Find all subgroups of the octic group

[Shackleford]

Here is my work thus far, and I included any pertinent notes.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110802_182239.jpg?t=1312327545

 

[micromass]

Looks good so far, shackleford!


I'll also assume that the "octic group" is just the dihedral group of order 8. I'm not familiar with the name octic group...

 

[Shackleford]

Looks good so far, shackleford!


I'll also assume that the "octic group" is just the dihedral group of order 8. I'm not familiar with the name octic group...

It's the operations on the square numbered

4 3
1 2.

I'm missing an element from the subgroup of order 4.

 

[micromass]

It's the operations on the square numbered

4 3
1 2.

Indeed, that would be the dihedral group.

I'm missing an element from the subgroup of order 4.

Well, what happens if you do (1 2 3 4)(1 2 3 4)?? Which familiar element do you get?

 

[Shackleford]

Indeed, that would be the dihedral group.
Well, what happens if you do (1 2 3 4)(1 2 3 4)?? Which familiar element do you get?

Well, I figured I could simply do another permutation different than alpha and alpha-squared.

(1,3,2,4)

 

[Shackleford]

Indeed, that would be the dihedral group.
Well, what happens if you do (1 2 3 4)(1 2 3 4)?? Which familiar element do you get?

(1 2 3 4)(1 2 3 4) = (1,3)(2,4) = α²

I already have (1,3)(2,4) in the subgroup of order 2.

 

[micromass]

(1 2 3 4)(1 2 3 4) = (1,3)(2,4) = α²

I already have (1,3)(2,4) in the subgroup of order 2.

Your elements can occur in multiple subgroups. That's not forbidden...

 

[Shackleford]

Your elements can occur in multiple subgroups. That's not forbidden...

I understand that, but the order is not four. The order of the permutation is the LCM of the order of the individual cycles.

 

[Shackleford]

Also, the next problem is Find all normal subgroups of the octic group.

Wouldn't that be all the subgroups since they each have the identity in them? The left and right cosets would be equal since the identity captures all of the elements in the octic group.

 

[micromass]

I understand that, but the order is not four. The order of the permutation is the LCM of the order of the individual cycles.

Yes, the order of a² is 2. But why would that prevent {1,a,a²,a³} to form a subgroup??

 

[micromass]

Also, the next problem is Find all normal subgroups of the octic group.

Wouldn't that be all the subgroups since they each have the identity in them? The left and right cosets would be equal since the identity captures all of the elements in the octic group.

What does having the identity has to do with normality??

Calculate the left and right cosets of \beta for example. You'll see that the left and right cosets will not be equal...

 

[Shackleford]

Yes, the order of a² is 2. But why would that prevent {1,a,a²,a³} to form a subgroup??

Oh, you're right. The order of that element divides 4. In general then, the subgroup is not unique then because I could stick another second-order element in its place.

 

[Shackleford]

What does having the identity has to do with normality??

Calculate the left and right cosets of \beta for example. You'll see that the left and right cosets will not be equal...

Look at my subgroups. Each of the subgroups have the identity.

Normality is set equality.

xH = Hx for all x in G.

H = G₁,...,G₈ = G.

 

[micromass]

Oh, you're right. The order of that element divides 4. In general then, the subgroup is not unique then because I could stick another second-order element in its place.

No, you couldn't

Having a in the subgroup will force a² in the subgroup, by definition.

 

[micromass]

Look at my subgroups. Each of the subgroups have the identity.

Yes, all subgroups of every group will contain the identity. This has nothing to do with normality.

 

[Shackleford]

No, you couldn't

Having a in the subgroup will force a² in the subgroup, by definition.

Oh. The subgroup has to be closed. That's why alpha-squared has to be in there.

 

[Shackleford]

Yes, all subgroups of every group will contain the identity. This has nothing to do with normality.

Duh. Sorry. I had a brain fart there. I need to look at the product of each subgroup with each element in G to determine if the subgroup is normal in G.

I have the handy group table in the back of the book.

 

[Shackleford]

Okay. I found that the only two normal subgroups of the octic group are, according to my notation, G₁ = {e} and G₂ = {e, alpha²}. G is not normal in G.

 

[micromass]

Okay. I found that the only two normal subgroups of the octic group are, according to my notation, G₁ = {e} and G₂ = {e, alpha²}. G is not normal in G.

These aren't all the normal subgroups yet, you're missing one. (Hint: subgroups of index 2 are always normal)

By the way, I'm also not convinced that you actually found all the subgroups either...

 

[Shackleford]

These aren't all the normal subgroups yet, you're missing one. (Hint: subgroups of index 2 are always normal)

By the way, I'm also not convinced that you actually found all the subgroups either...

Then I have no idea how to find them all. How do I find all subgroups of the octic group?

 

[micromass]

Then I have no idea how to find them all. How do I find all subgroups of the octic group?

Well, first you take an arbitrary element x and see what the subgroup generated by x is. This will give you all cyclic subgroups of the octic group. This is what you have done.

But you're not done yet. There-after, you will have to take two arbitrary elements x and y, and you'll have to see what kind of group that those two elements generate.

Then you'll have to take 3 elements and see what they generate. And so on...

This sounds like a lot of work, but it isn't. If you're smart, then you can cut a lot of work.

For example, let's say that you are in STEP 2 and you see what group is generated by 2 elements. Obviously, you don't need to check it for a and a², since these two elements will lie in the same cyclic subgroup. And obviously, the group generated by a and delta is the same as the group generated by a³ and delta.

 

[Shackleford]

It also looks like G₇ is a normal subgroup, too.

 

[Shackleford]

Okay. I generated the subgroups based on the distinct powers of the elements of G. The only nontrivial generated subgroups are

<α> = {α, α², α³, e}
<α³> = {α, α², α³, e}

I'm not smart. I'm not following your shortcut.

Do I multiply α, α², α³ by each of the other elements? I see they did that for A = {alpha, beta}.

If I do alpha with beta, gamma, theta, and delta I get {gamma, delta, beta, and theta}. It looks like I would get similar results with alpha squared and cubed.

 

[Shackleford]

Okay, I found that someone had already worked out this problem in response to a question on a different website.

If you adjoin any element with <a> or <a³> you get G. How do you quickly see this? Just by looking at the table?

Adjoining the other elements with <a²> gives you two distinct subgroups of order 4. I suppose I'm missing another subgroup of order 4. Well, now, I have these two new subgroups, the identity, but it would appear I'm missing one.

http://i111.photobucket.com/albums/n149/camarolt4z28/20110804161007509.jpg?t=1312492825

 

[micromass]

Okay, I found that someone had already worked out this problem in response to a question on a different website.

If you adjoin any element with <a> or <a³> you get G. How do you quickly see this? Just by looking at the table?

<a> has order 4, adjoining an element to <a> gives you a subgroup with order at least 5. But by Lagrange's theorem, the order must divide 8. So adjoining an element to <a> gives you a group of order 8: the entire group.

[/QUOTE]
Adjoining the other elements with <a²> gives you two distinct subgroups of order 4. I suppose I'm missing another subgroup of order 4. Well, now, I have these two new subgroups, the identity, but it would appear I'm missing one.

http://i111.photobucket.com/albums/n149/camarolt4z28/20110804161007509.jpg?t=1312492825 [/QUOTE]

Why do you get the impression that you're missing one?

 

[Shackleford]

<a> has order 4, adjoining an element to <a> gives you a subgroup with order at least 5. But by Lagrange's theorem, the order must divide 8. So adjoining an element to <a> gives you a group of order 8: the entire group.

Adjoining the other elements with <a²> gives you two distinct subgroups of order 4. I suppose I'm missing another subgroup of order 4. Well, now, I have these two new subgroups, the identity, but it would appear I'm missing one.

http://i111.photobucket.com/albums/n149/camarolt4z28/20110804161007509.jpg?t=1312492825 [/QUOTE]

Why do you get the impression that you're missing one?[/QUOTE]

Well, you told me I was missing a couple of subgroups. I found two new subgroups of order 4. I'm mistaken. I need to check that the elements of this subgroup divide 4. Let me check.

 

[micromass]

Adjoining the other elements with <a²> gives you two distinct subgroups of order 4. I suppose I'm missing another subgroup of order 4. Well, now, I have these two new subgroups, the identity, but it would appear I'm missing one.

http://i111.photobucket.com/albums/n149/camarolt4z28/20110804161007509.jpg?t=1312492825

Why do you get the impression that you're missing one?

Well, you told me I was missing a couple of subgroups. I found two new subgroups of order 4. I'm mistaken. I need to check that the elements of this subgroup divide 4. Let me check.

Yes, the subgroups you found are ok. In total, there are 10 subgroups! So you found them all!

 

[Shackleford]

Yes, the subgroups you found are ok. In total, there are 10 subgroups! So you found them all!

How do I know that's all? I'm trying to figure out the best approach.

In finding the distinct subgroups of a group, I know to look at the sets generated by the powers of each of the elements. I also need to look at the other possible combinations.

The set generated by powers of alpha yields a subgroup of order 4. It seems that the order of the subgroups in looking at the other possible combinations is important. Any other element adjoined with the set generated by alpha creates a subgroup that must jump up to 8 in order to divide 8. The other subgroups have order 2, so adjoining an additional element with each of them can generate a set that jumps to 4 or 8. How did we know to adjoin the other elements with alpha-squared? Is it because it's commutative?

 

[micromass]

How do I know that's all? I'm trying to figure out the best approach.

In finding the distinct subgroups of a group, I know to look at the sets generated by the powers of each of the elements. I also need to look at the other possible combinations.

The set generated by powers of alpha yields a subgroup of order 4. It seems that the order of the subgroups in looking at the other possible combinations is important. Any other element adjoined with the set generated by alpha creates a subgroup that must jump up to 8 in order to divide 8. The other subgroups have order 2, so adjoining an additional element with each of them can generate a set that jumps to 4 or 8. How did we know to adjoin the other elements with alpha-squared? Is it because it's commutative?

We didn't know that we had to adjoing a². We have to adjoin every possible element to our subgroups and show that you get nothing else.

So you have found all subgroups, but you might still want to show that you found them all.

 

[Shackleford]

We didn't know that we had to adjoing a². We have to adjoin every possible element to our subgroups and show that you get nothing else.

So you have found all subgroups, but you might still want to show that you found them all.

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