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Finite Groups –

killed by induction

Bulky volumes have been written on finite group theory. Here

is the way to make a clean sweep of that bulk of sophistication.

Lemma. Let G be a group, and let H1 , . . . , Hn be subgroups such

that G = H1 ∪ · · · ∪ Hn . Then G = Hi for some i ∈ {1, . . . , n}.

Proof. For n = 1, the lemma is trivial. Thus let us start

with the case n = 2. Assume that G = H1 ∪ H2 , but G 6= Hi

for i ∈ {1, 2}. Then there are elements g ∈ H1 r H2 and h ∈

H2 r H1 . By assumption, gh ∈ H1 ∪ H2 , say, gh ∈ H1 . Then

h = g −1 (gh) ∈ H1 , a contradiction. Now let us assume that the

lemma holds for a fixed n, and assume that G = H1 ∪ · · · ∪ Hn+1 .

By the inductive hypothesis, we can assume that H1 ∪· · ·∪Hn 6=

G and Hn+1 6= G. Arguing in the same way as before, we find

elements g ∈ (H1 ∪· · ·∪Hn )rHn+1 and h ∈ Hn+1 r(H1 ∪· · ·∪Hn ).

If gh ∈ H1 ∪ · · · ∪ Hn , we get h = g −1 (gh) ∈ H1 ∪ · · · ∪ Hn since

g ∈ Hi for some i ∈ {1, . . . , n}, and in case gh ∈ Hn+1 , we get

g ∈ Hn+1 , which is both impossible. The lemma is proved.

Recall that a group G is said to be cyclic if there is an element

g ∈ G such that no proper subgroup of G contains g.

Theorem. Every finite group G is cyclic.

Proof. For any S

g ∈ G, let Hg denote the subgroup generated

by g. Thus G = g∈G Hg . Since G is finite, we can apply the

lemma, which yields G = Hg for some g ∈ G. Hence G is cyclic.

Remark. To be sure, the above “theorem” also implies - ex

falso quodlibet - what your intuition prescribes: finite groups

need not be cyclic!

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