Find Subgroup of Order 4 in $\mathbb{Z}$ /13

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In summary, the conversation discusses finding a subgroup of the group of units modulo 13, denoted as (Z/13)^*, with an order of 4. The basic requirements for a subgroup are mentioned, such as satisfying the group operation and containing the identity element. The questioner expresses difficulty in applying the theory to a concrete example, but the responder suggests experimenting by trying different numbers and their powers. It is also clarified that there are no subgroups of (Z/13) under addition with an order of 4.
  • #1
ElDavidas
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The question reads as:

"Let [itex] G = ($\mathbb{Z}$ /13)^*. [/itex] Find a subgroup H of G such that |H| = 4. "

I think this means that you have to find a subgroup that has order 4. Although I'm not entirely sure what that means in this context.

Any help will be appreciated.
 
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  • #2
Find a subgroup of G with four elements.
 
  • #3
AKG said:
Find a subgroup of G with four elements.

Ok, I understand this.

I know that for a set to become a subgroup it has to satisfy the operation, contain the identity element and also have an inverse.

How do you go about writing what the possible subgroups are for (Z/13) with order equal to 4? (Which I think is all the congruence classes with mod 13)

I can understand the theory but don't really know how to apply it to a concrete example.
 
  • #4
I'm not sure you can claim to understand the theory if you cannot do this question.

Firstly you're talking about (Z/13)^* which is the group of units modulo 13, since 13 is a prime that is the numbers 1,2,3,...,11,12 with multiplcation mod 13.

We are not talking about "all the congruence classes with mod 13" if indeed you did understand the theory you'd know that there are no subgroups of Z/13 under addition (presumably you mean addition, as that is the group operation defined on all the congruence classes) with order 4 since 4 does not divide 13.

Secondly, why don't you just experiment? Pick a number, work out its square, cube, and 4th power and see if you get what you want. There are naively at most 12 things here for you to try, though you obviosuly won't pick 1 or 12, will you?
 

Related to Find Subgroup of Order 4 in $\mathbb{Z}$ /13

1. How do you find a subgroup of order 4 in $\mathbb{Z}$ /13?

To find a subgroup of order 4 in $\mathbb{Z}$ /13, you can start by listing out all the elements in the group, which in this case are the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12. Then, you can check which of these elements have an order of 4 when raised to different powers. The elements that have an order of 4 will form the subgroup of order 4.

2. What is the order of the subgroup of order 4 in $\mathbb{Z}$ /13?

The order of the subgroup of order 4 in $\mathbb{Z}$ /13 is 4, as it contains 4 elements.

3. Can there be more than one subgroup of order 4 in $\mathbb{Z}$ /13?

Yes, there can be more than one subgroup of order 4 in $\mathbb{Z}$ /13. In fact, there are exactly 2 subgroups of order 4 in this group, which are {0, 3, 6, 9} and {0, 2, 4, 6}.

4. How do you determine if a subgroup of order 4 in $\mathbb{Z}$ /13 is cyclic?

A subgroup of order 4 in $\mathbb{Z}$ /13 is cyclic if there exists an element in the subgroup that generates the entire subgroup. In other words, if you raise that element to different powers, you should get all the elements in the subgroup. If there is no such element, then the subgroup is not cyclic.

5. Is the subgroup of order 4 in $\mathbb{Z}$ /13 a normal subgroup?

Yes, the subgroup of order 4 in $\mathbb{Z}$ /13 is a normal subgroup. This is because in any group of prime order, all subgroups are normal subgroups. Therefore, since the order of $\mathbb{Z}$ /13 is 13, all its subgroups (including the subgroup of order 4) are normal subgroups.

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