“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 
