Subgroups of a cyclic group

Roxanne18

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Hi,

I was wondering if anyone could help with this question please

Let G=<x> be a cyclic group generated by x of order 21.
List the subgroups of G.
Which of them are proper subgroups of G.

I don't even know where to start with this question.
So I know G={ e, x , x^2, ..... , x^20}
and x^k is an element of G that has order 21/ gcd( 21, k )

But I dont know what or how to find the subgroup.

Any help would be appreciated (:

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ThomH97

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(Original post by Roxanne18)
and x^k is an element of G that has order 21/ gcd( 21, k )

So what possible orders could your generating elements have?

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3pointonefour

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(Original post by Roxanne18)
Hi,

I was wondering if anyone could help with this question please

Let G=<x> be a cyclic group generated by x of order 21.
List the subgroups of G.
Which of them are proper subgroups of G.

I don't even know where to start with this question.
So I know G={ e, x , x^2, ..... , x^20}
and x^k is an element of G that has order 21/ gcd( 21, k )

But I dont know what or how to find the subgroup.

Any help would be appreciated (:

Use lagrange's theorem to get you started

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Roxanne18

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(Original post by 3pointonefour)
Use lagrange's theorem to get you started

so by Lagrange's theorem if G is a subgroup of H then the order of g divides the order of H

So the order is 21
1, 3, 7 and 21 divide 21
So they are the order of my subgroups?
How am I meant to list them though ?

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RDKGames

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(Original post by Roxanne18)
Hi,

I was wondering if anyone could help with this question please

Let G=<x> be a cyclic group generated by x of order 21.
List the subgroups of G.
Which of them are proper subgroups of G.

I don't even know where to start with this question.
So I know G={ e, x , x^2, ..... , x^20}
and x^k is an element of G that has order 21/ gcd( 21, k )

But I dont know what or how to find the subgroup.

Any help would be appreciated (:

So $G = \{e,x,x^2, \ldots, x^{20} \}$ is your group.

Now to find the subgroups of

, you essentially take every element of $x^k \in G$ and produce a new group $\langle x^k \rangle$ . The order of this group must be either 1, 3, 7, 21 because it must divide the order of

.

Here's an example, let's pick $\langle x^3 \rangle = \{ e, x^3, x^6, x^9, x^{12}, x^{15}, x^{18}, x^{21}, ... \}$ but notice that the last element I've listed is just $x^{21}=e$ . So this just cycles back and we get that: $\langle x^3 \rangle = \{ e, x^3, x^6, x^9, x^{12}, x^{15}, x^{18} \}$ and this is one subgroup of

.

The two trivial subgroups are also just $\langle e \rangle = \{ e \}$ and

itself. Though I'm not sure if you need to list these.

There is an easy way to think about the other subgroups. I'll return to my previous example. Here we want 7 elements in our subgroup. So whatever

we pick, we want $(x^k)^7 = x^{7k}$ to be a multiple of $x^{21}$ . Clearly

is the smallest one to satisfy this, so $\langle x^3 \rangle$ is a subgroup and you can list it as I've done above.

This is all good if my fading knowledge of group theory isn't failing me here!

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Roxanne18

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(Original post by RDKGames)
So $G = \{e,x,x^2, \ldots, x^{20} \}$ is your group.

Now to find the subgroups of

, you essentially take every element of $x^k \in G$ and produce a new group $\langle x^k \rangle$ . The order of this group must be either 1, 3, 7, 21 because it must divide the order of

.

Here's an example, let's pick $\langle x^3 \rangle = \{ e, x^3, x^6, x^9, x^{12}, x^{15}, x^{18}, x^{21}, ... \}$ but notice that the last element I've listed is just $x^{21}=e$ . So this just cycles back and we get that: $\langle x^3 \rangle = \{ e, x^3, x^6, x^9, x^{12}, x^{15}, x^{18} \}$ and this is one subgroup of

.

The two trivial subgroups are also just $\langle e \rangle = \{ e \}$ and

itself. Though I'm not sure if you need to list these.

There is an easy way to think about the other subgroups. I'll return to my previous example. Here we want 7 elements in our subgroup. So whatever

we pick, we want $(x^k)^7 = x^{7k}$ to be a multiple of $x^{21}$ . Clearly

is the smallest one to satisfy this, so $\langle x^3 \rangle$ is a subgroup and you can list it as I've done above.

This is all good if my fading knowledge of group theory isn't failing me here!

Ah okay I think this makes sense thank you.
Can I just check though :
so is there only one subgroup of each size, as in am I expecting to get 4 sub groups, one of each order ?

and to find the subgroup of order 3 it would be $(x^k)^3 = x^{3k}$ so k = 7 then I put <x^7> and work it out the same way ?

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