On Semidirect Products

Outer Semidirect Product

Given any two groups N and H and a group homomorphism \phi:H\to\text{Aut}(N), we can construct a new group N\rtimes_\phi H, called the (outer) semidirect product of N and H with respect to \phi, defined as follows.
(i) The underlying set is the Cartesian product N\times H.
(ii) The operation, \bullet, is determined by the homomorphism \phi:

\bullet: (N\rtimes_\phi H)\times (N\rtimes_\phi H)\to N\rtimes_\phi H


for n_1,n_2\in N and h_1,h_2\in H.

This defines a group in which the identity element is (e_N, e_H) and the inverse of the element (n,h) is (\phi_{h^{-1}}(n^{-1}), h^{-1}).
Pairs (n, e_H) form a normal subgroup isomorphic to N, while pairs (e_N, h) form a subgroup isomorphic to H.

Inner Semidirect Product (Definition)

Given a group G with identity element e, a subgroup H, and a normal subgroup N\lhd G; then the following statements are equivalent:

(i) G is the product of subgroups, G=NH, where the subgroups have trivial intersection, N\cap H=\{e\}.
(ii) For every g\in G, there are unique n\in N and h\in H, such that g=nh.

If these statements hold, we define G to be the semidirect product of N and H, written G=N\rtimes H.

Inner Semidirect Product Implies Outer Semidirect Product

Suppose we have a group G with N\lhd G, H\leq G and every element g\in G can be written uniquely as g=nh where n\in N, h\in H.

Define \phi: H\to\text{Aut}(N) as the homomorphism given by \phi(h)=\phi_h, where \phi_h(n)=hnh^{-1} for all n\in N, h\in H.

Then G is isomorphic to the semidirect product N\rtimes_{\phi}H, and applying the isomorphism to the product, nh, gives the tuple, (n,h). In G, we have
\displaystyle (n_1h_1)(n_2h_2)=n_1h_1n_2(h_1^{-1}h_1)h_2=(n_1\phi_{h_1}(n_2))(h_1h_2)=(n_1,h_1)\cdot(n_2,h_2)
which shows that the above map is indeed an isomorphism.


About mathtuition88

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