Metsänkylä , Distribution of irregular prime numbers 127

Lemma 1. D^^ is the product of those distinct primes I for which I — 1 divides 2k.

Lemma 2. For every A: ^ 1, there is a unique decomposition к ^ /Cj/Cg such that ^1 ^ I7 ^2 ^ i? e^i? ^2) — 1? ^1 divides TVgjt and every prime in /Cg divides D^^. Lemma 3. For any odd prime /,

/ c h

if 2k^2h^0 (mod/ —1).

Note that, by Lemmas 1 and 2, the numbers on both sides of the above congruence mod / are integral with respect to l.

Lemma 4. For every positive integer t,

tN^ , ^D^^S^j^it) (mod ^2), where

S^^it ) = 1^^ + 2^"= + '" + (t — IfK

The next lemma is an easy consequence of Lemma 3 and the irregularity criterion, mentioned in the introduction (see, e. g., [6]).

Lemma 5. Let к =^ k^k^ be the decomposition according to Lemma 2. Every prime p which divides N^kl^^i is irregular (p is called a proper divisor of Л^гл)-

3 . Preparations for the proof

For any integer n, denote by [^г] the residue class mod m containing n. Moreover' denote by j^^, . . ., p^ the distinct odd prime factors of m (if there are any), and by К the greatest exponent in the canonical decomposition of m.

We shall need two more lemmas.

Lemma 6. For k^ K, к = l(mod 99(m^)), there is an integer A, relatively prime to m, such that

66'2fc ( ^ ) = Am (mod m^) (99 denotes Euler's function).

Proof . Let P be any prime factor of m, and let P^ \ m, P^^^ \ m. Because k'^h and к = i (mod (p{P^^)), we see that

S , , ( m ) - ^/* = ^f (mod P^"),

where the sums are extended over the numbers 1, 2, . . ., m — 1, except for multiples of P. Evaluating the last sum, we get

6 2: f = 2m^(m —^-] + (i — P)m,

Hence

( 1 ) GS^kim) ^ (1 — P)m (mod P^'^)

and the lemma follows.

Lemma 7. Let, as in the theorem, H be a proper subgroup of G. If m is odd, then G contains an element [f] such that

[ f ] iG — H, /'ф! (modi?,) {i = i,..,,s).

If i\m, 3 \ m, then the same is true with the additional condition f ф 1 (mod 4).

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