Semidirect Products and Group Homomorphisms
Let G be a semidirect product of the normal subgroup N and the subgroup H. Let Aut(N) denote the group of all automorphisms of N. The map φ : H → Aut(N) defined by φ(h) = φh, where φh(n) = hnh−1 for all h in H and n in N, is a group homomorphism. Together N, H and φ determine G up to isomorphism, as we show now.
Given any two groups N and H (not necessarily subgroups of a given group) and a group homomorphism : H → Aut(N), there is a new group (or simply ), called the semidirect product of N and H with respect to , defined as follows.
- As a set, is the cartesian product N × H.
- Multiplication of elements in is determined by the homomorphism . The operation is
-
- defined by
- for n1, n2 in N and h1, h2 in H.
This defines a group in which the identity element is (eN, eH) and the inverse of the element (n, h) is (h–1(n–1), h–1). Pairs (n,eH) form a normal subgroup isomorphic to N, while pairs (eN, h) form a subgroup isomorphic to H. The full group is a semidirect product of those two subgroups in the sense given above.
Conversely, suppose that we are given a group G with a normal subgroup N and a subgroup H, such that every element g of G may be written uniquely in the form g=nh where n lies in N and h lies in H. Let : H → Aut(N) be the homomorphism given by (h) = h, where
for all n in N and h in H. Then G is isomorphic to the semidirect product ; the isomorphism sends the product nh to the tuple (n,h). In G, we have the multiplication rule
A version of the splitting lemma for groups states that a group G is isomorphic to a semidirect product of the two groups N and H if and only if there exists a short exact sequence
and a group homomorphism γ : H → G such that, the identity map on H. In this case, : H → Aut(N) is given by (h) = h, where
If is the trivial homomorphism, sending every element of H to the identity automorphism of N, then is the direct product .
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