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Let $0 \to A \to B \to C \to 0$ be an exact sequence of finite abelian groups. Assume that $B$ and $C$ is a square (i.e. there are groups $D,E$ such that $B \cong D^2$, $C \cong E^2$). Does this imply that also $A$ is a square?

Of course we may assume that $A,B,C$ are finite abelian $p$-groups for some prime $p$.

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up vote 2 down vote accepted

No.

Let $B=\mathbb{Z}_4\oplus\mathbb{Z}_4\oplus\mathbb{Z}_8\oplus\mathbb{Z}_8$, and let $A=\mathbb{Z}_4\oplus\mathbb{Z}_2\oplus\mathbb{Z}_8$ map into $B$ via the monomorphism $(a,b,c)\mapsto (0,a,4b,c)$. The quotient is then $C=\mathbb{Z}_4\oplus\mathbb{Z}_4$ with the mapping from $B$ to $C$ being $(d,e,f,g)\mapsto (d,\pi(f))$, where $\pi:\mathbb{Z}_8\rightarrow\mathbb{Z}_4$ is the natural projection.


Generalizing this to primes $p>2$ is left as an exercise.

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The order of $A$ must be a square, though. –  Mariano Suárez-Alvarez Apr 30 '12 at 7:46
    
Yeah. I am just saying that while $A$ need not be a square, $|A|$ must be :) –  Mariano Suárez-Alvarez Apr 30 '12 at 7:48
    
Sry, I misunderstood. In my counter-example this condition is satisfied, but I thought you were criticizing the answer :-) –  Jyrki Lahtonen Apr 30 '12 at 7:49
    
Thank you ...... :) –  Martin Brandenburg Apr 30 '12 at 8:26
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