Since Z15 is cyclic, these subgroups must be cyclic. They are generated by 0 and the nonzero elements in Z15 which divide 15: 1, 3, and 5.

Every cyclic group is virtually cyclic, as is every finite group. An infinite group is virtually cyclic if and only if it is finitely generated and has exactly two ends; an example of such a group is the direct product of Z/nZ and Z, in which the factor Z has finite index n.

If the order of a group is 8 then the total number of generators of group G is equal to positive integers less than 8 and co-prime to 8. The numbers 1, 3, 5, 7 are less than 8 and co-prime to 8, therefore if a is the generator of G, then a3,a5,a7 are also generators of G.

We have already met examples of cyclic groups and subgroups: Problem 3. Show that Z8 = {0, 1, 2, , 7 } is a cyclic group under addition modulo 8, while C8 = {1, w, w2, , w7} is a cyclic group under multiplication when w = epi/4, by exhibiting elements m ∈ Z8 and ζ ∈ C8 such that |m| = |ζ| = 8.

Theorem: All subgroups of a cyclic group are cyclic. If G=⟨a⟩ is cyclic, then for every divisor d of |G| there exists exactly one subgroup of order d which may be generated by a|G|/d a | G | / d . Proof: Let |G|=dn | G | = d n .

All cyclic groups are Abelian, but an Abelian group is not necessarily cyclic. All subgroups of an Abelian group are normal. In an Abelian group, each element is in a conjugacy class by itself, and the character table involves powers of a single element known as a group generator.

It follows that any group of order 5 (and any group of prime order) must be generated by a single element and is hence, cyclic.

The group U10 = 11,3,7,9l is cyclic because U10 = <3>, that is 31 = 3, 32 = 9, 33 = 7, and 34 = 1.

Given that U(13) is a cyclic group of order 12, we can determine how many gener- ators the group has by taking φ(12) = 4. (Corresponding to 1,5,7,11; the numbers relatively prime to 12.) The cyclic group U(13) has 4 generators.

U8 = {1,3,5,7}. Observe, 32 ≡ 1,52 ≡ 1,72 ≡ 1 (mod 8) so U8 can not be a cyclic group of order 4. There is no element of order 4. Thus, U9 is cyclic of order 6 generated by the element 2.

Show that U(8) is not isomorphic to U(10). U(10) = {1,3,7,9} is a cyclic group generated by 3. So 3 is an element of order 4. But all non-identity elements of U(8) = {1,3,5,7} have order 2, so there is no element of order 4.

Problem Prove that the additive groups R and Q are not isomorphic. Solution By cantor’s diagonal argument, there is no possible bijection between Q and R. Since an isomorphism needs to be a bijection, there is no possible isomorphism between the additive groups R and Q.

Therefore U(10) is cyclic of order 4. Any cyclic group of order 4 is isomorphic to Z4. Therefore U(5) ∼ = Z4 ∼ = U(10).

9.8. Prove that Q is not isomorphic to Z. Since φ is surjective, there is an x ∈ Q with φ(x) = 1. Then 2φ(x/2) = φ(x) = 1, but there is no integer n with 2n = 1.

Since ab is arbitrary, this shows Q is not generated by any single element in Q, i.e., Q is not cyclic. The group Q has the property that for any x∈Q and any integer n⩾1, there exists y∈Q such that n⋅y=x. The group Z is not divisible, so since “being divisible” is invariant under isomorphism, Z≄Q.

The function / : Z ( 2Z is an isomorphism. Thus Z ‘φ 2Z. (Thus note that it is possible for a group to be isomorphic to a proper subgroup of itself Pbut this can only happen if the group is of infinite order).

R and C are both Q-vector spaces of continuum cardinality; since Q is countable, they must have continuum dimension. Therefore their additive groups are isomorphic. Q is a one-dimensional Q-vector space whereas Q[i] is a two-dimensional Q-vector space, so their additive groups are not isomorphic.

Using the axiom of choice, one can show that R and R2 are isomorphic as additive groups. In particular, they are both vector spaces over Q and AC gives bases of these two vector spaces of cardinalities c and c×c=c, so they are isomorphic as vector spaces over Q.

Group R consists of six classes, designated R1, R2, R3, R4, R5 and R-GT; some of these groups contain their own sub-groups, with cars allocated to each group based on their weight, engine size and powertrain.

It means they are exactly the same except for the names of the elements and the name of the binary operation. An isomorphism between groups is a function that renames all of the elements. (Hence, it is bijective… each element in the first group gets renamed to be exactly one element in the second group.)

The multiplicative groups Q*, Q*, R*, and R* are all isomorphic to proper subgroups of themselves.

Usually the easiest way to prove that two groups are not isomorphic is to show that they do not share some group property. For example, the group of nonzero complex numbers under multiplication has an element of order 4 (the square root of -1) but the group of nonzero real numbers do not have an element of order 4.

five properties

Proof: By definition, two groups are isomorphic if there exist a 1-1 onto mapping ϕ from one group to the other. In order for us to have 1-1 onto mapping we need that the number of elements in one group equal to the number of the elements of the other group. Thus, the two groups must have the same order.

Group G is isomorphic to itself. If ϕ is an isomorphism from G to group G′, then there exists an isomorphism from G′ to G. Hence, G≃G′ G ≃ G ′ if and only if G′≃G.