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I am doing some homework and it involves filling out a commutative diagram, I just want to discuss some of my thoughts and see what's wrong and most importantly, why.

One of the sequences in the diagram, we're given that they are all exact in rows and columns, is this one $$0\to \mathbb{Z}_3\to X \to \mathbb{Z}_2\to 0$$ Where $X$ is to be figured out, I have the general idea for something like this to work, in general, it'd be $\mathbb{Z}_6$ with multiplication by 2 from $\mathbb{Z}_3$ and natural surjection after that. I feel that it'd be the "natural" way to go about it but I am not entirely certain, some good resources if anything would be appreciated as well.

Hagen von Eitzen
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Zelos Malum
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1 Answers1

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In this diagram, $X$ could be either $\mathbb Z_6$ or $S_3$. (Except when this is supposed to be a diagram in the category of abelian groups, in which case only $\mathbb Z_6$ is possible).

More generally, in $0\to A\to X\to B\to 0$ you can always have the direct sum/product $X=A\oplus B$ as the simplest case, but other choices of $X$ may be possible. (In this specific example, $S_3$ is even not much "less" than the direct product: It is a (in fact the) semi-direct product of $\mathbb Z_3$ and $\mathbb Z_2$ induced by the only nontrivla action of the latter on the former; but that is probably a different story)

Hagen von Eitzen
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  • It is abelian groups, but could you go into how the reasoning for it goes? I am struggling a bit just to come ot grip with how one reason for these things. – Zelos Malum Sep 08 '15 at 06:30