All subgroups of {1, 2, 4, 8, 16, 32, 43, 64} under multiplication mod 85

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HMPARTICLE
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Homework Statement


Determine all the subgroups of (A,x_85) justify.
where A = {1, 2, 4, 8, 16, 32, 43, 64}.

The Attempt at a Solution


To determine all of the subgroups of A, we find the distinct subgroups of A.
<1> = {1}
<2> = {1,2,4..} and so on?
<4> = ...
...

is this true? are there any other possible subgroups, i know i havnt posted my full solution.
 
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HMPARTICLE said:

Homework Statement


Determine all the subgroups of (A,x_85) justify.
where A = {1, 2, 4, 8, 16, 32, 43, 64}.

The Attempt at a Solution


To determine all of the subgroups of A, we find the distinct subgroups of A.
<1> = {1}
<2> = {1,2,4..} and so on?
<4> = ...
...

is this true? are there any other possible subgroups, i know i havnt posted my full solution.

Ok, so you're dealing with a group of integers under multiplication mod 85. I think you should fill in the ...'s before anyone can figure out whether you have all of the subgroups.
 
<1> = {1}
<2> = {1,2,4, 8,16,32,43,64}
<4> = {4,16,64,1}
<8> = {1,2,4, 8,16,32,43,64}
<16> = {16,1}

Them are all the distinct subgroups of the group. for example. <2> = <43> .
 
HMPARTICLE said:
<1> = {1}
<2> = {1,2,4, 8,16,32,43,64}
<4> = {4,16,64,1}
<8> = {1,2,4, 8,16,32,43,64}
<16> = {16,1}

Them are all the distinct subgroups of the group. for example. <2> = <43> .

<2> and <8> aren't really distinct, are they? There are only four distinct subgroups. And I'm not sure what you are supposed to supply for justification. In a general group a subgroup might have more than one generator. But do you know what a cyclic group is?
 
Yes. A cyclic group is a group with order n which contains an element of order n. Or better still a cyclic group is a group which contains an element that generates the group. 2 and 8 are not distinct! Silly me!

I can't think of any other subgroups. I think I may have to show that <2> = <8> = <43> and so on?
 
HMPARTICLE said:
Yes. A cyclic group is a group with order n which contains an element of order n. Or better still a cyclic group is a group which contains an element that generates the group. 2 and 8 are not distinct! Silly me!

I can't think of any other subgroups. I think I may have to show that <2> = <8> = <43> and so on?

There is a theorem that if you have a cyclic group of order n then there is exactly one subgroup for each divisor of n. Since the four divisors of 8 are 1,2,4,8 then once you find four subgroups you know you are done.
 
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