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

Here is an example of a non-split extension. Although it involves an infinite group, it illustrates
the sort of obstructions that can prevent the existence of a subgroup that cross-sections
the N-cosets in G. The quaternion group Q8 to be discussed in Section 6.3 is a finite group
exhibiting similar behavior.
6.2.33 Example. If we take G = R and N = Z, the
extension e → Z → R → G/N → e does not split,
so G is not a semidirect product of N and H. To see
why, recall that G/N is isomorphic to the circle group
S 1 = {z ∈ C : |z| = 1}. In fact (as in 3.3.3), the homomorphism
φ : (R, +) → (S 1 , · ) given by the exponential
map φ(t) = e 2πit has kernel kerφ = Z, and hence by the
First Isomorphism Theorem 3.1.13 the map φ factors
through the quotient map π : R → R/Z to give an isomorphism
˜ φ : (R/Z, +) → (S 1 , · ) as shown in Figure 6.6
at right.
R
φ
−→ (S1 , · )
π ↓ ր
R/Z
˜φ
Figure 6.6. The induced map ˜ φ is
an isomorphism such that ˜ φ ◦ π = φ
(diagram commutes).
Arguing by contradiction, we now show that no subgroup H in R can cross-section the cosets
x+Z in R/Z. Suppose such an H actually exists. Consider any rational value 0 < θ < 1. Then
mθ ∈ Z for some m ∈ N, and hence 1 = φ(mθ) = φ(θ) m . Since H is a cross-section for R/Zcosets,
there is some x ∈ H such that x + Z = θ + Z. Therefore x − θ ∈ Z, mx − mθ ∈ Z, and
then φ(mx) = φ(mθ) = 1 because kerφ = Z. But if H is a cross-section we also know that the
restricted homomorphism φ H : H → S 1 is a bijection, and hence an isomorphism of groups.
Since x ∈ H ⇒ mx = x + . . . + x ∈ H, the only way to get φ(mx) = 1 is to have mx = 0
because φ : H → S 1 is one-to-one and we already have φ(0) = 1. This in turn implies that
x = 0, which is impossible because it would imply that θ ≡ x ≡ 0 (mod 1) and hence that θ is
an integer contrary to our choice of 0 < θ < 1. Conclusion: no subgroup can cross-section the
cosets in R/Z. �
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