Find all subgroups of the octic group

[micromass]

Really? Adjoin every element with the cyclic subgroups? This is a ridiculous, tedious problem.

It can be done very economically! You need to do very few calculations!

Let me take two arbitrary example

Take \theta and \gamma. Adjoining these together would give me a subgroup of at least order 4 (indeed, e needs to be in the subgroup, so the subgroup has at least order 3. So by Lagrange, it has at least order 4). But \{\theta,\gamma,e,a^2\} is such a subgroup.

Take a and \Delta. In the subgroup generated by these would have to be

e,a,a^2,a^3,\Delta

So the group has at least order 5. So by Lagrange, it has order 8.

These reasonings go very quickly. You'll end up with very few calculations!

 

[Shackleford]

I still have to look at the various combinations of alpha squared, beta, gamma, delta, and theta. I can reason they should yield subgroups of order 4. I would have to still look at each of them to find the distinct subgroups. There has to be a shortcut to know which is the magic element.

 

[amanda_ou812]

i think you are in my class. I emailed proof about this problem. I will copy and paste. If you are not in my class, then i hope this helps.

Me:#14) Find all subgroups of the octic group.

To understand this question, I am trying to understand example 5 which lists the subgroups of S3.

So, S3 has an order of 6 so all the subgroups will have the order of 1, 2, 3, or 6.

So, the subgroups with an order of 1 are only {(1)}. The subgroups with order 2 are {(1), (1, 2)}; {(1), (1,3)}, {(1), (2, 3)}. But why aren't other combinations of those element also subgroups ? Like, {(1), (1, 3), (2, 3)}? Doesn't this have an order of 2 as well? Continuing, the book lists the next subgroup as {(1), (1, 2, 3), (1, 3, 2)} which has order of 3. But, in the same vein as my question above, why are {(1), (1, 2, 3)} and {(1), (1, 3, 2)} not subgroups? Don't they have an order of three as well?

Proff:

Remember that a subgroup has to be closed with respect to multiplication. So if the subgroup contains (1 3 2) it must also contain its square which is (1 2 3)
If a subgroup contains (1 3 2) it must also contain ITS square which is (1 2 3) so you can't have a subgroup with one and not the other.

If a subgroup contains an element, it must also contain all powers of the element. That is why you can't have a subgroup with just (1) and (1 2 3)

If a subgroup contains 2 elements, it must also contain all possible products, so if a and b are in the set, so are a*a, a*b, b*a, b*b, a*b*a, a*b*b,...
With a small finite group there are only so many of these products that are actaually distinct.

Me:

I think i understand now.

So, the octic group has order of 8, so the subgroups have the order of 1, 2, 4, or 8.

The elements of the octic group by order are, e (order of 1), a^2, b, y, Delta, theta (all order of two), and a , a^3 (order of 4)

So H1 = {e} because it has an order of 1 and it is closed
H2 = {e, a^2} this has an order of 2 and it is closed bc e*e=e, e*a^2 =a^2*e=a^2, and a^2 *a^2=e
H3 = {e, b} same reason as above
H4 = {e, y} same reason
H5 = {e, Delta} same reason
H6= {e, theta) same reason

H7 = {e, a, a^2} this has order of 4, and a^2 was added so that the subgroup will be closed
H8 = {e, a^3, a^2} same reason

H9 = G

My method is to go element by element so long as it follows the order rule, and add the squares (or other elements) if needed.

Would you suppose this works?
Proff:
Almost,

remember if you have a, you also have a^2 and a^3 (and a^4, a^5,...

So, one subgroup is e, a, a^2, a^3
However, reading this post makes me think i have my subgroups incorrect. With my current subgroups, i got H1, H2, H9, H10 as normal

 

[amanda_ou812]

Oh, i changed H9 to = {e, a, a^2, a^3} and H10 =G

 

[amanda_ou812]

also, i am looking at the most recent pic that you posted and I don't understand how you got the two subsets on the right side of the page.

 

[Shackleford]

I understand now what to do. I just hate the tediousness of it.

Send me a message on here, and I'll tell you if we're in the same class.

 

[Shackleford]

also, i am looking at the most recent pic that you posted and I don't understand how you got the two subsets on the right side of the page.

I did a Google search for how to do this problem, and I saw that you had to adjoin the elements with alpha-squared to get those two subgroups.