[GAP Forum] listing all semidirect products

Benjamin Sambale bsambale at gmx.de
Tue Jun 16 08:14:40 BST 2009

There is only one semidirect product of S_3 with S_3 (up to isomorphism)
namely S_3\times S_3. This can easily be shown, since Out(S_3)=Z(S_3)=1
(where Out is the outer automorphism group and Z is the center).

And there are indeed 10 homomorphisms from S_3 to S_3. This can be shown
with the following GAP code (or by hand):


But all these homomorphisms lead to the same semidirect product (up to

Dinesh Krithivasan schrieb:
> Hi all,
>   I am trying to use GAP for the following: List all semidirect products of S_3 by S_3 (the symmetric group on 3 elements). This boils down to the choice of the homomorphism from S_3 to Aut(S_3) = S_3. While I could find some code that works for specific choices of this homomorphism, I am unclear as to how to obtain the entire list (I worked out that there are 10 homomorphisms from S_3 to Aut(S_3). Is this right? It would mean that there are 10 semidirect products some of which might be isomorphic to one another.). Further, since we know the result is a group of size 36 and there are only 14 groups of that order, could we get GAP to identify exactly which of these 14 groups the semidirect products are isomorphic to? Thanks a lot in advance for any help/suggestions.
> Regards,
> Dinesh.
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