n \gt 1$, then hypothesis says that for some$k$, index$[G:C_G(x)]=k$for all$x \in G-\{e\}$. That is, the order of all conjugacy classes of non-identity elements is constant. Suppose there are$l$conjugacy classes, outside$Cl_G(e)=\{e\}$. Then by the class-formula,$n=1+k\cdot l$. Hence$k \cdot l = n-1$, so$k | (n-1)$. But obviously also ... 1 Hint: $$N\le\Bbb Z^n\;\;\text{has finite index}\;\;\iff \text{rank}\,(N)=n\iff N\cong\Bbb Z^n$$ If yu have the above you have the answer of your question. Hint for the hint's proof: If$\;[\Bbb Z^n:N]=\infty\implies \exists\,a\in\Bbb Z^n\;\;s.t.\;\;\left|\langle a+N\rangle\right|=\infty\iff \exists\,\overline K\le\Bbb Z^n/N\;\;with\;\;\overline ...

1

Hopefully this isn't too similar to what you don't want to see. Here's a nice general result: in a PID, a submodule of a finitely generated free module is finitely generated of lesser or equal rank. The proof below I had written up earlier (and I hope it is sufficient/not too hand-wavy), so it uses $\mathbb Z$ instead of a general PID $R$. To see this, we ...

1

$H$ is normal, so commutes with everything in $G$. Then \begin{align*} & & abH &= baH \\ &\iff & abHa^{-1} &= baHa^{-1} \\ &\iff & abHa^{-1}b^{-1} &= baHa^{-1}b^{-1} \\ &\iff & aba^{-1}Hb^{-1} &= baa^{-1}Hb^{-1} \\ &\iff & aba^{-1}b^{-1}H &= baa^{-1}b^{-1}H \\ &\iff ...

1

Any abelian group can be split as $G=D\oplus M$, where $D$ is divisible and $M$ is reduced (that is, $\bigcap_{n>0}nM=\{0\}$). Of course, $G$ is artinian if and only if both $D$ and $M$ are artinian. Now, suppose $D$ is divisible artinian: then $D=t(D)\oplus D/t(D)$, where $t(D)$ is the torsion part of $D$. If $t(D)\ne D$, then $D$ contains an isomorphic ...

1

Since $G$ is abeleian it is not important the order of multiplication.Let $A$ be set of elements of with order $2$ and $B=G-A$.Notice that since $B$ has no element with order $2$,$b\neq b^{-1}$ for all $b\in B$ except $e$ which is not important in multiplication. Then,product of all elements in $G$ can be written as ...

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