Math 2.0

Group theory and generalizations Abelian groups Example

Prüfer group

Example

Let pp be a prime. The Prüfer pp-group is ℤ(p∞)=ℤ[1/p]/ℤ={m/pk+ℤ:m∈ℤ, k≥0},\Z(p^\infty) = \Z[1/p] / \Z = \set{m/p^k + \Z : m \in \Z, \ k \ge 0}, a subgroup of ℚ/ℤ\Q/\Z. Equivalently, it is the group of all complex roots of unity of order a power of pp, a subgroup of the circle group.

It is a divisible group. Let x=m/pk+ℤx = m/p^k + \Z and n≥1n \ge 1, and write n=pjrn = p^j r with p∤rp \nmid r. Choose an integer ss with rs≡1(modpk+j)rs \equiv 1 \pmod{p^{k+j}}, and put y=sm/pk+j+ℤy = sm/p^{k+j} + \Z. Then ny=rs m/pk+ℤny = rs \, m / p^k + \Z, and rs m/pk−m/pk=(rs−1)m/pkrs \, m/p^k - m/p^k = (rs - 1) m / p^k is an integer, because pk+jp^{k+j} divides rs−1rs - 1; so ny=xny = x.

Remarks

The Prüfer group is the union of the chain of cyclic groups ℤ/pℤ⊂ℤ/p2ℤ⊂⋯\Z/p\Z \subset \Z/p^2\Z \subset \cdots, where ℤ/pkℤ\Z/p^k\Z is identified with {m/pk+ℤ}\set{m/p^k + \Z}. Its proper subgroups are exactly these finite cyclic groups, so it is not finitely generated. It is a p-group in the sense that every element has order a power of pp. Together with ℚ\Q, the Prüfer groups are the building blocks of all divisible groups. Its Pontryagin dual is the group of p-adic integers.

Further reading

  1. Web https://en.wikipedia.org/wiki/Prüfer_group
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Changed2 November 2026

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