@TedShifrin : That's the case which I understand how to handle (in terms of finding no. of homomorphisms from G to G') but my next sentence asks a different case
my cat curled up on my daughter's bed on top of the sheet and blanket, so when we put our daughter to bed we didn't cover her with anything. it would have disturbed the cat
Let me build up the background just in case my question does not make sense.
I just want to say that the above example problem can be handled by me: For 1): I'll use first isomorphism theorem to conclude 0 and for 2) I'll see where $1\in Z_{10}$ will get mapped to in $Z_8$ and since mapping is onto I would wish that image to be generator of $Z_8$ so we have 4 possibilities (1,3,5,7) and hence 4 no. of such homomorphisms.
This was well-handled. Now, let's go to a different example: Let G be a group of order 60 and its homomorphic image is given to be cyclic and is of order 12. Then we are interested in finding order of all normal subgroups that G has. My idea is to somehow utilize the above idea (that was used in example)
I don't understand how to do so. I only know from first isomorphism theorem that G indeed will have a normal subgroup of order 5 (that is kernel of the homomorphism)
But apart from that, I don't know how to get the order of other normal subgroups of G.
Ted, I asked this question yesterday and I was suggested a solution also but I feel that to be complicated for a group theory beginner such as myself.
if you have a homomorphism from G to C_n and a homomorphism from C_n to C_m you can compose them to get a homomorphism from G to C_m which might have a larger kernel. that's one thought.
$\forall$ has $\exists$ as a "left adjoint" in Topos Theory.
Let $M \xrightarrow{f} M' \xrightarrow{g} M''$ be sequence of $A$-modules and $A$-module homomorphisms. Now suppose we don't have $gf = 0$ or in other words no ability to compute traditional Homology at $M'$, but instead we have sort...
I was actually trying to figure out how f(x), -f(x) and f(-x) are related generally. more specifically i wanted to know if there were graphically some way to plot -f(x) if i know f(x)
@hyper-neutrino like here, if you look
the cos and -cos are mirrored to each other along x axis
but if you look at sin and -sin x i thought they too were mirrored to each other along x axis... are you telling they are mirrored along y axis?
Without using polylogarithmic functions, how can I evaluate $\sum_{k=1}^\infty \frac{\sin(k\pi/2)}{k^3}$? Via, polylogs, I can show that this is equal to $\Im(\operatorname{Li_3}(i))=\beta(3)=\frac{\pi^3}{32}$. But can this be evaluated without Polylog?
I thought that $1/3k^3= \int_0^k \frac{1}{t^4}$. But then I cannot interchange the sum and integral since the upper limit of the integral is dependent on the lower limit of the sum
huh, it's mildly interesting that this series returns anything nice at all
Hey @Astyx, perhaps you know this: if I have a Dedekind domain $A$ and $0\neq f\in A$ and localize at $\{{1,f,f^2,\dots\}$, $A_f$ is Dedekind and I have an induced map $Cl(A)\rightarrow Cl(A_f)$ between the class groups. I have calculated that the kernel is generated by the residue classes of the prime factors of $(f)$. Do you know if this is correct? I'm just trying to confirm, but haven't been able to find this in the literature (it surely is there, but I don't know where).
For any normal Noetherian separated scheme X, there is a short exact sequence 0 -> Z -> Cl(X) -> Cl(X \ Y) for any irreducible codimension 1 closed subscheme Y of X
This is a well-known fact, see Hartshorne. Yours is a special case
$4 > p^2$ implies $2 > p \lor -2 < p$, which makes sense intuitively, but what would be the rigorous justification for getting $-2 < p$ from $\pm 2 > p$?
@AdilMohammed In general, there is no relation. Some functions have $f(x)=f(-x)$ and we call them even. Some functions have $f(-x)=-f(x)$ and we call them odd. Others, like $y=e^x$ have neither.
Yeah I know, I didn't find anything about them in the usual literature. I was told to look them up when I wanted to show a relationship between higher chern classes and the curvature forms on vector bundles
@SayanChattopadhyay This is the wrong name. If you want to understand the relationship between Chern classes and curvature forms, this is called Chern-Weil theory. It is explained in many references but I am fond of Morita's "Geometry of differential forms".
Secondary characteristic classes (like secondary cohomology operations) are defined in a setting when the characteristic classes you already know vanish. The standard example is the Chern-Simons invariant of a flat connection. The simplest case is really quite simple, so I'll explain it. Suppose one has a *flat complex line bundle*, so…
Secondary characteristic classes are more complicated and not seen as often. You probably do not actually want to look into those
you can think of it as taking a square root. if a and b are nonnegative numbers and a < b then sqrt(a) < sqrt(b). that's true and you can think of that as 'square rooting an inequality.'
but yeah, you run immediately into what sqrt(p^2) is and that's where the thinking cap has to be on.
Well, be careful. You need the conjunction ("and"). So $2-p>0$ (i.e., $p<2$) AND $2+p>0$ (i.e., $p>-2$) .... or $2-p<0$ and $2+p<0$. The latter says $p>2$ and $p<-2$, which of course cannot occur.
@copper.hat Well, even the unicode fonts here render those same symbols ugly.
Maybe I'm remembering wrongly (I distinctly recall that those symbols were available in some ASCII version), but anyway even the current unicode font here makes those two dual operations look terrible.
i'm very interested to know the cause of that. i don't know how florida works. the stereotype is shady developers putting stuff up overnight and lax building codes and minimal licensing requirements for contractors.
my most recent experience with building codes was a biotech startup. they were prototyping instruments and i asked, uh, are you plugging these into the wall? and they were, which was in violation of about a billion local ordinances. they thought it was OK because the product wasn't on the market.
strangely, a lot of regulation of what you can and can't plug in is decided on a city-by-city or county-by-county level. most refer to UL standards and such so it's almost the same thing as you go from place to place, but sometimes not.
i told them to put it on a battery at least when the inspector showed up
@leslietownes: my wife is an attorney and the firm she is working for specializes in construction defect. She was very interested in the paper this morning.
it's not my area but i'd love to know what went wrong here. something went horribly wrong. if it is regulatory failure there will be no recovery against the state but i would not want to have built that building right now.
my high school senior class loaded a waterslide improperly at a water park, causing it to collapse and kill somebody. there was a game where you'd try to see how many people you could put on the waterslide at once.
I don’t understand this: “if G is a finite group and $\phi$ is a homomorphism from G onto $Z_{10}$ . Then, $\phi^{-1}\langle 2\rangle$ is normal and of index 5.” Why? There’s absolutely no reason to believe that. All we should be able to say is $\phi^{-1}\langle 2\rangle$ is normal in G. Period.
Please let me know what I am
missing.
Is it that under homomorphisms indices of pullback of subgroups are preserved?
here's the general fact: if $f\colon G\rightarrow H$ is a surjective group homomorphism and $N\triangleleft H$ is normal, then $f^{-1}(N)\triangleleft G$ is normal and $(G\colon f^{-1}(N))=(H\colon N)$
given $f: G\to H$ surjective and $N \unlhd H$ normal, then $M := f^{-1}(N) \unlhd G$ (prove). now take $\pi\colon H \to H/N$ the canonical projection. prove that the map $\phi: G/M \to H/N$ given by $\phi(gM) = \pi(f(g))$ is well defined and a homomorphism. find its kernel (hint: trivial kernel!) and use the surjectivity of $f$ to conclude that $\phi$ is an isomorphism
@Koro he's messing around.
$\phi$ is an isomorphism $\implies |G/M| = |H:N| \implies (G:f^{-1}(N)) = (H:N)$
on this topic, here's a good (and, surprisingly, useful, yet not often mentioned) exercise, which I recommend: for a subgroup $H$ of a group $G$, compute the $G$-automorphisms of the $G$-set $G/H$
@ShaVuklia Look at the twisted cubic curve in $\Bbb{P}^3$, these are given be the vanishing of three $F_1,F_2,F_3$ hypersurfaces. Such things are called incomplete intersections
I think local complete intersections is the most general one can get if you deal with algebraic sets $X\subset\mathbb P^n$ (smooth of dimension 1), but I can't prove it
Hmm okay so if I am thinking this carefully, locally complete intersections are those subvarieties which at each local ring are generated by the right number of guys. The question we want to ask is if every incomplete intersection a locally complete intersection
The following is an Exercise 1.1.11 of Hartshorne's Algebraic Geometry.
Let $Y\subset \mathbb{A}^3$ be the curve given parametrically by $x=t^3, y=t^4, z=t^5$. Show that $I(Y)$ is a prime ideal of height $2$ in $k[x,y,z]$ which can not be generated by $2$ elements. We say $Y$ is "not a local com...
If, to each element of G, there is at least one element of F, and to each of F, also there is at least one G, is it enough to consider that both groups are homomorphic?
$\forall$ has $\exists$ as a "left adjoint" in Topos Theory.
Let $M \xrightarrow{f} M' \xrightarrow{g} M''$ be sequence of $A$-modules and $A$-module homomorphisms. Now suppose we don't have $gf = 0$ or in other words no ability to compute traditional Homology at $M'$, but instead we have sort...
