My Algebra Questions: Answers & Updates

[JasonRox]

OMG! Oh boy, did I ever overcomplicate a simple problem.

I'll post my solution in a few minutes. I had another student plus another who hasn't touched algebra in a long time trying to get it today. We must have all overcomplicated. Dang, I'm happy I got it now.

Thanks for pointing that out too morphism.

 

[JasonRox]

Ok, here is it...

We must show that G_y = g G_x g^-1 when y = g(x). (I use brackets to denote that it an action and not an operation.)

Let g_1 e G_y. Show that $g^{-1} g_1 g$ e G_x. If this is so then, g_1 e $g G_x g^{-1}$. Because that is $g g^{-1} g_1 g g^{-1} = g_1$.

$g^{-1} g_1 g (x) = g^{-1} g_1 (y) = g^{-1} y = g^{-1} y = x$

...which means $g^{-1} g_1 g$ e G_x and so g_1 e $g G_x g^{-1}$.

There reverse is done in a similar fashion.

Therefore, G_y = g G_x g^-1.

Now, let's show the second part.

Let's create a bijection from G_y to G_x. Let the bijection be as follows:

$f(g_y) = g^{-1} g_y g$

It is clear that f is well-defined. Now, let's show it is one-to-one.

$f(g_1) = f(g_2)$

$g^{-1} g_1 g = g^{-1} g_2 g$

$g_1 = g_2$

We have just shown that it is one-to-one. Now we are left to show it's onto. That is...

If g_3 e G_x, then...

$f(g g_3 g^{-1}) = g g^{-1} g_3 g g^{-1} = g_3$

Now, all we need to show is that $g g_3 g^{-1}$ is in fact in G_y.

Well, g g_3 g^{-1} (y) = g g_3 (x) = g (x) = y, so it is in G_y.

Therefore, there is a bijection from G_y to G_x, and hence |G_y| = |G_x|.

 

[JasonRox]

Although I made the dumbest errors, I learned a decent amount.

Therefore, I like that problem.

 

[Kummer]

I have a new question now. Be free to discuss previous problems if you like, or add extra notes regarding anything. I read all the posts and think about each one.

The question is...

Let X be a G-set, let x, y, e X, and let y = g*x (the group action) for some g e G. Prove that G_y = g G_x g^-1 (stabilizer); conclude that |G_y| = |G_x|.

I proved the second part without even using the first part. It's just a matter of using...

|O(x)| = [G:G_x] , where O(x) is the orbit of x, and showing O(x) = O(y).

Any help on starting the first part?

Here is how I would do it.

If we want to show $$G_y = g_0G_xg_0^{-1}$$ where $$y=g_0x$$ for some $$g_0\in G$$. We do it by show each if a subset of each other. I do it one way to show you the idea. Let $$a\in G_y$$ we want to show $$a\in g_0 G_x g_0^{-1}$$. By definition $$ay = y$$, so $$ag_0x=g_0x$$ that means $$g_0^{-1}ag_0 x = x$$ so it means $$g_0^{-1}ag_0 \in G_x$$ so $$g_0^{-1}ag_0= b$$ for some $$b\in G_x$$. That means $$g_0^{-1}ag_0 = b$$ thus $$a = g_0bg_0^{-1} \in g_0G_xg_0^{-1}$$.

Now we can show that $$|G_y|=|G_x|$$ but $$|G_x| = |g_0Gg_0^{-1}|$$ because you can define a bijection $$\phi : G_x \mapsto g_0Gg_0^{-1}$$ as $$\phi (c) = g_0cg_0^{-1}$$. Thus, $$|G_x|=|G_y|$$.

 

[morphism]

Here is how I would do it.

If we want to show $$G_y = g_0G_xg_0^{-1}$$ where $$y=g_0x$$ for some $$g_0\in G$$. We do it by show each if a subset of each other. I do it one way to show you the idea. Let $$a\in G_y$$ we want to show $$a\in g_0 G_x g_0^{-1}$$. By definition $$ay = y$$, so $$ag_0x=g_0x$$ that means $$g_0^{-1}ag_0 x = x$$ so it means $$g_0^{-1}ag_0 \in G_x$$ so $$g_0^{-1}ag_0= b$$ for some $$b\in G_x$$. That means $$g_0^{-1}ag_0 = b$$ thus $$a = g_0bg_0^{-1} \in g_0G_xg_0^{-1}$$.

Now we can show that $$|G_y|=|G_x|$$ but $$|G_x| = |g_0Gg_0^{-1}|$$ because you can define a bijection $$\phi : G_x \mapsto g_0Gg_0^{-1}$$ as $$\phi (c) = g_0cg_0^{-1}$$. Thus, $$|G_x|=|G_y|$$.

How is that any different from what Jason had done?

 

[Kummer]

How is that any different from what Jason had done?

I did not see what he posted. Anyway, mine is nice looking maybe he will like it more.

 

[JasonRox]

I did not see what he posted. Anyway, mine is nice looking maybe he will like it more.

Maybe. :P

Trying using the [itex ] command instead of [tex ]. It fits along the fonts a lot nicer.

For example: We have the polynomial $ax^3 + \frac{1}{2}x^2$ so find the derivative.

As opposted too...

We have the polynomial $$ax^3 + \frac{1}{2}x^2$$ so find the derivative.

I'll be posting more questions soon of course.

I have a whole pile of solutions to problems so I plan on making a website with all the solutions and hopefully people look over them and find mistakes and such.

 

[JasonRox]

Here is my next question...

It involves the proof of the theorem. I don't quite understand the step. There is a different proof of this, but I sure would like to understand this one too. There is a step I don't quite understand. Here it is...

Theorem 4.6 - Let G be a finite p-group. If H is a proper subgroup of G, then $H < N_{G}(H)$.

Proof: If H is normal in G, then $N_{G}(H) = G$ and the theorem is true. If X is the set of all the conjugates of H, then we may assume $|X| = [G : N_{G}(H)]$ is not equal to 1. Now H acts on X by conjugation and, since H is a p-group, every orbit of X has size of a power of p. As {H} is an orbit size of 1, there must be atleast p - 1 other orbits of size 1. And the proof continues...

My question is on the italic line. How does H have an orbit size of 1?

Oh crap, I think I got it now. Is it because the orbit of {H} is as follows...

Then the orbit of {H} is $|O(H)| = [H : H_{H}]$ and that is precisely 1?

Now since |X| is a power of p, it must certainly have atleast p-1 orbits of size 1.

So the proof continues as follows...

Thus there are atleast one conjugate gHg^(-1) =/= H with {gHg^(-1)} also of an orbit size of 1. Now $a g H g^{-1} a^{-1} = g H g^{-1}$ for all a e H, and so $g^{-1}ag is an element of N_{G}(H)$ for all a e H. But gHg^(-1) =/= H gives atleast one a e H with gag^(-1) not in H, and so H < N_G (H) (normalizer).

I think I got it now. Anyways, let me know what you think.

The prof. did the proof, but like it's barely explained. He just copies out of the textbook. Whenever I ask about details, he tells me not to worry about it because I'm just an undergrad. He doesn't seem to mind anymore since I go to his office and ask questions regardless.

 

[JasonRox]

Dang, I spent like an hour trying to fill in all the details.

 

[matt grime]

H acts on the set of G-conjugates of itself. Well, H is conjugate in G to itself, and H conjugates H into itself, hence the action of H acting by conjugation on the set of G-conjugates has at least one orbit of size 1, and hence at least p-1 of them.

I think I just wrote out what was already there, but then again, it was self explanatory.

 

[JasonRox]

H acts on the set of G-conjugates of itself. Well, H is conjugate in G to itself, and H conjugates H into itself, hence the action of H acting by conjugation on the set of G-conjugates has at least one orbit of size 1, and hence at least p-1 of them.

I think I just wrote out what was already there, but then again, it was self explanatory.

The textbook originally said G acts on X by conjugation. That was a typo for sure.

I don't really find all the details self-explanatory.

 

[JasonRox]

Ok, I need some help understand this passage or comment made in the textbook. The comments they make between the theorems.

Ok, here it goes.

Each term in the class equation of a finite group G is a divisor of |G|, so that multiplying by |G|^(-1) gives an equation of the form $1 = \sum_j \frac{1}{i_j}$ with each i_j a positive integer; moreover, |G| is the largest i_j occurring in this expression.

I understand that |G| is the largest possible i_j, but is there always such an i_j?

The only condition I can see that happening on is when G is non-abelian and the center of G is trivial.

Am I missing something here?

 

[morphism]

What you wrote down doesn't make sense. Did you forget a summation symbol somewhere? What are i and j, and what's S_j?

I'm going to assume you meant to write
$$1 = \sum_j S_j \frac{1}{i_j}$$

In which case what do you mean by "is there always such an i_j"? Why wouldn't there be one? If I understood all this correctly, then when|Z(G)|=1, we'll have that S_j=1 and i_j = |G|.

 

[JasonRox]

What you wrote down doesn't make sense. Did you forget a summation symbol somewhere? What are i and j, and what's S_j?

I'm going to assume you meant to write
$$1 = \sum_j S_j \frac{1}{i_j}$$

In which case what do you mean by "is there always such an i_j"? Why wouldn't there be one? If I understood all this correctly, then when|Z(G)|=1, we'll have that S_j=1 and i_j = |G|.

Ok, you came close! I totally forgot to write what S_j. I didn't know the LaTeX for it.

Here is the equation (which I'll edit in my last post):

$$1 = \sum_j \frac{1}{i_j}$$

So, my question is... will there always be an i_j = |G|? Or does the paragraph only mean that the largest i_j can be is |G| (which makes sense)?

I know |G| is the largest it can ever be. That's just stating the obvious. But I don't see how there will always be an i_j = |G|. In my opinion, if there is an i_j = |G|, then G is centerless.

 

[morphism]

Ok, you came close! I totally forgot to write what S_j. I didn't know the LaTeX for it.

Here is the equation (which I'll edit in my last post):

$$1 = \sum_j \frac{1}{i_j}$$

Ah, that makes more sense! I should have figure that S_j was meant to be summation.

I know |G| is the largest it can ever be. That's just stating the obvious. But I don't see how there will always be an i_j = |G|. In my opinion, if there is an i_j = |G|, then G is centerless.

That's absolutely right.

 

[matt grime]

Assuming you mean the standard class equation (the sum of the orders of the conjugacy classes) then you need to write down what your index set is otherwise the question is meaningless.

Properly, the class equation really should sum over all conjugacy classes, so of course there will always be an i_j equal to |G| - the one for the identity element.

Of course some people concatenate all the elements in the centre into a single term, and say it is

|Z(G)| + sum over remaining conjugacy classes.

 

[matt grime]

The textbook originally said G acts on X by conjugation. That was a typo for sure.

G does act on the set X by conjugation, since that is how it was defined. So does any subgroup of G, including H.

I don't really find all the details self-explanatory.

You don't find what part about your question self explanatory?

1. Let X be the set of G-conjugates of H, let [e],[g_1],..,[g_r] be a complete set of representatives of the conjugates (i.e. [g] is the conjugate gHg^-1, and inparticular [e] stands for the conjuate H of H).

2. H acts on X.

3. H sends the element [e] in X to [e], so there is at least one fixed point.

4. By the class equation for the action, this implies that there are at least p-1 fixed points.

 

[JasonRox]

I meant details to proofs in the textbook. It definitely doesn't read like a novel sometimes.