Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
I would not call it "salvaging intuition" so much as "salvaging the Freshman's Dream"... – Arturo Magidin Apr 4 '11 at 3:52
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. – Jack Schmidt Apr 4 '11 at 4:03
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. – Arturo Magidin Apr 4 '11 at 4:35
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! – Mariano Suárez-Alvarez Apr 10 '11 at 21:35

Another way to look at this is through exact sequences. Consider the exact sequence $$0 \longrightarrow B\longrightarrow A \longrightarrow C\longrightarrow 0.$$ You are asking when this thing splits. There are a bunch of equivalent conditions for exact sequences splitting. If these are modules, then certainly B being injective or C being projective will give it to you but those conditions are only sufficient, not necessary as far as I know. In the category of vector spaces over some fixed field, for example, the above sequence always splits because all vector spaces are free modules and hence projective. Note that the vector spaces don't even have to be finite dimensional. In this nice category, we even have that the dualizing functor (the functor sending a vector space to its dual, whatever you want to call that) is an exact functor (even though this fact may not be abundantly clear at first glance) which tells us that the dual of A is isomorphic to the product of the duals of B and C. I apologize for the lack of Tex. I'm gonna work on learning that. I would appreciate any Tex hints or advice anyone would be willing to give.

share|cite|improve this answer
\to gives $\to$; \longrightarrow gives $\longrightarrow$. Right click on the formula and ask "show source" to see the LaTeX code; it should be placed inside $ for in-line formulas, and surrounded by $$ of \[ and \] for displayed formulas. – Arturo Magidin Apr 10 '11 at 21:40
@Arturo: Thanks for Tex advice. Now I can learn Tex code by just inspecting other people's code. – bcwhite Apr 11 '11 at 21:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.