The order of an elementg in a group is the least positive integer k such that gk is the identity.Now look at the same group, we know there exists an element h such that gh=hg=e where e is the identity. This must be true because existence of inverses is one of the conditions required for a set to be a group. So if gk=e and gh=e, then gk =gh and we see the relation between k, the order and h the inverse in the group.
In abstract algebra, a generating set of a group is a subset of that group. In that subset, every element of the group can be expressed as the combination (under the group operation) of finitely many elements of the subset and their inverses.
A cyclic group of order 6 is isomorphic to that generated by elements a and b where a2 = 1, b3 = 1, or to the group generated by c where c6 = 1. So, find the identity element, 1. Next find an element which when operated on by itself, equals the identity. This element will correspond to a or c3. Finally find an element which when operated on by itself twice (so that it is cubed or multiplied by 3), equals the identity. This element will correspond to b or c2. The subgroups {1}, (1, a} = {1, c3} and {1, b, b2} = {1, c2, c4} will be proper subgroups.
In a group, the identity property is that each group contains an element, i, such that for all elements x, in the group, i*x = x*i = x. i is called the identity element.
(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.
The order of a group is the same as its cardinality - i.e. the number of elements the set contains. The order of a particular element is the order of the (cyclic) group generated by that element - i.e. the order of the group {...a-4, a-3, a-2, a-1, e, a, a2, a3, a4...}. If these powers do not go on forever, it will have a finite order; otherwise the order will be infinite.
The order of a group is the same as its cardinality - i.e. the number of elements the set contains. The order of a particular element is the order of the (cyclic) group generated by that element - i.e. the order of the group {...a-4, a-3, a-2, a-1, e, a, a2, a3, a4...}. If these powers do not go on forever, it will have a finite order; otherwise the order will be infinite.
The order of a group is the same as its cardinality - i.e. the number of elements the set contains. The order of a particular element is the order of the (cyclic) group generated by that element - i.e. the order of the group {...a-4, a-3, a-2, a-1, e, a, a2, a3, a4...}. If these powers do not go on forever, it will have a finite order; otherwise the order will be infinite.
"Carbonate" is not an element or an element group; instead, it is a polyatomic anion and is one of a large group of oxyanions.
chlorine, 17, 17 in order you asked.
The order of an elementg in a group is the least positive integer k such that gk is the identity.Now look at the same group, we know there exists an element h such that gh=hg=e where e is the identity. This must be true because existence of inverses is one of the conditions required for a set to be a group. So if gk=e and gh=e, then gk =gh and we see the relation between k, the order and h the inverse in the group.
In abstract algebra, a generating set of a group is a subset of that group. In that subset, every element of the group can be expressed as the combination (under the group operation) of finitely many elements of the subset and their inverses.
By LaGrange's Thm., the order of an element of a group must divide the order of the group. Since 3 is prime, up to isomorphism, the only group of order three is {1,x,x^2} where x^3=1. Note that this is a finite cyclic group. Since all cyclic groups are abelian, because they can be modeled by addition mod an integer, the group of order 3 is abelian.
The element "Cadmium" is in group number 12.
If we look at the periodic table, we can see that the first element in Group I is Hydrogen.
The group name for the element Pb is "group 14" or "group IV."
Group A sir.