Sylow Theorem Number One

**ThePerfectHacker** · October 22nd 2006, 05:36 PM

For anyone who has John Fraleigh "A First Course in Abstract Algebra" 7/e would be easier to follow. (I definitely know some of you have it, because many people referred to it already, very popular book).

Let $p$ be a prime.

On Page, 324-325
John is talking about the first Sylow theorem.
"Given a finite group $|G|=p^n m$ with $n\geq 1$ and $p\not | m$ , then there exists a subgroup of order $p^{i}$ for $0\leq i\leq n$ .

Furthermore, each subgroup of order $p^i$ is a normal subgroup of order $p^{i+1}$ . For $0\leq i<n$ .
----
The first part of the proof I understand, the second part ails my soul and makes my blood cold and so by degrees.
I believe that John wanted to write,
$p^i$ is a normal subgroup of some group of order $p^{i+1}$ . He makes it appear as though it is true for all subgroups, which is unlikely.

Anyone know what I am talking about?

**topsquark** · October 22nd 2006, 06:04 PM

Originally Posted by ThePerfectHacker

For anyone who has John Fraleigh "A First Course in Abstract Algebra" 7/e would be easier to follow. (I definitely know some of you have it, because many people referred to it already, very popular book).

Let $p$ be a prime.

On Page, 324-325
John is talking about the first Sylow theorem.
"Given a finite group $|G|=p^n m$ with $n\geq 1$ and $p\not | m$ , then there exists a subgroup of order $p^{i}$ for $0\leq i\leq n$ .

Furthermore, each subgroup of order $p^i$ is a normal subgroup of order $p^{i+1}$ . For $0\leq i<n$ .
----
The first part of the proof I understand, the second part ails my soul and makes my blood cold and so by degrees.
I believe that John wanted to write,
$p^i$ is a normal subgroup of some group of order $p^{i+1}$ . He makes it appear as though it is true for all subgroups, which is unlikely.

Anyone know what I am talking about?

If I understand which part you are asking about you are correct. My book: "Algebra" by Hungerford writes the 1st Sylow theorem as:
Let G be a group of order $p^nm$ with $n \geq 1$ , p prime, and (p,m) = 1. Then G contains a subgroup of order $p^i$ for each $1 \leq i \leq n$ and every subgroup of G of order $p^i$ (i < n) is normal in some subgroup of order $p^{i+1}$ .

I can provide you with this book's proof, but you are on your own for about half of it. I'm too rusty to follow the whole thing anymore.

-Dan

**ThePerfectHacker** · October 22nd 2006, 06:35 PM

Originally Posted by topsquark

If I understand which part you are asking about you are correct. My book: "Algebra" by Hungerford writes the 1st Sylow theorem as:
Let G be a group of order $p^nm$ with $n \geq 1$ , p prime, and (p,m) = 1. Then G contains a subgroup of order $p^i$ for each $1 \leq i \leq n$ and every subgroup of G of order $p^i$ (i < n) is normal in some subgroup of order $p^{i+1}$ .

Thank you, that is what I needed.
I think I seen people refer to Hungerford. Is it good?

Originally Posted by topsquark

I can provide you with this book's proof, but you are on your own for about half of it.

That is okay, I perfectly understand the proof in my book. Tell me, does it use Cauchy's theorem and the fact that:
$|X_G|\equiv |X| (\mbox{mod }p)$ ?
The reason why I am asking is because John says that is a new way and more elegant way to appraoch Sylow theorems.

Originally Posted by topsquark

...to follow the proof anymore

Did you consider re-reading it? It should go fast and smooth a second time if you knew if well the first time.

One last question. Did you read study Group theory before or after you got a Ph.D in Physics?

**topsquark** · October 23rd 2006, 04:50 AM

Originally Posted by ThePerfectHacker

Thank you, that is what I needed.
I think I seen people refer to Hungerford. Is it good?

It seems fairly clear and well organized. My only trouble with it is that I've had to teach myself all the background and thus I have trouble with some of the details.

Originally Posted by ThePerfectHacker

That is okay, I perfectly understand the proof in my book. Tell me, does it use Cauchy's theorem and the fact that:
$|X_G|\equiv |X| (\mbox{mod }p)$ ?
The reason why I am asking is because John says that is a new way and more elegant way to appraoch Sylow theorems.

It uses Cauchy's theorem, but I don't think it uses the equivalence. (Presumably to save space it refers to a number of lemmas and corrolaries, which I haven't looked up, so it may be in there.)

Originally Posted by ThePerfectHacker

Did you consider re-reading it? It should go fast and smooth a second time if you knew if well the first time.

Actually I have approached it twice now. Each time I get a bit further into it. There always seems to be a point where it just seems that I'm reading gobbledegook so I put it down for a couple of months and come back to it. The last time I got up to chapter 4, which is on modules, so I got all the way through the Sylow theorems (which are in chapter 2). The problem is that I've been working with QFT for the last 6 months or so and I've apparently forgotten some of the notation.

Originally Posted by ThePerfectHacker

One last question. Did you read study Group theory before or after you got a Ph.D in Physics?

I would say that is a "yes" because I don't have a PhD yet... Group theory (The Physics version: it was rushed and didn't cover details) was a requirement at Purdue for the Graduate program, meaning the Master's level students also took it. But it wasn't offered at Binghamton for their Master's program. I have a Mathematics "level" text on group theory and we really didn't go that deep in the Physics course. It seems that the goal of the Physics course was more or less character tables and how to use them to determine the irreducible representations of a group. (Which is the big deal for Quantum Mechanics, but I don't know where else in Physics you might use them. Presumably Classical Mechanics as well, but I haven't done and in-depth study of that field.)

-Dan

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