“Historically,” note Strade and Farnsteiner in [1] , “Lie algebras emerged from the study of Lie groups.” In Section 1.1 of [1] , they give a simple example of the close connection between Lie algebras and Lie groups. In prime characteristic, David Winter [2] has defined maps which mimic the zero-characteristic exponential maps. See also Lemma 1.2 of [3] . In this paper, we focus on the following “Winter maps”: if
is an element of a characteristic-
Lie algebra
such that
we set

where
is the identity transformation of
. Such ad-nilpotent elements of degree less than
do exist in some graded Lie algebras, as can be seen from Lemma 2.3 and Proposition 2.7 of Chapter 4 of [1] , as well as from Lemma 1 of [4] ; of course, it is well known that non-zero-root vectors of simple classical-type Lie algebras are ad-nilpotent of degree less than or equal to four.
We will show here that for
such that
the inverse of
as a linear transformation of
is
, so that such transformations generate a group
of linear transformations of
. We will also show that
where, for
a linear transformation of
, and
as above, we define
(1)
Thus, like
and
,
is, in a sense, the functional inverse of
.
Lemma 1 If
and
are elements of
such that
and
then

Proof. We group terms with respect to total degree in
and

Lemma 2 Let
, and suppose that
is an element of
such that
then

Proof. We have by Lemma 1 that
equals

which we can write in terms of binomial coefficients as

By the Binomial Theorem, the above expression is equal to

which we can rewrite as

and recognize as
. 
Lemma 3 For any integer
and any integer
,
, we have

Proof. We proceed by induction on
and
. When
, we must have
, and we have
For any
, when
, we have

Now, for any
and any positive integer
less than
, suppose that
for all positive
less than
Then we have

by induction, and the fact that
(the “
case”). 
Lemma 4 Let
be an element of
such that
. Define
(2)
Then for any positive integer
less than
,
(3)
Proof. We proceed by induction on
. Since when
, (3) is just (2), the initial step of the induction proof is established. Suppose (3) is true for
. Then
equals

We group terms with respect to total degree (
, in this case) in
and get that
.
Rewriting the above expression using another binomial coefficient, we get that
equals

We change the order of summation to get

We replace the index of summation
by
to get
.
Adding and subtracting terms, we get

Setting
, we see, as in the proof of Lemma 3, that when r ≥ 1,

by that same Lemma 3. Thus,

so from the Binomial Theorem, we get that
equals
.
We now distribute to get that
equals

We replace the latter index of summation
by
to get that
equals

We change the order of summation and factor to get that
equals

By binomial arithmetic
equals

The above displayed formula is just (3) for
; i.e.,
equals
.
Thus, the induction step is complete. 
Theorem The linear transformation
of
has
as its inverse, whereas the map
of
to the group of non-singular linear transformations of
has
as its inverse, in the sense that
(a).
, and
(b).
.
Proof. (a) If, in Lemma 2, we let
and
, we see that (a) is true.
(b) Since
equals the
of Lemma 4, we have that
equals

which, by Lemma 4 equals

We replace the index
by
to get that

We change the order of summation to get that

We replace the index
by
to get that

We cancel an
and a
and combine the
factors to get that

We replace the index
by
and we replace the index
by
, and we get that

We change the order of summation to get that

We now appeal to a little more binomial arithmetic to observe that since
and
, it follows by induction that

from which we obtain that

We replace the index
by
to get that

Finally, we use Lemma 3 to see that we are left with
