For example, armed with JUST THIS INFORMATION, prove there is only ONE homomorphism $\Bbb Z_3 \to \Bbb Z_5$.
Step one: identify the normal subgroups of $\Bbb Z_3$.
Well, this is easy, there's just two: $\{0\}$ and $\Bbb Z_3$ (because $\Bbb Z_3$ is abelian, ALL subgroups are normal, and those are the ONLY TWO subgroups)
(Side-note-someone posted this to an answer I gave: If I were a dictator, I would demand that a statue of your likeness be built upon all public grounds in my fine nation. Bless your soul.) Moments like this make it all worthwhile.
It's called the "free group generated by $X$", where $X$ is the set.
So there are MORE groups than sets, and there's a LOT of sets.
The typical way you create this group, is you regard the elements of $X$ as "letters" and you make "words" out of them (multiplication is concatenation).
And you introduce a "rule of reduction": replace any occurence of $aa'$ with the empty word, any occurence of $a'a$ with the empty word, any occurence of $bb'$ with the empty word, etc.
The elements of the free group are "reduced words" in the alphabet $X \cup X'$
I wouldn't say that it turns a set into a group it more like the set becomes a generating set (in a very important way). Although assuming the axiom of choice for every set there is a binary operation that makes that set a group.
Vertical lines in the plane are not functions. Their slope, as defined by $\dfrac{y_2-y_1}{x_2-x_1}$ is undefined. But geomterically, sure, they go straight up.
@DavidWheeler how can I control... like you know with cup and big-cup with big cup the thing will appear above and below when you use ^ and _ but with just cup it'll be sub and super script
How can I control that with sums, limits, so forth?
Hi everyone! Just a quick question about formalisation. Let $A=aI_n$, where $I_n$ denotes the identity matrix of order $n$. Then I need to use a function $f(A)$, but $f$ is given in terms of $a$, not $A$. Is it correct to use the form $f(A)$? For instance, say $f$ is equal to the quantity $2a+1$, is it correct (rigor, or what else) to write $f(A)=2a+1$? Thank you very much!
@PedroTamaroff, what do you think about my question above? And apologies for interrupting you...
@nullgeppetto Define your terms! What is $a$? what is $A$?
On the one hand, you are claiming that $a \in \Bbb R^{n \times n}$, since $f(a)$ is defined, on the other hand, by claiming that $f(a) = 2a + 1$, you are claiming that $a = \dfrac{f(a) - 1}{2} \in \Bbb R$
@PedroTamaroff, @DavidWheeler this is what I am asking actually. So I need to write $f(a)$ instead? I mean, $f$ is given matrices (but a priori multiples of the idednity matrix) and gives numbers in terms of $a$. Is it wrong to write $f(A)$? Or I necessarily need to write $f(a)$ instead?
So, let's see, that gives us: $\dfrac{\partial^2f}{\partial x^2},\dfrac{\partial^2f}{\partial x\partial y}, \dfrac{\partial^2f}{\partial x\partial z}$ (three so far....
The Hessian is the matrix consisting of all SECOND-ORDER partial derivatives. SECOND ORDER, meaning we differentiate twice. You are somehow insisting it is merely all the first-order partial derivatives. History of mathematics disagrees.
@DavidWheeler I have this theorem in my book which is the key to understanding the solution to classification problem I saw the proof but it doesn't provide any insight very bad proof. If T is any subgroup of G/N, then T = H/N, where H is a subgroup of G that contains N.
the way they proved it is they let H = {A $\in$ G | Na $\in$ T}
Easier proof: let $a,b \in H$, so that $Na,Nb \in T$. Since $T$ is a group, $Na(Nb)^{-1} \in T$. But $Na(Nb)^{-1} = Nab^{-1}$, so $ab^{-1} \in H$. Done.
Writing the cosets as $k + \text{ker }\pi = k + n\Bbb Z$ is just adding "notational baggage", we can just write $[k]_n$ to convey the same information.
Note that the "$k$" in $[k]_n$ is not UNIQUE, for example, $[k+n]_n = [k]_n$, even though, as INTEGERS, $k \neq k+n$.
I can see now for example going back to our proof the reason we get that it must of that form is that we are trying to make a coarser partition of gN so to do that we must have a group H such that $N \subset H$ correct @DavidWheeler
This is usually written as $D_4 = \{e,r,r^2,r^3,s,sr,sr^2,sr^3\}$ where $r$ is a rotation of 90 degrees (clockwise, let's say), and $s$ is reflection about the $x$-axis.
I am confused due to yesterday, is a mobius transformation ever $\Bbb C \to \Bbb C$ or always $\hat {\Bbb C} \to \hat {\Bbb C}$
The wiki page makes me thing it is always the extended plane, but Mike made it seem like it can be non-extended
Unfortunately I don't have time before my assignment is due to read the 100 pages or so that would get me ready for the chapter we are being tested on[via assignment] (...)
Is a mobius transformation ever defined on $f:\Bbb C \to \Bbb C$ or is it always $f:\hat {\Bbb C} \to \hat {\Bbb C}$? The wikipedia page makes me believe it is the latter, but my assignment has the first written(perhaps lazy notation) and someone has led me to believe these will yield different r...
