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July 2nd, 2014, 04:59 AM   #1
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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!
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July 5th, 2014, 04:42 AM   #2
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I know that the order of an element divides the order of the group.
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July 5th, 2014, 05:55 AM   #3
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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...
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July 11th, 2014, 10:39 PM   #4
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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.
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