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One equation that every first-year student wants to be true is $A/B\cong C\implies A\cong B\times C$, where $A$, $B$, and $C$ are algebraic structures of some kind (modules, groups, rngs, . . .). However, this is not true: set $A=\mathbb{Z}$, $B=2\mathbb{Z}$, $C=\mathbb{Z}/2\mathbb{Z}$.

In the above case, though, $B\times C$ is not too far away from $A$: there is some $D$ such that $(B\times C)/D\cong A$. In this case, $D$ can be taken to be one of the copies of $B$.

My question is, when does this hold in general? One way (although certainly not the only way) to phrase this precisely would be to ask, what are those algebraic structures $A$ such that for all $B$, there is some $C$ such that $A\cong (B\times (A/B))/C$? (This is somewhat vague, as exactly what substructures $B$ are allowed here is not explicit, but I was thinking that this data might be incorporated into the data describing $A$; e.g., $A$ as a group, or $A$ as a ring without unity, etc.)

Thank you very much in advance!

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    I would not call it "salvaging intuition" so much as "salvaging the Freshman's Dream"...2011-04-04
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    This is not true for (finite) (abelian) groups. The cyclic group of order 4 is not a quotient of the direct product of two cyclic groups of order 2. I guess in general it will not be true for any algebraic structure that includes finite-but-not-semisimple objects.2011-04-04
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    The fact that it works in the example given is really a bit more "accidental": it's because it just so happens that any subgroup of $\mathbb{Z}$ is isomorphic to $\mathbb{Z}$ itself.2011-04-04
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    Why would you want to *salvage an intuition* which is simply misleading in most usual situations? Something can very well be intuitive and wrong... You should probably instead be looking for ways to develop correct intuitions to replace it!2011-04-10

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