MHB Question on subgroup and order of the elements

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The discussion revolves around the group of symmetries of a regular heptagon, specifically focusing on the cyclic subgroup generated by a rotation. The subgroup R is confirmed to be represented as ⟨(1 2 3 4 5 6 7)⟩, with members having an order of 7. Participants clarify that since the group G has 14 elements, any subgroup H, other than G itself, must have an order of 1 or 2, making it cyclic. The conversation also addresses the need to identify the specific subgroups corresponding to these orders. The thread concludes with inquiries about subgroup properties and their implications in group theory.
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Let G be the group of symmetries (including flips) of the regular heptagon (7-gon).

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As usual, we regard the elements of G as permutations of the set of vertex labels; thus, G ≤ S7.

(a) Let σ denote the rotation of the 7-gon that takes the vertex 1 to the vertex 2. Write down the cyclic subgroup R := ⟨σ⟩ as a set of elements of S7 in cycle notation.

(b) What are the orders of each of the elements of R?

does this mean R := ⟨( 1 2 3 4 5 6 7 )⟩?
and I am unsure how to answer part b)

thanks
 

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himynameJEF said:
Let G be the group of symmetries (including flips) of the regular heptagon (7-gon).
As usual, we regard the elements of G as permutations of the set of vertex labels; thus, G ≤ S7.

(a) Let σ denote the rotation of the 7-gon that takes the vertex 1 to the vertex 2. Write down the cyclic subgroup R := ⟨σ⟩ as a set of elements of S7 in cycle notation.

(b) What are the orders of each of the elements of R?

does this mean R := ⟨( 1 2 3 4 5 6 7 )⟩?
and I am unsure how to answer part b)

thanks

Hi himynameJEF, welcome to MHB! ;)

Yes, that is correct for what $R$ stands for.
Which elements exactly are in $R$ though?
That is, what is for instance ( 1 2 3 4 5 6 7 )2?

As for (b), let's start with the order of ( 1 2 3 4 5 6 7 ).
What is it?
 
I like Serena said:
Hi himynameJEF, welcome to MHB! ;)

Yes, that is correct for what $R$ stands for.
Which elements exactly are in $R$ though?
That is, what is for instance ( 1 2 3 4 5 6 7 )2?

As for (b), let's start with the order of ( 1 2 3 4 5 6 7 ).
What is it?

hi! :)

( 1 2 3 4 5 6 7 )2 would be ( 1 3 5 7 2 4 6 )?

and that would be an order of 7?

thanks!
 
himynameJEF said:
hi! :)

( 1 2 3 4 5 6 7 )2 would be ( 1 3 5 7 2 4 6 )?

and that would be an order of 7?

thanks!

Yep. (Nod)

So now we have 2 elements in $R$ that both have order 7.
How about ( 1 2 3 4 5 6 7 )3? (Wondering)
 
I like Serena said:
Yep. (Nod)

So now we have 2 elements in $R$ that both have order 7.
How about ( 1 2 3 4 5 6 7 )3? (Wondering)

thanks i understand it now! :)

also another question

Let H be any subgroup of G other than G itself. explain why H is cyclic?

since G is prime then |H| is 1 or 7. then H must equal G and it would be cyclic but the question says any other subgroup other than G so H must be {e}? is this cyclic? I am confused
 
himynameJEF said:
thanks i understand it now! :)

also another question

Let H be any subgroup of G other than G itself. explain why H is cyclic?

since G is prime then |H| is 1 or 7. then H must equal G and it would be cyclic but the question says any other subgroup other than G so H must be {e}? is this cyclic? I am confused

Isn't the order of G 14?
That is, aren't there 14 symmetries with rotations and flips?
 
I like Serena said:
Isn't the order of G 14?
That is, aren't there 14 symmetries with rotations and flips?

oh! yes sorry your right
so how would i explain this? and then determine the number of subgroups?

thanks!
 
himynameJEF said:
oh! yes sorry your right
so how would i explain this? and then determine the number of subgroups?

thanks!

Since H is a subgroup of G, it means that we must have |H|=1, |H|=2, |H|=7, or |H|=14, mustn't we?
Which are the subgroups that correspond to those?
 

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