2
$\begingroup$

Cancellation theorem in group theory (for direct product) says that if $B$ is a finite group and $A \times B \simeq A_1 \times B_1$ and $B \simeq B_1$ then $A \simeq A_1.$

Of course, if $B$ is not finite, the result is absurd, even for finitely presented groups (Here is an example by Steve)

I wonder whether the cancellation theorem holds for different products (in finite or infinite cases), such as semi-direct product, free product, fiber product over a given group, Zappa-Szep product (knit product), Wreath product.

$\endgroup$
8
  • 1
    $\begingroup$ Which of these various examples have you already tried? $\endgroup$
    – Yemon Choi
    Dec 14, 2011 at 5:58
  • 1
    $\begingroup$ There's an inherent lack of symmetry in other products, like the semidirect product or wreath product. Which of the two subgroups involved would you like to cancel? I'm pretty sure there are counterexamples for either one though. $\endgroup$
    – Steve D
    Dec 14, 2011 at 6:03
  • 1
    $\begingroup$ Consider the maps of short exact sequences in the case of semidirect products... $\endgroup$ Dec 14, 2011 at 6:43
  • 1
    $\begingroup$ Here's a counterexample for knit products: mathforum.org/kb/… Of course, knit products and semidirect products are more general than direct products, so the counterexample you cited above works in both those cases as well. $\endgroup$
    – Steve D
    Dec 14, 2011 at 8:04
  • 3
    $\begingroup$ I think the question is a little bit to vague and unfocused. $\endgroup$ Dec 14, 2011 at 11:44

1 Answer 1

3
$\begingroup$

For a straightforward example that the subgroup doing the acting in a semidirect product cannot automatically be cancelled (and as semidirect products are knit products, also addresses that case), the dihedral group of order $8$ can be written as a semidirect product in two different ways: $C_4\rtimes C_2$ (with the action being inversion) and $(C_2\times C_2)\rtimes C_2$ (with the action being swap coordinates).

$\endgroup$
1
  • $\begingroup$ FWIW, the second of these is even a wreath product: $(C_2\times C_2)\rtimes C_2 = C_2\wr C_2$, though I admit this doesn't compare a wreath product to a wreath product. $\endgroup$
    – John McVey
    Apr 19, 2019 at 14:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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