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Math Help - what is semidirect product of groups?

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    what is semidirect product of groups?

    what is semidirect product of groups?
    Last edited by mr fantastic; June 15th 2009 at 04:55 AM. Reason: Re-titled the post
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    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by ikramullahgul View Post
    what is semidirect product of groups?

    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 C_2 = <a> with C_n = <b> such that the elements of the group interact according to the rule b^{a} = b^{-1}.
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    Quote Originally Posted by Swlabr View Post
    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 C_2 = <a> with C_n = <b> such that the elements of the group interact according to the rule b^{a} = b^{-1}.
    actually both are not necessarily normal but one of them will be normal. internally, we say that a group G is a semidirect product of two subgroups H and N and we write G=N \rtimes H if:

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

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

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

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

    N=G_1 \times 1_{G_2} and H=1_{G_1} \times G_2.
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    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    actually both are not necessarily normal but one of them will be normal.
    Ah yes, my bad.
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