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How do we prove that a finite group G of order p prime is cyclic using Lagrange?
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∙ 16y ago
Lagrange theorem states that the order of any subgroup of a group G must divide order of the group G. If order p of the group G is prime the only divisors are 1 and p, therefore the only subgroups of G are {e} and G itself. Take any a not equal e. Then the set of all integer powers of a is by definition a cyclic subgroup of G, but the only subgroup of G with more then 1 element is G itself, therefore G is cyclic. QED.
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∙ 16y ago
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How do you prove that order of a group G is finite only if G is finite and vice versa?
(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.
How do you prove that a group of order 3 is cyclic?
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:abb-1 = bb-1, so ae = e, or a = e.This contradicts the givens, so ab != b. Similarly, ab != a, leaving only possibility: ab = e. Multiplying by a-1, b = a-1. So our group has three elements: e, a, a-1.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 = a-1, and these were shown to be distinct. One possibility remains: a2 = a-1.That means that a3 = e, and the powers of a are: a0 = e, a, a2 = a-1, a3 = e, a4 = a, etc. Thus, the cyclic group generated by a is given by: = {e, a, a-1}.QED.Let g be any element other than the identity. Consider , the subgroup generated by g. By Lagrange's Theorem, the order of is either 1 or 3. Which is it? contains at least two distinct elements (e and a). Therefore it has 3 elements, and so is the whole group. In other words, g generates the group.QEDIn 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 contains more than one element and ||1 such that gn =1 if such an n exists.
How to determine the number of subsets of the given sets?
If the set is of finite order, that is, it has a finite number of elements, n, then the number of subsets is 2n.
In group theory what is a group generator?
In abstract algebra, a generating set of a group Gis a subset S such that every element of G can be expressed as the product of finitely many elements of S and their inverses.More generally, if S is a subset of a group G, then , the subgroup generated by S, is the smallest subgroup of G containing every element of S, meaning the intersection over all subgroups containing the elements of S; equivalently, is the subgroup of all elements of G that can be expressed as the finite product of elements in S and their inverses.If G = , then we say S generatesG; and the elements in S are called generators or group generators. If S is the empty set, then is the trivial group {e}, since we consider the empty product to be the identity.When there is only a single element x in S, is usually written as . In this case, is the cyclic subgroup of the powers of x, a cyclic group, and we say this group is generated by x. Equivalent to saying an element x generates a group is saying that it has order |G|, or that equals the entire group G.My source is linked below.
How do you find proper subgroups of a cyclic group of order 6?
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.
Prove that a group of order three is abelian?
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.
What is finite and infinite cyclic group?
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.
What is the order of a group?
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.
What is the order of an element of a group?
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.
What is the order of grouping?
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.
What is the order of the cyclic group mean?
The order of a cyclic group is the number of distinct elements in the group. It is also the smallest power, k, such that xk = i for all elements x in the group (i is the identity).
How do you find the order a factor group?
Let G be a finite group and H be a normal subgroup. G/H is the set of all co-sets of H forming a group known as factor group. By Lagrange's theorem the number of cosests (denoted by (G:H)) of H under G is |G|/|H|.
Is every group whose order is less than or equal to 4 a cyclic group?
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.
Prove that a group of order 5 must be cyclic?
There's a theorem to the effect that every group of prime order is cyclic. Since 5 is prime, the assertion in the question follows from the said theorem.
A cyclic group of length 2 is called identity?
A cyclic group of order two looks like this.It has two elements e and x such that ex = xe = x and e2 = x2 = e.So it is clear how it relates to the identity.In a cyclic group of order 2, every element is its own inverse.
Let G be a cyclic group of order 8 then how many of the elements of G are generators of this group?
Four of them.
How do you determine number of isomorphic groups of order 10?
There are two: the cyclic group (C10) and the dihedral group (D10).
