By Miller G. A.

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R could be Z or Zp ). Then one easily checks that the following groups (with topologies naturally induced from R) are proﬁnite groups: – R× , the group of units of R [one can verify the compactness of R× as follows: consider the multiplication mapping μ : R × R −→ R; then μ−1 {1} is compact; on the other hand, R× is the image of μ−1 {1} under one of the projections R × R −→ R]. , the group of units of the ring Mn (R) of all n×n matrices over R). [One can verify this as in the previous case, eventhough Mn (R) is not commutative: just observe that, for matrices over R, having a left inverse is equivalent to being invertible].

Proof. Consider the set P of all pairs (N, XN ), where N such that cG and XN ⊆ G−N (i) for every open subgroup U of G containing N , XN − U is a ﬁnite set; and (ii) G = XN , N . ˜ N = {xN | x ∈ XN } is a set Note that these two conditions imply that X of generators of G/N converging to 1. Clearly P = ∅. Deﬁne a partial ordering on P by (N, XN ) (M, XM ) if N ≥ M , XN ⊆ XM and XM − XN ⊆ N . We ﬁrst check that the hypotheses of Zorn’s Lemma are met. Let {(Ni , Xi ) | i ∈ I} be a linearly ordered subset of P; put K = i∈I Ni and XK = i∈I Xi .

2 Let G be a ﬁnitely generated proﬁnite group and let ϕ : G −→ G be a continuous epimorphism. Then ϕ is an isomorphism. 5 Finitely Generated Proﬁnite Groups 45 Proof. We claim that ϕ is an injection.