Recent content by Swlabr1

The point of Linear Algebra Done Right is to get you to think about Linear Algebra in a very different way from "usual" Linear Algebra texts. So, don't just do the questions - you want to do them they way they the author would do them! (That is, if you ever use a determinant in an argument, then...

This is the thread I was thinking of. I think it would be important to save the un-rendered files also, incase of an incorrectly-rendered symbol and you're not sure what it is meant to be...and that is all my ideas...good luck!

Using his notation allows you to separate the set from the group. I mean, a binary operation *is* a function taking an ordered pair to a single element. Thinking about it this way allows you to define other structures on a well-known set with perhaps less confusion (and encourages students to...

It depends if it works or not... I would like to say that I think it would be wonderful if you got the touchscreen version working so that I could take notes on a tablet. That would be very convenient! (Although, perhaps a thing exists? I have maybe heard tale of it on MathOverflow.)

I believe the level of stress for actuaries depend on their specific field. I have a friend who is an actuary in a bank, and he thinks it is a cushy job. However, I have met actuaries who work for consultancies who have to really work for their paycheck! To echo pickslides answer, I know in...

Well, what is a Sylow $p$-subgroup of a given group? What do you know about it? Well, it has order $p^n$ where $p^n$ divides the order of the group. So, what is the prime decomposition of $6!$? This will give you the list of possible primes $p$. You now need to find an $n$ for each prime $p$...

I presume the definition of $p$-group the OP has is "$G$ is a $p$-group if the order of $G$ is $p^n$ for some prime $p$". Then the result follows from Lagrange's theorem, which says that if $H$ is a subgroup of $G$ then $|H|$ divides $|G|=p^n$. If $g\in G$ then the set $\{g, g^2, \ldots...

I should learn to read questions fully...

Basically, yeah. The coset is "what is happening", but for all practical purposes your ring is the set of all polynomials of degree $1$ where multplication is modulo $x^2+1$. I mean, when you are working in the integers mod $n$ you are really working with cosets of numbers, but you don't ever...

What you suggest doing will find the coset representative of your given polynomial. Remember, you are no longer dealing with polynomials, but with cosets! So, the elements of your ring are the cosets $[r(x)]=\{f(x): (x^2+2)q(r)+r(x)\}$. (As a LaTeX side-point, instead of writing deg, write...

A statement like $2x+2y+2z=0$ in $\langle\mathbb{Z}, +\rangle$ is equivalent to $x^2y^2z^2=1$ when we write the group multiplicatively. Equations like this have been much studied, and quite frankly the solutions to these problems, in the 'easiest' case (of free groups), go way over my head! (The...

Talking about Serge Lang, have you ever seen this? Ken Ribet is the guy who proved the connection between Fermat's Last Theorem and elliptical curves. Also, Huppert wrote an influential book called "finite groups", which spawned two further volumes with him and Blackburn. I haven't managed to...

HINT: Start by proving that if $G/Z(G)$ is cyclic then $G=Z(G)$. This means that if $|G|=8$ then $|Z(G)|=1$, $2$ or $8$. Why can't it be $1$?

I have heard tale of permanent jobs (in arbitrary universities) in France without teaching. However, they do not pay very well...

Re: Ring and cosest Essentially, yes. For example, $(x+I)\cdot (x+1+I)=x^2+x+I=x-1+I=x+3+I$, as you know that $x^2=-1\text{ mod }I$ because $x^2+1\in I$