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 G, 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 G.

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 G.

The two trivial subgroups are also just \langle e \rangle = \{ e \} and G 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 x^k we pick, we want (x^k)^7 = x^{7k} to be a multiple of x^{21}. Clearly k=3 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 G, 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 G.

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 G.

The two trivial subgroups are also just \langle e \rangle = \{ e \} and G 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 x^k we pick, we want (x^k)^7 = x^{7k} to be a multiple of x^{21}. Clearly k=3 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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