Math Is Fun Forum

The definition of an isomorphism is that there exists a bijective homomorphism, yes.  However, this does not mean one has to find such a map.  If you can prove the map exists (without finding it), then that is enough.

Let me give you a simpler example.  Prove to me that 189518123599594883469824968269829864264098262498494898487654654186546542 is even.  Well, an even number by definition is a number n where n = k + k for some integer k.  But you can conclude the above number is even without ever finding what k is by just looking at the 1's digit: it's 2, and therefore the above number is indeed even.  It would be a complete disaster to try to find k without an infinite-precision calculator to help you.

So if we can show the existence of a map without actually finding a map, then we're done.  It turns out that a lot of times, showing the existence of a map is actually easier than finding the map itself.

You may have heard of them as "the integers modulo 4", also written as:

When I say Z/2Z x Z/2Z, I mean:

Where addition is component-wise modulo 2 (i.e (0, 1) + (1, 0) = (1, 1) and (1, 1) + (1, 0) = (0, 1)).

Now as for the claim I made, perhaps that is too advanced for you.  Let me offer an alternate solution to your problem:

You should know that isomorphisms preserves the order of elements: if f is an isomorphism from G to H, and f(g) = h, then the order of g is equal to the order of h.

So if H were isomorphic to V, there would exist an isomorphism f from V to H.  Now f((12)(34)) is an element of H, and by the above it must have order 2.  So must all the other nonidentity elements of V.

What are the elements of order 2 in S_4?  As I said before, they are all elements that you can represent by disjoint transpositions:

(12), (13), (14), (23), (24), (34), (12)(34), (13)(24), (14)(23)

Well we want H to not be equal to V, so it's best to start by choosing something such as (12).  From here, you have severely limited choices for what the other elements are mapped to.  A little playing around and you should come across a solution in no time (considering that multiple solutions exist).

For the first, you want to be looking at generators of the group.  If you define where generators are sent, you define the entire isomorphism.

For the second, write x = g^k * z_1 and y = g^l * z_2 where z_1 and z_2 are in Z(G).  Now compute xy and yx.

I'm going to assume the question is stating that V is a group.  A quick check on the Cayley Table will verify this if not.  There is also a much nicer construction for what I'm about to say, but it requires use of Van Dyck's theorem, which I don't believe you would have seen yet.

V is a group of order 4, and so it is either Z/4Z or Z/2Z x Z/2Z.  Taking a glance at the structure of the elements will indeed tell you that it is Z/2Z x Z/2Z.  With the isomorphism type found, what we are really doing is finding another copy of the Klein-4 group in S4.

The Klein-4 group is described by 4 elements where all nonidentity elements have order 2.  But looking at elements in S4, an element has order two if and only if it can be represented by disjoint 2-cycles (called transpositions).  So let's look at all the elements that have 2 disjoint transpositions:

(12)(34), (13)(24), (14)(23)

Certainly an element of S4 can't have more than 2 disjoint transpositions.  So there is only one other possibility.  Figure this out, and it will give you a very good idea of what the elements of H have to be.

Compute it directly.

Now compute N(ab), N(a), and N(b).

You don't need to conclude that.  All you need to conclude is that g1(x) <= h(x).

Set h(x) = max(g1(x), g2(x)).  Now how can you relate f1 and f2 to h?

You want to assume (without loss of generality) that O(g1(n)) <= O(g2(n)).

Lets take a look at your first integral:

You use x twice for two completely different things.  Inside the integral, we think of x as varying over the region of integration.  On the other hand, x as a limit of integration must be fixed.  You can't use the same symbol for multiple meanings.

What's in a name?  zeta here is just a dummy variable, and nothing more.  You need to integrate with respect to something.  Cauchy's integral formula says (under the proper restrictions):

Again, w is just a dummy variable, it could be any letter (one of my friends likes to integrate with respect to symbols such as a smiley face or a boat).

Cauchy's integral formula tells you a rather startling fact: the value of an analytic function is determined by an integral around a curve.  In order to integrate around a curve, we introduce a variable that takes on the values of Gamma, in my post it's w, in yours its zeta.

Does that clear things up?

My issue is not the same as your issue.  My issue is with how you handled your issue.  While related, they are two entirely different things.  As far as I'm concerned, my issue is not resolved.

Because I need to know that you understand what you did was rude, insulting, and downright wrong.  If you don't know that, then your behavior will continue, as it has been for the past n months.

Edited to add: If this was a one time incident, I wouldn't have posted anything at all.  But you seem to do this every week to a different member.  And from what I've seen, it's been getting worse overtime.  It needs to stop.

I was making a parallel to your interactions with people in person rather than across the internet.  You're stupid for not getting that.

Only a stupid person would think my reply was about the issue.  In reality, my post was about how you handled the issue.  Stupid stupid stupid.

Do you see my point?

You should know how to derive "common" values for sine and cosine.  For this problem, you use a right triangle with legs of length 1.  Find the length of the hypotenuse, and since the triangle is isosceles, the other two angles are pi/2.  Now you can find sine and cosine of pi/2.   But you must figure out the sign of the angles.  Use the mnemonic:

ASTC - All students take calculus

In quadrant 1, All values of trig functions are positive.  In quadrant 2, only sine is positive.  In quadrant 3, only tangent is positive, and in quadrant 4, only cosine is positive.

No calculator needed.