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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: $|B| = [B:C]\ast|C| = |A|\ast|C|$ , 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: $|A|\ast|C| = |B|\ast|D|$ Last edited by Deveno; July 11th, 2014 at 10:42 PM. Tags exact, order, sequence ### algebra/exact sequence

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