Why is the short exact sequence of abelian groups not split exact?

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SUMMARY

The short exact sequence of abelian groups given by $$0 \rightarrow \bigoplus_p \Bbb Z/(p) \rightarrow \prod_p \Bbb Z/(p) \rightarrow \frac{\prod_p \Bbb Z/(p)}{\bigoplus_p \Bbb Z /(p)} \rightarrow 0$$ is definitively not split exact. This conclusion is drawn from the properties of direct sums and products of groups, particularly in the context of prime numbers. The sequence illustrates the failure of the splitting property due to the inherent structure of the involved groups.

PREREQUISITES
  • Understanding of abelian groups
  • Familiarity with exact sequences in homological algebra
  • Knowledge of direct sums and products of groups
  • Basic concepts of prime numbers and their properties
NEXT STEPS
  • Study the properties of exact sequences in homological algebra
  • Explore the implications of direct sums versus direct products in group theory
  • Investigate the role of prime numbers in the structure of abelian groups
  • Learn about the concept of splitting in exact sequences and its applications
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Mathematicians, particularly those specializing in algebra and group theory, as well as students studying homological algebra and exact sequences in their coursework.

Euge
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Here's this week's problem!

_____________

Problem. Show that the short exact sequence of abelian groups

$$0 \rightarrow \bigoplus_p \Bbb Z/(p) \rightarrow \prod_p \Bbb Z/(p) \rightarrow \frac{\prod_p \Bbb Z/(p)}{\bigoplus_p \Bbb Z /(p)} \rightarrow 0$$

is not split exact. (The sums and products are extended over all prime numbers $p$.)
_____________Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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No one answered this week's problem. Here is my solution.

Let

$$A = \bigoplus_p \Bbb Z/(p) \quad \text{and} \quad B = \prod_p \Bbb Z/(p).$$

It is enough to show that there is no isomorphism from $A \oplus B/A$ onto $B$. Suppose to the contrary that there is such an isomorphism, call it $f$. Then the composition $B/A \xrightarrow{\iota} A \oplus B/A \xrightarrow{f} B$ is zero, which contradicts the fact that $f$ is one-to-one.
 
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