@KarlKronenfeld "In a category $\mathscr C$ a morphism $f:A\to B$ is called an equivalence if there is in $\mathscr C$ a morphism $g:B\to A$ such that $gf=1_A$ and $fg=1_B$"
@PedroTamaroff yeah, I have never seen that called an equivalence. Equivalence is often reserved for relating two categories (in a way that is not necessarily an isomorphism), in my experience.
Also, I got one ugly question. When you prepare for a test and there's a question you know the answer to, yet it's ugly and long, do you usually go through it either-way or skip?
Here's a nice one. Let $\lambda_1$, $\lambda_2$ and $\lambda_3$ be Cardinal numbers, so that $\lambda_1$<$\lambda_2$. Prove that $\lambda_1*\lambda_3$<$\lambda_2^\lambda3$.
It's extremely intutive but getting the proof around is a bit harder
I have the recurrence relation:
$a_{n+2} = \dfrac{-a_{n}}{(n+2)(n+1)}$.
I have done some work to identify that two cases emerge: one is n is positive, and one if n is negative.
If n = 2m (even)
$a_{n} = a_{2m} = \dfrac{(-1)^m a_0}{(2m)!}$.
If n = 2m+1 (odd)
$a_{n} = a_{2m + 1} = \dfrac{(-1)^m ...
okay. so, you would assume the even case, then show k+1 and then you would separately show the odd case, then induct k+1 for the odds? So wouldn't that mean you are going from an odd to an even? Or would you just do for example n = 2k and n = 2k+2?
Oh..so for example, if I have for the even case, n = 2m, then I induct on m, so I assume a_n for n = 2m and use the inductive hypothesis to induct a_n for n = 2(m+1) = 2m + 2. Then if I have an odd case, n= 2m+1, then I assume n = 2m+1, and induct n = 2(m+1) + 1 = 2m +3 .. right?
So a product P for a family of objects A_i is defined as another object in the category together with a set of maps p_i (projections?) with the following property.
Now, assuming <P,p_i> and <Q,q_i> are two products, we know there is a map f : Q --> P such that p_i f = q_i and a map g: P --> Q for which q_i g= p_i.
This means q_i (gf)= q_i and p_i(fg)=p_i for each i.
@PedroTamaroff by universal property there is a unique map u:P->P (or v:Q->Q) such that p_i (u) = p_i for all i. observe u=id_P and v=id_Q work. use this to show that fg is id_P and gf is id_Q
I explained why the equations q_i(gf)=q_i and p_i(fg)=p_i mean gf and fg are the identities. which subsequently means indeed that x(gf)=x and y(fg)=y for an x in End(Q) or fg in End(P).
@PedroTamaroff you can't say "they only work for the p_i,q_i's" (note you said something contradictory above: "shouldn't the above be true for any map?"), all you can do is ask "how does just knowing these are true for the p_i,q_i's entail that they hold true for all x,y's, or equivalently that fg is id_P and gf is id_Q?" which is what I answered
What universal property are you referring to here "by universal property there is a unique map u:P->P (or v:Q->Q) such that p_i (u) = p_i for all i. observe u=id_P and v=id_Q work. use this to show that fg is id_P and gf is id_Q"?
what are the big three things covered in a first abstract algebra course? you got the first two right, then suddenly veered off into modules at precisely the wrong moment to veer...
I was the one who said it was weird, not Mike. I also said first course, and usually first courses focus more on abelian/commutative than nonabelian/noncommutative (for obvious reasons).
for elevated perspective I would form a graph whose vertices are configurations and whose edges are moves from one configuration to another. I think it's a rooted tree or somesuch.
Would it be unfair to say that if an infinite group had a unique non-identity automorphism, then that group must be abelian and the automorphism must be the sending of every element to its inverse?
If you conjugate by 2 different elements, and they both map one thing to the same place, we must have that every element is invariant under conjugation.
@MWarsi by "if you conjugate by 2 different elements and they both map one thing to the same place" do you mean "any two conjugation maps $x\mapsto axa^{-1}$ and $x\mapsto bxb^{-1}$ are the same automorphism"?
Essentially, what I'm trying to do is find an infinite group, other than the integers, that has a unique, non-identity automorphism. I know, in a sense, now, that it must be abelian, and that the automorphism in question is inversion, and also that there exists an element in it not of order 2.
I know that if we have an element of infinite order, than one can embed Z in there.
So, I'm actually a little curious as to whether this is possible with an infinite group where every element has finite order.
If you conjugate by 2 different elements, and they both map one thing to the same place, we must have that every element is invariant under conjugation.
@MWarsi by "if you conjugate by 2 different elements and they both map one thing to the same place" do you mean "any two conjugation maps $x\mapsto axa^{-1}$ and $x\mapsto bxb^{-1}$ are the same automorphism"?
@MWarsi so you're saying "if any two conjugations are identical then everything is conjugation-invariant" ... but how do you first prove that any two conjugations must be the same automorphism?
I thought your assumption was that there was only one nontrivial auto? Now it's your task to go from that assumption to "any two conjugations define the same auto"
@PedroTamaroff I see they're trying to brainwash you into believing fields are among the big three. Anyway, just take the category with two objects and no non-identity morphisms.
Assume, that they aren't the identity, ie, that the group is non-abelian. Then, they must be that unique automorphism, from which it follows that it is abelian.
Conradiction.
Thus, they must be the identity and the group must be abelian.
for example, you say "assume that [there are conjugation maps which] aren't the identity, i.e. that the group is nonabelian. then, the nontrivial conjugations must be the nontrivial automorphism." how does it follow from that that the group must be abelian?
I have the recurrence relation:
$a_{n+2} = \dfrac{-a_{n}}{(n+2)(n+1)}$.
I have done some work to identify that two cases emerge: one is n is positive, and one if n is negative.
If n = 2m (even)
$a_{n} = a_{2m} = \dfrac{(-1)^m a_0}{(2m)!}$.
If n = 2m+1 (odd)
$a_{n} = a_{2m + 1} = \dfrac{(-1)^m ...
