Group, DYK Fact #100
 algebraic topology Posted: 23:37 Tuesday 20 February 2007 Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007 Did you know? A set G is said to form a group under the binary operation ∘ iff: ∘ is associative in G. ∃ e ∊ G such that ∀ g ∊ G, g∘e = e∘g = g. ∀ g ∊ G, ∃ g−1 ∊ G such that g∘g−1 = g−1∘g = e. Notes: A further axiom is sometimes stated: ∀ g1, g2 ∊ G, g1∘g2 ∊ G. This “closure property”, however, is already implicit in the definition of a binary operation. (A binary operation on a set S is defined as a function from S×S to S.) It follows that the element e is unique: it is called the identity element of G under ∘. Similarly, each g ∊ G has a unique g−1, called the inverse element of g in G under ∘ ∘ does not have to be commutative in G. If it is, G under the binary operation is said to be an Abelian group. For convenience, the symbol for the binary operation in a group is often left unwritten if it is clear which binary operation the group is associated with – thus g1∘g2 in G is simply written g1g2.
 JaneFairfax Posted: 08:05 Saturday 28 March 2009 The Enlightenment Group: Moderators Posts: 673 Member No.: 20 Joined: 03 Mar 2007 Did you know? The first mathematician to use the term “group” as it is used in mathematics today was Évariste Galois! At about that same time, Augustin Louis Cauchy was independently studying these structures, but did not use the same terminology as Galois. This is information I gather from the book A Course in Group Theory by John F. Humphreys.
 algebraic topology Posted: 15:31 Saturday 28 March 2009 Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007 Did you know? The first groups to be studied were permutation groups.
 JaneFairfax Posted: 23:37 Sunday 12 April 2009 The Enlightenment Group: Moderators Posts: 673 Member No.: 20 Joined: 03 Mar 2007 Did you know? A permutation on $\{1,\,\ldots,\,n\}$ is a bijection from that set to itself. Under composition of mappings, the set of all these permutations forms a group of order n!, called the symmetric group of degree n, Sn. Did you also know? Permutations can be odd or even. A permutation $\pi\in{S_n}$ is odd if there are an odd number of ordered pairs in the set $\{(i,j)\,:\,1\le{i}\,<\,j\le{n},\,\pi(i)\,>\,\pi(j)\,\}$ and is even if the set above contains an even number of elements. Using this definition, it is easy to determine that the identity permutation is even and any transposition (permutation interchanging two distinct elements in $\{1,\,\ldots,\,n\}$ and leaving everything else fixed) is odd. While this definition of odd and even permutations is intuitively easy to grasp, it can unfortunately be mathematically unwieldy. A much better definition, more elegant in an abstract sense and more useful in a practical way, is given in the book by Humphreys – which contains an excellent treatment of the subject of permutations.
 algebraic topology Posted: 06:51 Monday 13 April 2009 Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007 Did you know? A k-cycle is even if and only if k is odd. This a permutation $\sigma$ for which there exist k distinct integers $i_1,\,\ldots,\,i_k$ such that $\sigma(i_r)=i_{r+1}$ for $1\le{r}\le{k-1}$, $\sigma(i_k)=i_1$ and $\sigma(i)=i$ for any other i.
 JaneFairfax Posted: 11:20 Tuesday 14 April 2009 The Enlightenment Group: Moderators Posts: 673 Member No.: 20 Joined: 03 Mar 2007 Did you know? If k is prime, every group of order k is cyclic. However, the converse is false: if every group of order k is cyclic, k is not necessarily a prime. For example, although 15 is not a prime, there are no noncyclic groups of order 15. This can be proved using the Sylow theorems. For if G has order 15, the number of its Sylow 5-subgroups must divide 15 and be congruent to 1 modulo 5. This means that G has a unique Sylow 5-subgroup. Similarly G must have a unique Sylow 3-subgroup. Thus G has 1 element of order 1, 2 elements of order 3, and 4 elements of order 5. The other 8 elements must have order 15, and are therefore generators of the group G.
 algebraic topology Posted: 13:59 Monday 20 April 2009 Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007 Did you know? Although the pioneering group theorists only studied permutation groups, the results they developed are equally applicable to abstract groups – because of Cayley’s theorem: Every group is isomorphic to a subgroup of a symmetric group.
 JaneFairfax Posted: 17:15 Sunday 26 April 2009 The Enlightenment Group: Moderators Posts: 673 Member No.: 20 Joined: 03 Mar 2007 Did you know? If H is subgroup of a group G, we can say that H is a maximal normal subgroup of G iff H is a proper subgroup of G and is normal in G, such that for any subgroup K, if H ◁ K and K ◁ G, then either H = K or K = G. A composition series for G is a finite chain of subgroups G0, G1, …, Gr with G0 = G and Gr = <1G> such that Gi is a maximal normal subgroup of Gi−1 for each i = 1, …, r.
 algebraic topology Posted: 18:11 Sunday 26 April 2009 Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007 Did you know? H is a maximal normal subgroup of G if and only if the factor group G/H is nontrivial and simple. A simple group is a group having no normal subgroups other than itself and the trivial subgroup. Hence the composition factors of a composition series are nontrivial and simple.
 JaneFairfax Posted: 19:44 Sunday 26 April 2009 The Enlightenment Group: Moderators Posts: 673 Member No.: 20 Joined: 03 Mar 2007 Did you know? If G0 ▷ G1 ▷ … ▷ Gr and H0 ▷ H1 ▷ … ▷ Hs are composition series of a group G, then r = s and there exists a bijection $\varphi\,:\,\{G_{i-1}/G_i\,:\,1\le{i}\le{r}\}\to\{H_{j-1}/H_j\,:\,1\le{j}\le{s}\}$ such that $G_{i-1}/G_i$ is isomorphic to $\varphi\left(G_{i-1}/G_i\right)$ for each i. This result is known as the Jordan–Hölder theorem.
 algebraic topology Posted: 21:35 Sunday 26 April 2009 Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007 Did you know? It is important to note that the Jordan–Hölder theorem does not say that every group has a composition series – only that if it has a composition series, then that composition series is unique up to isomorphism of the composition factors. Indeed, there are groups with no composition series at all, an example being the infinite cyclic group.
 JaneFairfax Posted: 09:53 Monday 27 April 2009 The Enlightenment Group: Moderators Posts: 673 Member No.: 20 Joined: 03 Mar 2007 Did you know? The theorem was first discovered by the French mathematician Camille Jordan in 1870 and proved rigorously by the German mathematician Otto Hölder in 1889. Nowadays, however, the theorem can be easily proved using Schreier’s refinement theorem, which in turn can be easily proved using a result called the Zassenhaus lemma. This is how the Jordan–Hölder theorem is tackled in Chapter 15 of A Course in Number Theory by John F. Humphreys.
 algebraic topology Posted: 11:52 Monday 27 April 2009 Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007 Did you know? Hans Zassenhaus was a German mathematician while Otto Schreier was an Austrian mathematician.
 JaneFairfax Posted: 16:40 Tuesday 28 April 2009 The Enlightenment Group: Moderators Posts: 673 Member No.: 20 Joined: 03 Mar 2007 Did you know? If a group G has subgroups G = G0, G1, …, Gr = <1G> such that Gi is a normal subgroup of Gi−1 and Gi−1/Gi is Abelian for all 1 ≤ i ≤ r, then G is said to be a soluble group.
 algebraic topology Posted: 10:50 Wednesday 29 April 2009 Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007 Did you know? Americans call such a group “solvable” rather than “soluble”.
 JaneFairfax Posted: 18:29 Thursday 30 April 2009 The Enlightenment Group: Moderators Posts: 673 Member No.: 20 Joined: 03 Mar 2007 Did you know? If G is a soluble group, then any subgroup H of G is soluble, and for any normal subgroup N of G, G/N is soluble.
 algebraic topology Posted: 14:47 Monday 04 May 2009 Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007 Did you know? If N is a normal subgroup of G such that N and G/N are both solvable, then G is solvable.
 JaneFairfax Posted: 10:36 Tuesday 05 May 2009 The Enlightenment Group: Moderators Posts: 673 Member No.: 20 Joined: 03 Mar 2007 Did you know? Another way of looking at soluble groups G is to consider commutators in G. These are elements of the form xyx−1y−1 where x, y ∊ G. The derived group G′ of G is the subgroup generated by all the commutators in G. Now if we set G(0) = G and G(1) = G′, the nth derived group of G is defined inductively by G(n) = G(n−1)′ = Then G is soluble if and only if G(r) = <1G> for some positive integer r.
 algebraic topology Posted: 13:25 Wednesday 06 May 2009 Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007 Did you know? Every nilpotent group is solvable. A nilpotent group is a group with a terminating upper central series.
 JaneFairfax Posted: 18:56 Wednesday 06 May 2009 The Enlightenment Group: Moderators Posts: 673 Member No.: 20 Joined: 03 Mar 2007 Did you know? A finite group is nilpotent if and only if all its Sylow subgroups are normal and it is the internal direct product of all its Sylow subgroups.
 algebraic topology Posted: 08:47 Tuesday 12 May 2009 Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007 Did you know? $\left\{\begin{tabular}{c}\mbox{cyclic}\\\\\mbox{groups}\end{tabular}\right}\quad\subseteq\quad\left\{\begin{tabular}{c}\mbox{abelian}\\\\\mbox{groups}\end{tabular}\right}\quad\subseteq\quad\left\{\begin{tabular}{c}\mbox{nilpotent}\\\\\mbox{groups}\end{tabular}\right}\quad\subseteq\quad\left\{\begin{tabular}{c}\mbox{solvable}\\\\\mbox{groups}\end{tabular}\right}\,.$
 JaneFairfax Posted: 17:54 Tuesday 12 May 2009 The Enlightenment Group: Moderators Posts: 673 Member No.: 20 Joined: 03 Mar 2007 Did you know? There is another interesting class of groups between the nilpotent and soluble ones: the supersoluble groups. A supersoluble group is like a soluble group, except that the factor groups in its normal series are cyclic rather than Abelian. All supersoluble groups are soluble, and all finitely generated nilpotent groups are supersoluble. Supersoluble groups are not mentioned in John F. Humphreys’s book A Course in Group Theory (at least not in what I have read of it so far) but I came across them today while browsing at Foyles in Central London in a book on finite groups written by a former lecturer of mine: Prof B.A.F. Wehrfritz of Queen Mary, University of London!
 algebraic topology Posted: 10:06 Wednesday 13 May 2009 Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007 Did you know? All supersolvable groups are polycyclic, and all polycylic groups are solvable. A polycyclic group is like a supersolvable group, except that it only needs to have a subnormal rather than normal series with cyclic factors.
 JaneFairfax Posted: 12:43 Wednesday 13 May 2009 The Enlightenment Group: Moderators Posts: 673 Member No.: 20 Joined: 03 Mar 2007 Did you know?
 algebraic topology Posted: 16:37 Tuesday 19 May 2009 Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007 Did you know? The Feit–Thompson theorem states that all finite groups of odd order are solvable. This implies that every finite group of odd non-prime order is not simple.
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