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## Hot answers tagged group-theory

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$, the total number of maximal subgroup ... 1 In fact, it turns out that there are only finitely many finite groups with a given number of conjugacy classes. This is not hard to prove and it's not too hard to find all of them with up to four conjugacy classes, say: http://groupprops.subwiki.org/wiki/There_are_finitely_many_finite_groups_with_bounded_number_of_conjugacy_classes 1 Given$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$2^{m\cdot n}$. 1 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. 1 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. 1 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 ... 1 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$2-subgroups. But a ... 1 Your step (1) looks good; after that, I would look at a factor group: Assume that$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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