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July 2nd, 2014, 04:59 AM  #1 
Newbie Joined: Jun 2014 From: / Posts: 3 Thanks: 0  Order in exact sequence
Hello!, As far as I know the order in exact sequences is multiplicative? Consider an exact sequence of, let's say, finite abelian groups: 0>A>B>C>D>0 What can I say about the orders (cardinality) of A,B,C,D? Do you know any references? Thanks a lot! 
July 5th, 2014, 04:42 AM  #2 
Newbie Joined: Apr 2014 From: USA Posts: 24 Thanks: 1 
I know that the order of an element divides the order of the group.

July 5th, 2014, 05:55 AM  #3 
Newbie Joined: Jun 2014 From: / Posts: 3 Thanks: 0 
Thanks. But I rather meant something like ord(A)*ord(C)=ord(B)*ord(D). I don't know how to use the exactness of the sequence for the orders... 
July 11th, 2014, 10:39 PM  #4 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88 
For a short exact sequence: $0 \to A \to B \to C \to 0$ by exactness, $A$ divides $B$ (since $A \to B$ is an injection). Also by exactness, $C$ divides $B$ (since $B \to C$ is a surjection). Now exactness dictates that the kernel of $B \to C$ is the image of $A \to B$, that is to say: $B = [B:C]\astC = A\astC$ , since if: $A'$ is the image of $A$ under the morphism $A \to B$, then: $C \cong B/A'$, so $C = \dfrac{B}{A'}$, and $A = A'$. ******************* Now, if we have: $0 \to A \to B \to C \to D \to 0$ we no longer know that $B \to C$ is surjective or injective, but as before, $A \to B$ is injective, and $C \to D$ is surjective. If we label the maps like so: $0 \to A \stackrel{i}{\to} B \stackrel{f}{\to} C \stackrel{g}{\to} D \to 0$, we see that $g$ is surjective, so that: $D \cong C/\text{ker }g$. Since $\text{ker }g = \text{im }f$, we can write: $D \cong C/f(B)$. Since $\text{ker }f = \text{im }i$, we have: $f(B) \cong B/i(A)$ so that: $D = \dfrac{C}{\left(\dfrac{B}{A}\right)}$ that is: $A\astC = B\astD$ Last edited by Deveno; July 11th, 2014 at 10:42 PM. 

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