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2

Yes, quaternion group is non abelian and not isomorphic to the direct product of $\Bbb{Z/2Z\times Z/2Z\times Z/2Z}$ or $\Bbb{Z/2Z\times Z/4Z}$, and is the one you are looking for.


2

An element of an additive group has finite order if there is a non null $n$ integer such that $n(a+ Z)= Z$. In the case of $1/2$, we have that $2(1/2 + Z)=(1+Z)=Z$. Therefore the order of $1/2$ if finite.


2

Elements of $\widehat{\mathbb{Z}}$ are called profinite integers. The profinite integers have a universal property in the category of profinite groups in exactly the same way that the integers have a universal property in the category of groups: namely, $\widehat{\mathbb{Z}}$ is the free profinite group on one generator. This means precisely that elements $g ...


2

Finite p-group $|G|=p^r$ always has subgroup of order $|H|=p^{r-1}$. So $|M_i|\geqslant |H|$. Moreover, since $|M_i|$ divides $|G|$, $|M_i|=|H|=p^{r-1}$. Since sylow p-group of $|G|$ is of order $p^{r-1}$, the maximal subgroup of $|G|$ is sylow p-group of $|G|$. Since the number of all sylow p-groups is $s_p=1 \mod p$, the total number of maximal subgroup ...


1

In fact, it turns out that there are only finitely many finite groups with a given number of conjugacy classes. This is not hard to prove and it's not too hard to find all of them with up to four conjugacy classes, say: http://groupprops.subwiki.org/wiki/There_are_finitely_many_finite_groups_with_bounded_number_of_conjugacy_classes


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Given $m,n\ge 1$, the set of all the $m\times n$ matrices with entries in the finite field $\Bbb F_2$ is, under addition, an abelian group which is also unipotent ($x+x=0$ for all $x$). The order of this group is $2^{m\cdot n}$.


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By taking the quotient, we are effectively treating any integer as the identity element. Hence an element in $\mathbb{R}/\mathbb{Z}$ has finite order if some multiple of the coset representative is an integer.But this is precisely the definition of a rational number.


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The quaternion group is indeed the minimal counter-example. Clearly, any group of orders $1,2,3,5$ or $7$ is cyclic, and the two (non-isomorphic) groups of order $4$ are: $\Bbb Z_4$ and $\Bbb Z_2 \times \Bbb Z_2$. There are likewise just two non-isomorphic groups of order $6$: $\Bbb Z_6$ and $S_3 \cong \Bbb Z_3 \rtimes \Bbb Z_2$ (this is the only possible ...


1

Not sure but a good candidate should be the quaternion group. If not, the alternate group $A_5$ (much bigger), as it is a simple group.


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As an online resource, you could have a look at this website: http://www.uwyo.edu/moorhouse/pub/bol/. This is about the weaker (not necessarily associative) structure of some finite loops but, among these, you can search the isotopy classes of groups. For example http://www.uwyo.edu/moorhouse/pub/bol/htmlfiles8/8_5_2_0.html or ...


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If you have a picture in mind of the action groupoid then the Burnside's lemma is quite intuitive. Truth is, you can see it as a 'global' version of the orbit-stabilizer theorem, in fact in any orbit the automorphisms are in bijection with the group and since the orbits form a partition of the set you have a bijection between all the automorphisms and the ...


1

There are $n=2^{a−1}k$ elements of $D_{2n}$ lying outside of the cyclic subgroup $\langle r \rangle$, all of order 2. (These are usually called reflections.) Each such reflection is contained in at least one Sylow $2$-subgroup. A Sylow $2$-subgroup has order $2^a$ and contains $2^{a-1}$ reflections. So there must be at least $k$ Sylow $2$2-subgroups. But a ...


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Your step (1) looks good; after that, I would look at a factor group: Assume that $a>1$ and let $N:=\langle r\rangle$ and $P\in Syl_2(D_{2n})$. Then $N$ is cyclic and therefore abelian, and $Q:=P\cap N$ has index $2$ in $P$, so $Q$ is normalized by $\langle P, N\rangle = D_{2n}$. By looking at the relations in the definition of $D_{2n}$, it's easy to see ...



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