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## My Algebra Questions

That's not gonna work out very well. Instead, remember (or prove) that a subgroup has the same cardinality as any of its conjugates (hint: write down the obvious bijection).
 Recognitions: Gold Member Homework Help 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.
 Recognitions: Gold Member Homework Help 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|.
 Recognitions: Gold Member Homework Help Although I made the dumbest errors, I learned a decent amount. Therefore, I like that problem.

 Quote by JasonRox 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 eachother. 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, $$|G_x|=|G_y|$$.

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 Quote by Kummer 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 eachother. 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?

 Quote by morphism 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.

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 Quote by Kummer 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.
 Recognitions: Gold Member Homework Help 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.
 Recognitions: Gold Member Homework Help Dang, I spent like an hour trying to fill in all the details.
 Recognitions: Homework Help Science Advisor 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.

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 Quote by 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.
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.
 Recognitions: Gold Member Homework Help 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 occuring 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?
 Recognitions: Homework Help Science Advisor 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|.

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 Quote by 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|.
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.

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 Quote by JasonRox 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): $$1 = \sum_j \frac{1}{i_j}$$