From the definitions it is clear that condition 41

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From the definitions, it is clear that condition (4.1) above is equivalent to the condition p e i = ord( a i ) = ord i - 1 ( a i ) = max { ord i - 1 ( a ) : a G } . (4.2) Assume now that k > 0, and consider expressions of the form x 1 a 1 + · · · + x k a k , for x 1 , . . . , x k Z . It is clear from the definition that every element of H k can be expressed in this way. Claim 1: x 1 a 1 + · · · + x k a k = 0 G implies p e i | x i for 1 i k . To prove this claim, assume that x 1 a 1 + · · · + x k a k = 0 G , and p e j - x j for some j . Moreover, assume that the index j is maximal, i.e., p e j 0 | x j 0 for j < j 0 k . Write x j = p f y , where p - y and 0 f < e j , and let y 0 be a multiplicative inverse of y modulo p e j . Then we have p f a j = - ( y 0 x 1 a 1 + · · · + y 0 x j - 1 a j - 1 ) H j - 1 , contradicting the assumption that p e j - 1 a j / H j - 1 . That proves the claim. From this claim, it is easy to see that the map sending ([ x 1 mod p e 1 ] , . . . , [ x k mod p e k ]) to x 1 a 1 + · · · + x k a k is an isomorphism of Z p e 1 × · · · × Z p e k with H k . That the definition of this map is unambiguous follows from the fact that ord( a i ) = p e i for 1 i k . Moreover, it is clear that the the map is a surjective homomorphism, and by Claim 1, the kernel is trivial. It is also clear that under this isomorphism, the subgroup H i of H k , for 1 i k , corresponds to the subgroup of Z p e 1 × · · · × Z p e k consisting of all k -tuples whose last k - i components are zero. Next, assume that H k ( G . We show how to extend a good sequence ( a 1 , . . . , a k ) to a good sequence ( a 1 , . . . , a k , a k +1 ) for some a k +1 G . If k = 0, this is trivial: simply choose a 1 to be an element of maximal order in G . Now assume k > 0. Let us choose b G such that ord k ( b ) is maximal. Let ord k ( b ) = p f . Note that f e i for 1 i k , since by definition, p e i b H i - 1 H k . In general, we have p f | ord( b ), and if p f = ord( b ), then we can set a k +1 := b , and we are done. However, in general, we cannot expect that p f = ord( b ). Note, however, that ord k ( b + h ) = ord k ( b ) for all h H k , so if we can find h H k such that p f ( b + h ) = 0 G , we will also be done. Write p f b = k i =1 x i a i for some integers x i . If we can find integers z 1 , . . . , z k such that p f z i + x i 0 (mod p e i ) for 1 i k , then setting h := k i =1 z i a i , we see that p f ( b + h ) = k X i =1 ( p f z i + x i ) a i = 0 G , and we will be done. Moreover, by Theorem 2.5, such integers z 1 , . . . , z k exist provided p f | x i for 1 i k . Claim 2: If p f b = k i =1 x i a i as above, then p f | x i for all 1 i k . To prove this claim, assume that p f - x j for some j . Multiplying the equation p f b = k i =1 x i a i by p e j - f , we see that p e j b can be expressed as k i =1 x 0 i a i , where p e j - x 0 j . By Claim 1, it follows 33
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that p e j b / H j - 1 , which contradicts the assumption that p e j a H j - 1 for all a G . That proves the claim.
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