3.2. SEMIDIRECT PRODUCTS
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(a) Suppose G is a group and 'i W G ! Ai is a homomorphism for i D 1; 2; : : : ; r. Suppose ker.'1/ \ : : : ker.'r / D feg. Show that ' W x 7! .'1.x/; : : : ; 'r .x// is an injective homomorphism into A1 Ar .
(b) Explore conditions for ' to be surjective.
3.1.19. Let N1; N2; : : : ; Nr be normal subgroups of a group G. Show that Ni \ .N1 : : : Ni 1NiC1 : : : Nr/ D feg for all i if, and only if, whenever xi 2 Ni for 1 i r and x1x2 xr D e, then x1 D x2 D D xr D e.
3.2. Semidirect Products
We now consider a slightly more complicated way in which two groups can be fit together to form a larger group.
Example 3.2.1. Consider the dihedral group Dn of order 2n, the rotation group of the regular n-gon. Dn has a normal subgroup N of index 2 consisting of rotations about the axis through the centroid of the faces of the n-gon. N is cyclic of order n, generated by the rotation through an angle of 2=n. Dn also has a subgroup A of order 2, generated by a rotation j through an angle about an axis through the centers of opposite edges (if n is even), or through a vertex and the opposite edge (if n is odd). We have Dn D NA, N \ A D feg, and A Š Dn=N , just as in a direct product. But A is not normal, and the nonidentity element j of A does not commute with N . Instead, we have a commutation relation j r D r 1j for r 2 N .
Example 3.2.2. Recall the affine or “AxCb” group Aff.n/ from Exercises 2.4.20 and 2.7.3 and Examples 2.7.5 and 2.7.9. It has a normal subgroup N consisting of transformations Tb W x 7! x C b for b 2 Rn. And it has the subgroup GL.n; R/, which is not normal. We have N GL.n; R/ D Aff.n/, N \GL.n; R/ D fEg, and Aff.n/=N Š GL.n; R/. GL.n; R/ does not commute with N ; instead, we have the commutation relation ATb D TAbA for A 2 GL.n; R/ and b 2 Rn.
Both of the last examples are instances of the following situation: A group G has a normal subgroup N and another subgroup A, which is not normal. We have G D NA D AN , A\N D feg, and A Š G=N . Since N is normal, for each a 2 A, the inner automorphism ca of G restricts to an automorphism of N , and we have the commutation relation an D ca.n/a for a 2 A and n 2 N . (Recall that the inner automorphism ca is defined by ca.x/ D axa 1.)
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3. PRODUCTS OF GROUPS Now, if we have groups N and A, and we have a homomorphism ˛ W a 7! ˛a from A into the automorphism group Aut.N / of N , we can build from these data a new group N Ì A, called the semidirect product
˛
of A and N . The semidirect product N Ì A has the following features: It
˛
contains (isomorphic copies of) A and N as subgroups, with N normal; the intersection of these subgroups is the identity, and the product of these subgroups is N Ì A; and we have the commutation relation an D ˛a.n/a
˛
for a 2 A and n 2 N .
The construction is straightforward. As a set, N Ì A is N A, but now
˛
the product is defined by .n; a/.n0; a0/ D .n˛a.n0/; aa0/.
Proposition 3.2.3. Let N and A be groups, and ˛ W A ! Aut.N / a homomorphism of A into the automorphism group of N . The Cartesian product N A is a group under the multiplication .n; a/.n0; a0/ D .n˛a.n0/; aa0/.
This group is denoted N Ì A. Q N D f.n; e/ W n 2 N g and Q A D f.e; a/ W
˛
a 2 Ag are subgroups of N Ì A, with Q N Š N and Q A Š A, and Q N is
˛
normal in N Ì A. We have .e; a/.n; e/ D .˛a.n/; e/.e; a/ D .˛a.n/; a/
˛
for all n 2 N and a 2 A.
Proof. We first have to check the associativity of the product on N Ì A.
˛
Let .n; a/, .n0; a0/, and .n00; a00/ be elements of N Ì A. We compute
˛
..n; a/.n0; a0//.n00; a00/ D .n˛a.n0/; aa0/.n00; a00/ D .n˛a.n0/˛aa0.n00/; aa0a00/:
On the other hand, .n; a/..n0; a0/.n00; a00// D .n; a/.n0˛a0.n00/; a0a00/ D .n˛a.n0˛a0.n00//; aa0a00/:
We have n˛a.n0˛a0.n00// D n˛a.n0/˛a.˛a0.n00// D n˛a.n0/˛aa0.n00/; where the first equality uses that ˛a is an automorphism of N , and the second uses that ˛ is a homomorphism of A into Aut.N /. This proves associativity of the product.
It is easy to check that .e; e/ serves as the identity of N Ì A.
˛
Finally, we have to find the inverse of an element of N Ì A. If .n0; a0/
˛
is to be the inverse of .n; a/, we must have aa0 D e and n˛a.n0/ D e.
Thus a0 D a 1 and n0 D ˛ 1 a .n 1/ D ˛a 1 .n 1/. Thus our candidate i
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for .n; a/ 1 is .˛a 1.n 1/; a 1/. Now we have to check this candidate by multiplying by .n; a/ on either side. This is left as an exercise.
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Remark 3.2.4. The direct product is a special case of the semidirect product, with the homomorphism ˛ trivial, ˛.a/ D idN for all a 2 A.
Corollary 3.2.5. Suppose G is a group, N and A are subgroups with N normal, G D NA D AN , and A \ N D feg. Then there is a homomorphism ˛ W A ! Aut.N / such that G is isomorphic to the semidirect product N Ì A.
˛ Proof. We have a homomorphism ˛ from A into Aut.N / given by ˛.a/.n/ D ana 1.
Since G D NA and N \ A D feg, every element g 2 G can be written in exactly one way as a product g D na, with n 2 N and a 2 A. Furthermore, .n1a1/.n2a2/ D Œn1.a1n2a 1/Œa
1
1a2 D Œn1˛.a1/.n2/Œa1a2. Therefore, the map .n; a/ 7! na is an isomorphism from N Ì A to G.
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˛
Example 3.2.6. Z7 has an automorphism ' of order 3, '.Œx/ D Œ2x; this gives a homomorphism ˛ W Z3 ! Aut.Z7/, defined by ˛.Œk/ D 'k. The semidirect product Z7 Ì Z3 is a nonabelian group of order 21. This group
˛
is generated by two elements a and b satisfying the relations a7 D b3 D e, and bab 1 D a2.
Exercises 3.2 3.2.1. Complete the proof that N Ì A (as defined in the text) is a group by
˛
verifying that .˛a 1.n 1/; a 1/ is the inverse of .n; a/.
3.2.2. Show that j W Œx 7! Œ x defines an order 2 automorphism Zn.
Conclude that ˛ W Œ12 7! j determines a homomorphism of Z2 into Aut.Zn/. Prove that Zn Ì Z2 is isomorphic to Dn .
˛
3.2.3. Show that the affine group Aff.n/ is isomorphic to a semidirect product of GL.n; R/ and the additive group Rn.
3.2.4. Show that the permutation group Sn is a semidirect product of Z2 and the group of even permutations An.
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