Is Z12 isomorphic to Z3 Z4?

Only groups that have the same order and same properties (either both abelian or both nonabelian; either both cyclic or both noncyclic) can be isomorphic. Thus among the above groups, only Z12 and Z3 × Z4 be isomorphic. This two groups are indeed isomorphic by Proposition 3.4.

Are the groups Z2 Z12 and Z4 Z6 are isomorphic Why or why not?

By the fun- damental theorem of finitely generated abelian groups, there are three abelian groups of order 24 up to isomorphism: Z2 × Z2 × Z2 × Z3, Z2 × Z4 × Z3 and Z8 × Z3. Z4 × Z6 is also isomorphic to the second group in the list. But then Z2 × Z12 and Z4 × Z6 are isomorphic.

Is Z12 isomorphic to Z6 Z2?

Z12,Z6 ⊕ Z2,D6,A4 The first two groups are Abelian so they are not isomorphic to the latter two groups. Z12 has an order 12 element, but on Z6 ⊕ Z2, the maximum order of an element is lcm(6,2) = 6. So Z12 ≈ Z6 ⊕ Z2.

Is Z4 a subgroup of Z12?

Which one is it? Proof: Note that Z4⊕Z12 is abelian, so any subgroup is normal. Also note that |Z4⊕Z12| = 48 and ⟨(2,2)⟩ = {(2,2),(0,4),(2,6),(0,8),(2,10),(0,0)}.

Is Z4 isomorphic to Z4?

Is there an “nontrivial” isomorphism from Z4 to Z4 that preserves the order of elements? Yes, the map g : Z4 → Z4 defined by g(0) = 0, g(1) = 3, g(2) = 2, and g(3) = 1 is an isomorphism of G. Definition. An isomorphism f : G → G for a group G is called an automorphism of G.

Z2⨁Z3 is isomorphic to Z6 Z3⨁Z3 is isomorphic to Z9 IIT Jam 2015 group theory gate Mathematics

Is Z2 Z4 isomorphic to Z8?

There's an element of Z2 × Z4 of order 4, namely ([0]2,[1]4), but there's no element of order 8 in Z2 × Z4. The element [1]8 of Z8 has order 8. Hence the three groups Z2 × Z2 × Z2, Z2 × Z4 and Z8 are not isomorphic, by Theorem 41(d).

What groups are isomorphic to Z4?

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

Is Z12 isomorphic to D6?

Solution: four non-isomorphic groups of order 12 are A4,D6,Z12,Z2 ⊕ Z6. The first two are non-Abelian, but D6 contains an element of order 6 while A4 doesn't. The last two are Abelian, but Z12 contains an element of order 12 while Z2 ⊕ Z6 doesn't.

Is Z3 a subgroup of Z12?

1 Answer. Show activity on this post. For your information, Z12 is the integers 0,1,…,11 under addition mod 12, while Z3 is the integers 0,1,2 under addition mod 3; the latter is not a subgroup of the former.

Is Z15 isomorphic to Z3 Z5?

Or We accepted the less complete answer that both Z15 and Z5 ⊕ Z3 are cyclic groups, the latter since GCD(5,3) = 1, hence they are isomorphic since there is a unique cyclic group of given order, up to isomorphism.

Is Z3 Z3 isomorphic to Z9?

Thus, Z9 an Z3*Z3 are not isomorphic.

Is Z6 isomorphic to Z2 Z3?

This re-labelling is just a mapping phi : Z2 x Z3 -> Z6, which is one-to-one and onto and so is an isomorphism. Note that, in this case, phi is a discrete mapping. Hence, Z2 x Z3 is isomorphic to Z6.

Is Z12 abelian?

The group S3 ⊕ Z2 is not abelian, but Z12 and Z6 ⊕ Z2 are.

How can you prove two groups are isomorphic?

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.

Is Z4 isomorphic to U 8 )? Justify?

Suppose that φ : U(8) → Z4 is an isomorphism. By Theorem 6.3, since Z4 is cyclic, then so is U(8), which is false. Hence, there is no isomorphism from U(8) to Z4.

What is the group Z12?

Z12 is a cyclic group, generated by 1, so need to determine image of 1. In order to have isomorphism, need to find all elements of order 12 in Z4 ⊕ Z3. 5. Consider group homomorphism f : Z12 → Z4 ⊕ Z3 given by f(a)=(a,0).

Is Z4 a subgroup of Z8?

The subgroup is a normal subgroup and the quotient group is isomorphic to cyclic group:Z4. is the group direct product of Z8 and Z2, written for convenience using ordered pairs with the first element an integer mod 8 (coming from cyclic group:Z8) and the second element an integer mod 2. The addition is coordinate-wise.

Is Z2 a subgroup of Z4?

Z2 × Z4 itself is a subgroup. Any other subgroup must have order 4, since the order of any sub- group must divide 8 and: • The subgroup containing just the identity is the only group of order 1. Every subgroup of order 2 must be cyclic.

Is S4 isomorphic to Z24?

But the maximum order of an element in S4 is 4 as you can check . Since the order of an element is preserved under isomorphism, there can't be any isomorphism between these two groups even though both groups have the same number of elements, 4! = 24 = 12 2. b Prove that neither is isomorphic to Z24.

Is U 16 isomorphic to Z8?

To be isomorphic to Z8, U(16) must have an element of order 8. To be isomorphic to Z4 ⊕ Z2, it must have only elts of order 1,2, or 4. To be isomorphic to Z2 ⊕ Z2 ⊕ Z2, it must have elts only of order 1 or 2. Thus U(16) ≈ Z4 ⊕ Z2.

How many Homomorphisms are there from Z20 to Z8?

There is no homomorpphism from Z20 onto Z8. If φ : Z20 → Z8 is a homomorphism then the order of φ(1) divides gcd(8,20) = 4 so φ(1) is in a unique subgroup of order 4 which is 2Z8.

Is U 10 and Z4 isomorphic?

Examples and Notes: (a) The mapping φ : Z4 → U(10) given by φ(0) = 1, φ(1) = 3, φ(2) = 9 and φ(3) = 7 is an isomorphism as the table suggests. Thus Z4 ≈ U(10). Note the different operations in each group.

What is the order of Z4 Z4?

Z4 × Z4: The elements have orders 1, 2, or 4. The elements of order 2 are (2, 0), (2, 2), and (0, 2). Thus, there is 1 element of order 1 (identity), 3 elements of order 2, and the remainder have order 4, so there are 12 elements of order 4.

What are the generators of Z12?

(a) The generators of Z12 are 1, 5, 7, and 11. These are the elements of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} which are relatively prime to 12.

How do you find the subgroup of Z12?

Z12 is cyclic, so the subgroups are cyclic and are in one-to-one correspon- dence with the divisors of 12. Thus, the subgroups are: H1 = 〈0〉 = {0} H2 = 〈1〉 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} H3 = 〈2〉 = {0, 2, 4, 6, 8, 10} H4 = 〈3〉 = {0, 3, 6, 9} H5 = 〈4〉 = {0, 4, 8} H6 = 〈6〉 = {0, 6}.