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