Currently i am trying to understand the concept of Semi direct product of groups from Abstract Algebra text of Dummit and Foote.The discussion given in the same book is bit confusing to me and i am unable to understand this concept properly.

What is the difference between internal and external semi direct product ? I found different definitions one by using the group action and other by using the splitting of exact sequence but i could not figure out how these definitions are equivalent.

Definition: Consider $G$,$H$ groups and $\varphi : G\to \operatorname{Aut}(H)$ a homomorphism, we define the semidirect product group as the cartesian product $G\times H$ together with the operation


and we denote the resulting group as $G\ltimes H$.

At some places i found the following definition also:

Definition: Existence of a split short exact sequence $1 \rightarrow N \rightarrow G \rightarrow H \rightarrow 1$ gives that $G$ is isomorphic to a semidirect of $N$ and $H$

Please suggest any book/notes to understand this concept properly.

Thank you

  • $\begingroup$ Since you found several definitions, I don't see why you are asking for a definition. Are you asking why the various definitions are equivalent? $\endgroup$ – Tobias Kildetoft Oct 3 '16 at 7:06
  • $\begingroup$ @TobiasKildetoft: yeah i would like to see why these are equivalent.I'll edit the question thanks! $\endgroup$ – Math Lover Oct 3 '16 at 7:09
  • $\begingroup$ Then please write out in full the relevant definitions (there are several ways to phrase them). Also, it might be helpful to clear up the distinction between internal and external first (basically, the first is a structure, the second is a construction). $\endgroup$ – Tobias Kildetoft Oct 3 '16 at 7:11
  • $\begingroup$ @TobiasKildetoft: Edited,thanks. $\endgroup$ – Math Lover Oct 3 '16 at 7:18
  • $\begingroup$ Note that the second definition is not really a definition but a statement, namely that if such a sequence exists then $G$ has the form given by the definition above (the statement is unfortunately also false in the current form, as one would need to require the sequence to split). $\endgroup$ – Tobias Kildetoft Oct 3 '16 at 7:20

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