Abstract Algebra Abstract Algebra Math Forum

January 2nd, 2012, 03:03 PM   #1
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Hello all-- I've been writing up some solutions for problems in algebra and analysis I've been doing.. If anyone has some extra time to read through one, or more, of my solutions and give me feedback whether or not I'm missing anything, I'd be GREATLY appreciative!

I will attach the file for some of the algebra solutions I have written up so far.. I'll be adding to it and editing, but I appreciate any comments as is (whether typos, more serious mistakes, or additions). Thanks!

[attachment=0:380wzkmi]Algebra Problems and Solutions.pdf[/attachment:380wzkmi]
Attached Files
 Algebra Problems and Solutions.pdf (215.9 KB, 17 views)

 January 5th, 2012, 08:50 AM #2 Senior Member   Joined: Sep 2008 Posts: 150 Thanks: 5 Re: Proof-read / give comments on algebra solutions? Ok, I checked the first few; here are my comments: 1.1: Claim 1: Looks good enough, but in your proof you should replace |G| by |N| twice. (Seems to be a simple typo.) Claim 2: Good. Claim 3: That is a correct description of the orbits, but it is not very far from the definition of "orbit". In fact, one can describe the orbits very explicitly. 1.2: Claim 1: Good. Claim 2: Good. 1.3: Claim 1:Also note that $ker \phi \neq G$ as $ker \phi \subset P$. That is not apparent! It is in fact wrong for every non-simple group G of that order. However, the fact that $ker \phi \neq G$ can be deduced quite simply. That's how far I got for now. Maybe I have more time later. rgds Peter
 January 5th, 2012, 09:19 AM #3 Senior Member   Joined: Nov 2011 Posts: 100 Thanks: 0 Re: Proof-read / give comments on algebra solutions? Thank you for your comments so far! 1.1 Claim 3: I realize your point, and I was worried that was the case; can you give me a hint how I might actually describe the orbits explicitly? 1.3 Claim 1: Yes, I realize that I can't just note $\ker\subseteq P\not=G$, I was just being a bit lazy I guess! To see why it's true, one can observe the following: Recall $X=\{gPg^{-1}:g\in G\}$ and that $\phi: G\to \Sigma_X$, where $\phi(x)=\sigma_g$, defined as $\sigma_g(xPx^{-1})=gxPx^{-1}g^{-1}=(gx)P(gx)^{-1}$. Let $g\in\ker\phi$. Then $\phi(g)=\sigma_g(xPx^{-1})=xPx^{-1}$ for every such conjugate $xPx^{-1}$ of $P$. But in particular, we have that $\phi(g)=\sigma_g(P)=P$, i.e., $gPg^{-1}=P$ for all $g\in \ker\phi$, and so (since we are assuming $|X|>1$), we get $g\in P$.. I think.. Or maybe this is a better approach: Define$X=\{gP:g\in G\}$ and $\phi: G\to \Sigma_X$, where $\phi(x)=\sigma_g$, defined as $\sigma_g(xP)=gxP=(gx)P$. Let $g\in\ker\phi$. Then $\phi(g)=\sigma_g(xP)=(gx)P$ for every such left coset $xP$ of $P$. But in particular, we have that $\phi(g)=\sigma_g(P)=P$, i.e., $gP=P$ for all $g\in \ker\phi$, and so we must have $g\in P$. Thanks again!
January 5th, 2012, 10:23 AM   #4
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Quote:
 Originally Posted by watson Thank you for your comments so far! 1.1 Claim 3: I realize your point, and I was worried that was the case; can you give me a hint how I might actually describe the orbits explicitly?
Hint 1: There aren't to many.
Hint 2: Take a vector of your liking and try a bit where you can map it.
Hint 3: As matrices in SL_n are invertible the column-vectors form a basis. Given any basis, how can you change it, so that it makes up the column-vectors of an element of Sl_n?
Quote:
 1.3 Claim 1: Yes, I realize that I can't just note $\ker\subseteq P\not=G$, I was just being a bit lazy I guess! To see why it's true, one can observe the following: Recall $X=\{gPg^{-1}:g\in G\}$ and that $\phi: G\to \Sigma_X$, where $\phi(x)=\sigma_g$, defined as $\sigma_g(xPx^{-1})=gxPx^{-1}g^{-1}=(gx)P(gx)^{-1}$. Let $g\in\ker\phi$. Then $\phi(g)=\sigma_g(xPx^{-1})=xPx^{-1}$ for every such conjugate $xPx^{-1}$ of $P$. But in particular, we have that $\phi(g)=\sigma_g(P)=P$, i.e., $gPg^{-1}=P$ for all $g\in \ker\phi$, and so (since we are assuming $|X|>1$), we get $g\in P$.. I think.. Or maybe this is a better approach: Define$X=\{gP:g\in G\}$ and $\phi: G\to \Sigma_X$, where $\phi(x)=\sigma_g$, defined as $\sigma_g(xP)=gxP=(gx)P$. Let $g\in\ker\phi$. Then $\phi(g)=\sigma_g(xP)=(gx)P$ for every such left coset $xP$ of $P$. But in particular, we have that $\phi(g)=\sigma_g(P)=P$, i.e., $gP=P$ for all $g\in \ker\phi$, and so we must have $g\in P$. Thanks again!
I actually prefer your first approach. Maybe if you drop the calculations, it looks nicer. Let me rephrase it: G operates transitive on X. By definition the kernel operates trivial on X. So if the Kernel was G, then G operates trivial and transitive, i.e. X has only one element, contrary to the assumption. (Well, now that I write it, it might be that for it to look really easy, one has to be a bit used to operations.)

January 5th, 2012, 10:33 AM   #5
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Quote:
 Originally Posted by Peter G operates transitive on X. By definition the kernel operates trivial on X. So if the Kernel was G, then G operates trivial and transitive, i.e. X has only one element, contrary to the assumption.
This is nice, thanks.

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