Characterizing subgroups of a cyclic group

[Mr Davis 97]

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.

 

[fresh_42]

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?

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.

 

[Mr Davis 97]

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?

 

[fresh_42]

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 $|H|=k\,?$

 

[Mr Davis 97]

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 $n ~|~ lk$.

 

[fresh_42]

All I can see is that if $g^l \in H$, then $g^{lk} = 1$, and so $n ~|~ lk$.

Or $lk=a\cdot n = amk$ ...

 

[Mr Davis 97]

Or $lk=a\cdot n = amk$ ...

Is the point that since $l = am$, we then know that $g^l = (g^m)^a$, so $g^l \in H'$?

 

[fresh_42]

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 $|H'|=k=|H|$.

 

[Mr Davis 97]

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.

 

[fresh_42]

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.