what is semidirect product of groups?

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- Jun 14th 2009, 10:19 PMikramullahgulwhat is semidirect product of groups?
what is semidirect product of groups?

- Jun 15th 2009, 02:51 AMRaoh
- Jun 15th 2009, 04:25 AMSwlabr

Essentially, it is when you pin together two groups (such that they are still groups in their own right - they are both normal in the new group and their intersection is trivial), and then you say how they interact with one another. For instance, with the direct product you say that they commute with one another, while the dihedral group is the semidirect product of $\displaystyle C_2 = <a>$ with $\displaystyle C_n = <b>$ such that the elements of the group interact according to the rule $\displaystyle b^{a} = b^{-1}$. - Jun 15th 2009, 03:02 PMNonCommAlg
actually both are not necessarily normal but one of them will be normal.

__internally__, we say that a group $\displaystyle G$ is a semidirect product of two__subgroups__$\displaystyle H$ and $\displaystyle N$ and we write $\displaystyle G=N \rtimes H$ if:

$\displaystyle H \cap N=\{1\}, \ HN = G,$ and $\displaystyle N \lhd G.$ a more general definition is to say that $\displaystyle G$ is a semidirect product of two__groups__$\displaystyle G_1,G_2$ and we write $\displaystyle G=G_1 \rtimes G_2$ if there exists an exact sequence

$\displaystyle 1 \longrightarrow G_1 \overset{f}\longrightarrow G \overset{g}\longrightarrow G_2 \longrightarrow 1$ which is split, i.e. there is a map $\displaystyle h: G_2 \longrightarrow G$ such that $\displaystyle gh=\text{id}_{G_2}.$ Then it's easy to see that $\displaystyle G$ is an internal semidirect product of subgroups $\displaystyle N=f(G_1)$

and $\displaystyle H=h(G_2).$ an equivalent definition is to assume that there is a group homomorphism $\displaystyle \varphi: G_2 \longrightarrow \text{Aut}(G_1).$ then define the semidirect product of $\displaystyle G_1,G_2$ with respect to $\displaystyle \varphi$ to be the set

$\displaystyle G=G_1 \times G_2$ with multiplication defined by: $\displaystyle (x,y)*(z,t)=(x \varphi(y)(z),yt).$ then we write $\displaystyle G=G_1 \rtimes_{\varphi} G_2.$ again it's easy to see that $\displaystyle G$ is an internal semidirect product of two subgroups

$\displaystyle N=G_1 \times 1_{G_2}$ and $\displaystyle H=1_{G_1} \times G_2.$ - Jun 15th 2009, 10:18 PMSwlabr