Any group must have an identity element e. As it has order 3, it must have two other elements, a and b. Now, clearly, ab = e, for if ab = b, then a = e:
abb1 = bb1, so ae = e, or a = e.
This contradicts the givens, so ab != b. Similarly, ab != a, leaving only possibility: ab = e. Multiplying by a1, b = a1. So our group has three elements: e, a, a1.
What is a2? It cannot be a, because that would imply a = e, a contradiction of the givens. Nor can it be e, because then a = a1, and these were shown to be distinct. One possibility remains: a2 = a1.
That means that a3 = e, and the powers of a are: a0 = e, a, a2 = a1, a3 = e, a4 = a, etc. Thus, the cyclic group generated by a is given by: = {e, a, a1}.
QED.
Let g be any element other than the identity. Consider
QED
In fact here is the proof that any group of order p where p is a Prime number is cyclic. It follow precisely from the proof given for order 3.
Let p be any prime number and let the order of a group G be p. We denote this as
G=p. We know G has more than one element, so let g be an element of the group and g is not the identity element in G. We also know


QED
Notation...Order of an element
A cyclic group, by definition, has only one generator. An example of an infinite cyclic group is the integers with addition. This group is generated by 1.
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.
There are 5 groups of order 8 up to isomorphism. 3 abelian ones (C8, C4xC2, C2xC2xC2) and 2 nonabelian ones (dihedral group D8 and quaternion group Q)
There are five groups of order 8: three of them are Abelian and the other two are not. These are 1. C8, the group generated by a where a8 = 1 2. C4xC2, the group generated by a and b where a4 = b2 = 1 3. C2xC2xC2, the group generated by a, b and c where a2 = b2 = c2= 1 4. the dihedral group 5. the quaternion group
In order for 4 red marbles to be onethird of a group and 3 blue marbles to make up onefourth of the same group the number has to equal 12.
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.
Yes. The only group of order 1 is the trivial group containing only the identity element. All groups of orders 2 or 3 are cyclic since 2 and 3 are both prime numbers. Therefore, any group of order less than or equal to four must be a cyclic group.
A cyclic group, by definition, has only one generator. An example of an infinite cyclic group is the integers with addition. This group is generated by 1.
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 {...a4, a3, a2, a1, 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 {...a4, a3, a2, a1, 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 {...a4, a3, a2, a1, 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.
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.
A cyclic set of order three, under multiplication, consists of three element, i, x, and x^2 such that x^3 = i where i is the identity.For the set to be a group it must satisfy four axioms: closure, associativity, identity and invertibility. i*i = i, i*x = x, i*x^2 = x^2,x*i = x, x*x = x^2, x*x^2 = x^3 = ix^2*i = x^2, x^2*x = x^3 = ix^2*x^2 = x^4 = x^3*x = i*x = x Since each of the elements on the right hand side belongs to the set, closure is established.It can, similarly be shown that the elements of the set satisfy the associative property.As can be seen from the entries for closure, i is the identity.Also, the inverse of i is i,the inverse of x is x^2 andthe inverse of x^2 is xand therefore, the set has invertibiity. It is, therefore a group.
in a circle 1\3 is quadrilateral
The amino acid proline is the only amino acid that has a secondary amine functional group. This is because proline is a cyclic amino acid that links the 3carbon Rgroup back to the amine group, resulting in a secondary amine.
A group is when you have a number of people or things together. In order for it to be a group it has to be 3 or more.
Proposed since September 2009.A cyclic number is an integer in which cyclic permutations of the digits are successive multiples of the number. The most widely known is 142857:142857 × 1 = 142857142857 × 2 = 285714142857 × 3 = 428571142857 × 4 = 571428142857 × 5 = 714285142857 × 6 = 857142