Actually a stronger statement can be made:A group G is finite if and only if the number of its subgroups is finiteLet G be a group. If G is finite there is only a finite number of subsets of G, so clearlya finite number of subgroups.Now suppose G is infinite , let'ssuppose one element has infinite order. The this element generates an infinite cyclicgroup which in turn contains infinitely many subgroups.Now suppose all the subgroups have finite order Take some element of G and let it generate a finite group H. Now take another element of G not in H and let it generate a finite group I. Keep doing this by next picking an element of G not H or I. You can continue this way.
Turquoise is a member of the turquoise group and is classed as a phosphate. Phosphates are a class of minerals that is part of a large and diverse group of minerals.
no. it must be class
You don't need a syllabus for a play group. It is play and not a class. A syllabus is used for a college class to tell the students what chapters they need to read by certain dates and when the tests are.
middle class
False. G may be a finite group without sub-groups.
an equation in the form of a polynomial having a finite number of terms and equated to zeroan equation in the form of a polynomial having a finite number of terms and equated to zero
yes form cayleys theorem . every group is isomorphic to groups of permutation and finite groups are not an exception.
In group theory, an alternating group is a group of even permutations of a finite set.
Wolfgang Hamernik has written: 'Group algebras of finite groups' -- subject(s): Finite groups, Group algebras
Michael Aschbacher has written: '3-transposition groups' -- subject(s): Finite groups 'The classification of finite simple groups' -- subject(s): Group theory and generalizations -- Abstract finite groups -- Finite simple groups and their classification, Finite simple groups, Representations of groups, Group theory and generalizations -- Representation theory of groups -- Modular representations and characters 'Fusion systems in algebra and topology' -- subject(s): Combinatorial group theory, Topological groups, Algebraic topology 'The classification of quasithin groups' -- subject(s): Classification, Finite simple groups 'Finite group theory' -- subject(s): Finite groups
Actually a stronger statement can be made:A group G is finite if and only if the number of its subgroups is finiteLet G be a group. If G is finite there is only a finite number of subsets of G, so clearlya finite number of subgroups.Now suppose G is infinite , let'ssuppose one element has infinite order. The this element generates an infinite cyclicgroup which in turn contains infinitely many subgroups.Now suppose all the subgroups have finite order Take some element of G and let it generate a finite group H. Now take another element of G not in H and let it generate a finite group I. Keep doing this by next picking an element of G not H or I. You can continue this way.
An affine group is the group of all affine transformations of a finite-dimensional vector space.
No, for instance the Klein group is finite and abelian but not cyclic. Even more groups can be found having this chariacteristic for instance Z9 x Z9 is abelian but not cyclic
Normally, a cyclic group is defined as a set of numbers generated by repeated use of an operator on a single element which is called the generator and is denoted by g.If the operation is multiplicative then the elements are g0, g1, g2, ...Such a group may be finite or infinite. If for some integer k, gk = g0 then the cyclic group is finite, of order k. If there is no such k, then it is infinite - and is isomorphic to Z(integers) with the operation being addition.
(1). G is is finite implies o(G) is finite.Let G be a finite group of order n and let e be the identity element in G. Then the elements of G may be written as e, g1, g2, ... gn-1. We prove that the order of each element is finite, thereby proving that G is finite implies that each element in G has finite order. Let gkbe an element in G which does not have a finite order. Since (gk)r is in G for each value of r = 0, 1, 2, ... then we conclude that we may find p, q positive integers such that (gk)p = (gk)q . Without loss of generality we may assume that p> q. Hence(gk)p-q = e. Thus p - q is the order of gk in G and is finite.(2). o(G) is finite implies G is finite.This follows from the definition of order of a group, that is, the order of a group is the number of members which the underlying set contains. In defining the order we are hence assuming that G is finite. Otherwise we cannot speak about quantity.Hope that this helps.
A permutation group is a group of permutations, or bijections (one-to-one, onto functions) between a finite set and itself.