How do you determine the number of subgroups in a cyclic group?

A finite cyclic group has exactly one subgroup for each divisor of the order, so if the order is 6, that makes 4 subgroups….

1. Let Zn be a cyclic group of order n.
2. For each divisor of n there exists unique subgroup of Zn.
3. Hence number of distinct subgroups is equal to number of divisors of n.

How many groups are there of order 12 upto isomorphism?

So there are two abelian groups of order 12, up to isomorphism, Z2 × Z2 × Z3 and Z4 × Z3.

How do you find the number of non isomorphic subgroups?

In a cyclic group of order n, there is exactly one subgroup of order d for each divisor d of n. Groups of different order cannot be isomorphic. Therefore, the number of non-isomorphic subgroups of G is the number of divisors of n.

How do you find the number of subgroups of order?

In order to determine the number of subgroups of a given order in an abelian group, one needs to know more than the order of the group, since for example there are two different groups of order 4, and one of them has one subgroup of order 2, which the other has 3.

How many subgroups does a cyclic group of order 12 have?

The answer is there are 6 non- isomorphic subgroups.

How many cyclic subgroups does Z 16 have?

one subgroup
Z16: A cyclic group has a unique subgroup of order dividing the order of the group. Thus, Z16 has one subgroup of order 2, namely 〈8〉, which gives the only element of order 2, namely 8. There is one subgroup of order 4, namely 〈4〉, and this subgroup has 2 generators, each of order 4.

What is a group of Order 12?

There are five groups of order 12. We denote the cyclic group of order n by Cn. The abelian groups of order 12 are C12 and C2 × C3 × C2. The non-abelian groups are the dihedral group D6, the alternating group A4 and the dicyclic group Q6.

How many non-Abelian group of order 12 are there?

3 non-abelian groups
We conclude that in addition to the two abelian groups Z12 and Z2 × Z6, there are 3 non-abelian groups of order 12, A4, Dic3 ≃ Q12 and D6.

How many different non isomorphic groups of order 30 are there?

4 non-isomorphic groups
Note that the centers of these 4 groups are non-isomorphic. So these are non-isomorphic groups and there are exactly 4 non-isomorphic groups of order 30. 2.12 #8 Let G be a group of order 231 = 3 × 7 × 11. Let sp be the number of p-Sylow subgroups of G.

What is non isomorphic group?

If we look at 2 element groups, one of the elements is identity element and the other one has to have its inverse. Therefore there are no 2 element groups. 2) As a group doesn’t have to be commutative, there’s quite a lot of non isomorphic groups.

How many subgroups are there in a group of order 13?

We know that there is only one subgroup of order 13(By Sylow’s thm) which implies there are exactly 12 elements of order 13 (precisely the non-identity elements of the subgroup of order 13). Now every element has either order=3 or order=13 or order=1 (by Lagrange’s thm).

How many subgroups does G have?

Note: Every group G has at least two subgroups: G itself and the subgroup {e}, containing only the identity element.

How many non-isomorphic subgroups are there of order 12?

The answer is there are 6 non- isomorphic subgroups. They are of course all cyclic subgroups. Here is how you write the down. Since G is cyclic of order 12 let x be generator of G. Then the subgroup generated by x, has order 12, the subgroup generated by has order 4, has order 3, and

How do you find the Order of a cyclic group with order 12?

Since G is cyclic of order 12 let x be generator of G. Then the subgroup generated by x, has order 12, the subgroup generated by has order 4, has order 3, and

How many different subgroups are there in a cyclic group?

To help you understand where you’re going wrong, why not try writing out these “six different subgroups”: if $G$ is a cyclic group of order $7$, and $a$ is a generator of $G$, then

How to find the number of generators of an automorphism?

Since an automorphism of G should map a generator of G to a generator of G it’s enough to know how many generators does G have. ( i, m) = 1. ( G) | = ϕ ( m) where ϕ ( m) is Euler’s function. ( G). Then f ( g) = g i for some i.