Algebra I: Section 6. The structure of groups. 6.1 Direct products of ...

To see why the group described in (35) is isomorphic
to A4, regarded as the symmetry group of the tetrahedron
T, consider the tetrahedron shown in Figure 6.8
with vertices labeled 1, 2, 3, 4. Any 2,2-cycle, such as
(12)(34), corresponds to a 180 ◦ rotation about an axis
passing through the midpoints of opposite edges. We
have previously shown (Chapter 5) that the Klein Viergroup
N = {e} ∪ {all three 2,2-cycles}
is a normal subgroup in S4 isomorphic to Z2 × Z2. For
each axis passing through a vertex and the center of the
opposite face, we get a subgroup of order 3 (one of the
four Sylow subgroups H3) consisting of 0 ◦ , 120 ◦ , and
240 ◦ rotations about this axis. For example the subgroup
{e, (123), (132)} corresponds to rotations about
the axis passing through vertex 4 and the center of
the opposite face. With these geometric descriptions
in mind, it is not hard to see how to identify elements
in our abstract model (35) with the correct geometric
operations on T. �
Figure 6.8. A regular tetrahedron
centered at the origin. We show:
rotation A1 by 120 ◦ about an axis
through vertex 1, and B13 by 180 ◦
about an axis through midpoints of
opposite edges.
6.3.11 Exercise. Explain why the five groups G (1) , . . . , G (5) of order 12 are pairwise nonisomorphic.
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We take up one last example to show that the Sylow theorems do not always yield a definitive
analysis, and that as the order of G increases we begin to encounter groups that are not
semidirect products – i.e. they arise as extensions e → N → G → G/N → e that do not split.
6.3.12 Example (Groups of order 8). These are all p-groups so the Sylow theorems are
not much help; however by (3.2.5, 3.2.6) we know G has nontrivial center Z = Z(G), which
can only have order 2, 4, 8. If |Z| = 8 the group is abelian and the possibilities are G =
Z8, Z4 × Z2, Z2 × Z2 × Z2, in view of the following result.
6.3.13 Lemma. If G is abelian with |G| = 8 then G is isomorphic to Z8, Z4 × Z2, or
Z2 × Z2 × Z2.
Proof: If the maximum order of an element is o(x) = 8 then G ∼ = Z8. If the maximum is 4,
say for x = a, then A = 〈a〉 ∼ = Z4. Suppose there exists some b /∈ A such that o(b) = 2. Then
B = 〈b〉 ∼ = Z2 is transverse to A, A ∩ B = (e), AB = G, and since G is abelian it is a direct
product ∼ = Z4 × Z2.
Otherwise we have o(b) = 4 for all b /∈ A and B ∼ = Z4, but now we must have |A ∩ B| = 2
(because |G| = |A|·|B| = 16 if A∩B = (e)). The only subgroup of Z4 with order 2 is {[0] 4 , [2] 4 },
whose nontrivial element has order 2. That means a 2 and b 2 are the only elements of A and B
of order 2, and we must have a 2 = b 2 , a −2 = a 2 , b −2 = b 2 , and A ∩ B = {e, a 2 }. Now look at
the powers of the element ab: we get e, ab, (ab) 2 = a 2 b 2 = a 2 a −2 = e. Since b /∈ A ⇒ ab /∈ A we
have found an element outside of A with order 2. Contradiction. This case cannot arise. �
If |Z| = 4 then G/Z ∼ = Z2. Since G/Z is cyclic and Z is central in G, G must be abelian
(why?). Contradiction. Thus |Z| cannot equal 4.
6.3.14 Exercise. If G is a finite group and Z a subgroup such that (i) Z is central in G
(zg = gz for all z ∈ Z, g ∈ G), and (ii) G/Z is cyclic, prove that G must be abelian. �
Assuming |Z| = 2, let x be an element of highest order in G. We can only have o(x) = 2
or 4 (G is abelian if o(x) = 8). In the first case all elements y �= e would have order 2 and
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