Subgroups of a symmetric group

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RDKGames
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#1
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#1
Got two questions on this that I can't seem to get my head around, then again maybe doing this with a headache isn't the best of scenarios :ninja:

First one is below. How do I go about getting the elements for part (c) ??

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Zacken
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#2
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(Original post by RDKGames)
Got two questions on this that I can't seem to get my head around, then again maybe doing this with a headache isn't the best of scenarios :ninja:

First one is below. How do I go about getting the elements for part (c) ??
It's just a cyclic subgroup, so keep calculating powers of \alpha till you reach \alpha^n = e for some (small) n. Then your subgroup is \{e,\alpha,\alpha^2 ,\ldots, \alpha^{n-1}\}.
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RDKGames
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(Original post by Zacken)
It's just a cyclic subgroup, so keep calculating powers of \alpha till you reach \alpha^n = e for some (small) n. Then your subgroup is \{e,\alpha,\alpha^2 ,\ldots, \alpha^{n-1}\}.
Thanks, got it now

What about part (d) of this? How can I go about finding a subgroup here??
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DFranklin
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(Original post by RDKGames)
Thanks, got it now

What about part (d) of this? How can I go about finding a subgroup here??
Well, what is Cayley's theorem (and what does the standard proof look like)? How might this relate to the rest of the question?
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RichE
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(Original post by RDKGames)
Thanks, got it now

What about part (d) of this? How can I go about finding a subgroup here??
What was your answer to (c)? If no, do you know a group that G is isomorphic to?
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RDKGames
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(Original post by RichE)
What was your answer to (c)?
I said that they are indeed a isomorphic as there exists a bijection \phi : G \rightarrow \mathbb{Z}_8 defined by the map below but I'm unsure if this can help in part (d) which I'm currently trying to follow from Cayley Th. proof.
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RichE
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(Original post by RDKGames)
I said that there is indeed a bijection \phi : G \rightarrow \mathbb{Z}_8 defined by the map below but I'm unsure if this can help in part (d) which I'm currently trying to follow from Cayley Th. proof.
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A bijection isn't immediately an isomorphism (and typically won't be). In any case finding an explicit isomorphism is probably unnecessary here. It would be enough to find an order 8 element in G. Did you consider the elements' orders?
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RDKGames
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(Original post by RichE)
A bijection isn't immediately an isomorphism (and typically won't be). In any case finding an explicit isomorphism is probably unnecessary here. It would be enough to find an order 8 element in G. Did you consider the elements' orders?
I did not consider the orders. I don't think there is an order 8 element in G otherwise I've done part (b) wrong...

ord(1)= 1
ord(2)=ord(7)=ord(8)=ord(13)= 4
ord(4)=ord(11)=ord(13)= 2

and I can see the orders are different for \mathbb{Z}_8.
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RichE
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(Original post by RDKGames)
I did not consider the orders. I don't think there is an order 8 element in G otherwise I've done part (b) wrong...

ord(1)= 1
ord(2)=ord(7)=ord(8)=ord(13)= 4
ord(4)=ord(11)=ord(13)= 2

and I can see the orders are different for \mathbb{Z}_8.
Correct - so you've shown G isn't isomorphic to Z_8. Can you tell what order 8 group it is isomorphic to?
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RDKGames
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(Original post by RichE)
Correct - so you've shown G isn't isomorphic to Z_8. Can you tell what order 8 group it is isomorphic to?
No I cannot unfortunately

Also, I have followed Cayley's theorem, and so is it enough to say that if I use the mult. table's each row as a permutation \sigma_i \in S_8, then \{ \sigma_1, \sigma_2, ..., \sigma_8 \}=G' \subset S_8 is isomorphic to G ??
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Zacken
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(Original post by RDKGames)
No I cannot unfortunately
Hint: If it's not C_8 then it's a direct product of (two, in this case) cyclic groups.
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RichE
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(Original post by RDKGames)
No I cannot unfortunately

Also, I have followed Cayley's theorem, and so is it enough to say that if I use the mult. table's each row as a permutation \sigma_i \in S_8, then \{ \sigma_1, \sigma_2, ..., \sigma_8 \}=G' \subset S_8 is isomorphic to G ??
Sure, but that might give you a rather involved and opaque view of G's structure.

Do note that G is abelian.
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DFranklin
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(Original post by RichE)
Sure, but that might give you a rather involved and opaque view of G's structure.
I have to say that given the explicit mention of Cayley's theorem before I'd have assumed this is what they're actually after, although I agree with your point that in many ways it's not the most informative way of looking at things.

I do think that it's good to do an explicit "build an isomorphism using the methodology in Cayley's theorem" once (or even twice!). So I'm partly thinking this may be that one time...
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RichE
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(Original post by DFranklin)
I have to say that given the explicit mention of Cayley's theorem before I'd have assumed this is what they're actually after, although I agree with your point that in many ways it's not the most informative way of looking at things.

I do think that it's good to do an explicit "build an isomorphism using the methodology in Cayley's theorem" once (or even twice!). So I'm partly thinking this may be that one time...
Re-reading the question I'm fairly certain you're right, though I think my point stands that this isn't typically a helpful way into appreciating group structure (as you yourself say above) but most likely what is expected here.
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