Characterizing subgroups of a cyclic group

In summary: Don't forget that ##k##, resp. ##m##, is fixed so that this notation is the correct oneIn summary, we want to show that for every subgroup H of cyclic group G, H is equal to <g^(|G|/|H|)>, or the subgroup generated by g^(|G|/|H|). We can see that |H| is equal to the order of <g^(|G|/|H|)> because if (g^(|G|/|H|))^p = e for some value of p < |H|, we would have that g^q = e for some q < |G|, which is a contradiction. However, to show that
  • #1
Mr Davis 97
1,462
44

Homework Statement


Show that for every subgroup ##H## of cyclic group ##G##, ##H = \langle g^{\frac{|G|}{|H|}}\rangle##.

Homework Equations

The Attempt at a Solution


At the moment the most I can see is that ##|H| = |\langle g^{\frac{|G|}{|H|}}\rangle|##. This is because if ##(g^{\frac{|G|}{|H|}})^p = e## for some value of ##p < |H|##, we would have that ##g^q = e## for some ##q < |G|##, which is a contradiction.

But I'm not seeing why they must be the same subgroup. I was going to try to prove that if two subgroups of cyclic group have the same order, then they must be the same group. If I did this I think I would first need to prove that every subgroup of a cyclic group is cyclic.
 
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  • #2
Mr Davis 97 said:

Homework Statement


Show that for every subgroup ##H## of cyclic group ##G##, ##H = \langle g^{\frac{|G|}{|H|}}\rangle##.

Homework Equations

The Attempt at a Solution


At the moment the most I can see is that ##|H| = |\langle g^{\frac{|G|}{|H|}}\rangle|##. This is because if ##(g^{\frac{|G|}{|H|}})^p = e## for some value of ##p < |H|##, we would have that ##g^q = e## for some ##q < |G|##, which is a contradiction.

But I'm not seeing why they must be the same subgroup. I was going to try to prove that if two subgroups of cyclic group have the same order, then they must be the same group, but I think that's basically what I'm already trying to prove
Haven't you done this already?
Mr Davis 97 said:
Don't they not generate the group because they have a different order than the generator ##g##, since ##|g^k| = \frac{n}{\operatorname{gcd} (k,n)}##?
With ##|G|=n## and ##k=|H|## we have ##k\,|\,n##, say ##n=k\cdot m## and we are interested in the order of ##g^m##. Now apply your formula.
 
  • #3
fresh_42 said:
Haven't you done this already?

With ##|G|=n## and ##k=|H|## we have ##k\,|\,n##, say ##n=k\cdot m## and we are interested in the order of ##g^m##. Now apply your formula.
Well what we have is that ##|\langle g^{\frac{|G|}{|H|}} \rangle| = |g^{\frac{n}{k}}| = |g^m| = \frac{n}{\gcd (m,n)} = \frac{mk}{\gcd (m,mk)}= \frac{mk}{m} = k = |H|##. So now I have that the orders are equal. But how I should I go about showing that they are the same group?
 
  • #4
Mr Davis 97 said:
Well what we have is that ##|\langle g^{\frac{|G|}{|H|}} \rangle| = |g^{\frac{n}{k}}| = |g^m| = \frac{n}{\gcd (m,n)} = \frac{mk}{\gcd (m,mk)}= \frac{mk}{m} = k = |H|##. So now I have that the orders are equal. But how I should I go about showing that they are the same group?
Let ##H':=\langle \,g^m\,|\,g^n=1\, , \,n=mk \,\rangle## and ##g^l \in H##. Now what does the last property mean, if we only know that ##|H|=k\,?##
 
  • #5
fresh_42 said:
Let ##H':=\langle \,g^m\,|\,g^n=1\, , \,n=mk \,\rangle## and ##g^l \in H##. Now what does the last property mean, if we only know that ##|H|=k\,?##
All I can see is that if ##g^l \in H##, then ##g^{lk} = 1##, and so ##n ~|~ lk##.
 
  • #6
Mr Davis 97 said:
All I can see is that if ##g^l \in H##, then ##g^{lk} = 1##, and so ##n ~|~ lk##.
Or ##lk=a\cdot n = amk## ...
 
  • #7
fresh_42 said:
Or ##lk=a\cdot n = amk## ...
Is the point that since ##l = am##, we then know that ##g^l = (g^m)^a##, so ##g^l \in H'##?
 
  • #8
Mr Davis 97 said:
Is the point that since ##l = am##, we then know that ##g^l = (g^m)^a##, so ##g^l \in H'##?
Yes, and this is true for all elements of ##H## since we didn't set any conditions on the element - don't forget that ##|H'|=k=|H|##.
 
  • #9
fresh_42 said:
Yes, and this is true for all elements of ##H## since we didn't set any conditions on the element - don't forget that ##|H'|=k=|H|##.
So since ##H \le H'## and ##|H| = |H'|##, we have ##H = H'##?

Could you explain a little bit why ##\langle g^{\frac{n}{k}}\rangle = \langle g^m~|~g^n=1, n = mk \rangle##? And also what the notation on the RHS means exactly.
 
  • #10
Mr Davis 97 said:
So since ##H \le H'## and ##|H| = |H'|##, we have ##H = H'##?
Yes.
Could you explain a little bit why ##\langle g^{\frac{n}{k}}\rangle = \langle g^m~|~g^n=1, n = mk \rangle##? And also what the notation on the RHS means exactly.
##\langle g^{\frac{n}{k}}\rangle = \langle g^m \rangle## by the definitions of ##k,m,n##

However, this notation is strictly seen not the group we deal with, because here ##g## would be of infinite order as the order isn't mentioned. The correct way to write this group is ##\langle g^{\frac{n}{k}}\rangle = \langle g^m~|~g^n=1, n = mk \rangle##, because it lists all relevant data. In general the notation is
$$
G = \langle \, g_1, \ldots ,g_n \,|\, r_1,\ldots ,r_m \,\rangle \text{ or } G = \langle \, g_1, \ldots ,g_n \,|\, r_1=\ldots =r_m=1 \,\rangle
$$
where the ##g_i## are generators of the group, possibly infinitely many, and ##r_j## are the relations, i.e. words built over the alphabet ##g_i## which equal ##1##. The relation, in our case only one, namely ##r_1=g^n=1##, makes the group cyclic and finite. ##n=mk## could have been omitted so that ##H=\langle \,g^m\,|\,g^{mk}\,\rangle## fits the just defined notation. ##r_j=1## is usually omitted, as the position behind ##"|"## indicates that it is a relation, so they are simply listed.
 

Related to Characterizing subgroups of a cyclic group

1. What is a cyclic group?

A cyclic group is a mathematical structure that is generated by a single element. This means that the group can be formed by repeatedly applying the group operation (such as addition or multiplication) to the generator element.

2. How do you characterize subgroups of a cyclic group?

To characterize subgroups of a cyclic group, you need to determine the elements that are generated by the subgroup. This can be done by finding the powers of the generator element that form the subgroup, and then using these elements to determine the subgroup's structure.

3. What is the order of a subgroup in a cyclic group?

The order of a subgroup in a cyclic group is the number of elements in the subgroup. This can be found by taking the highest power of the generator element that is contained in the subgroup.

4. Can a cyclic group have more than one subgroup?

Yes, a cyclic group can have multiple subgroups. In fact, any element in a cyclic group can be used as a generator for a subgroup, so the number of subgroups in a cyclic group is infinite.

5. How are subgroups of a cyclic group related to its generator?

The generator of a cyclic group is the element that generates all of its subgroups. This means that every subgroup of a cyclic group can be formed by repeatedly applying the group operation to the generator element. Additionally, the order of a subgroup is always a factor of the order of the cyclic group's generator.

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