Is the multiplicative Group of surcomplex numbers of modulus 1 isomorphic to the additive Group of the surreal numbers modulo the sub-Group of surreal integers? And, do Norman Alling's surreal extensions of sine and cosine (defined in section 7.5 of his book "Foundations of analysis over surreal number fields") accomplish the isomorphism?



Uniformizing the surcomplex unit circle
Let me say at least this: the usual series for sine and cosine "converge" for the finite surreals, and provide an isomorphism from (the finite surreals modulo the standard integers) onto (the surcomplex unit circle). 


An alternate for the sine on the finite surreals, write $x = a+z$ where $a$ is a standard real and $z$ is infinitesimal, then use the addition formulas for $\sin(a+z)$ and $\cos(a+z)$.


**added March 18**
Extension to all surreals depends on the choice for the complementary subgroup of the finite surreals. What (beyond the usual $\mathbb Z$) should be called an "integer". Conway has such a choice in his formulation, called $\mathbf{Oz}$.


surjective ... Conway emphasizes more the algebraic and combinatorial side, less the analytic side. But, in fact, this same thing will work in all the usual canonical ways of constructing nonarchimedean extensions of the reals. 


In nonstandard analysis, $\sin$ and $\cos$ have corresponding nonstandard versions, and surjectivity is a first-order property, so it transfers.


In transseries, there are many possibilities: series expansion for $\arcsin$; an integral; a solution of a differential equation; ...


In the surreals, Erlich [[LINK](http://www.ohio.edu/people/ehrlich/Alternative.pdf)] showed $\mathbf{No}$ can be realized as a space of Hahn series, and after that it will be the same as for transseries. It does seem less convenient in Conway's original formulation, admittedly.


**added March 19**  

Here is how we do it when using Hahn series. Once you reach a certain point in Conway's book ONAG, you can do this also for surreals, using his Theorem 23 with his "normal forms". 


Hahn series look like $\sum\_{i \in I} c\_i g\_i$, where the coefficients $c\_i$ are real, and the "monomials" $g\_i$ are reverse well-ordered. One possible monomial is $1$; monomials larger than that are "infinite", those smaller are "infinitesimal". The set of possible monomials is an ordered abelian group under multiplication.


Given a general element $A$ of our field of Hahn series, we write it as $A = L + t + S$, 
where every monomial in $L$ is infinite, $t \in \mathbb R$, and every monomial in $S$ is infinitesimal. Define
$$\begin{align}
\sin A &= \sin t \cos S + \cos t \sin S,
\cr
\cos A &= \cos t \cos S - \sin t \sin S
\end{align}$$
and for infinitesimal $S$,
$$\begin{align\*}
\sin S &= S - \frac{1}{6} S^3 + \frac{1}{5!} S^5 + \dots,
\cr
\cos S &= 1 - \frac{1}{2} S^2 + \frac{1}{4!}S^4 + \dots,
\end{align\*}$$
with convergence in the most trivial sense: each monomial occurs in only finitely many terms of the expansion, so you just collect terms. Then observe that there is an inverse series:
$$
\arcsin T = T + \frac{1}{6} T^3 + \frac{3}{40} T^5 + \dots
$$
with convergence in the same sense. Actually, for the surjectivity in this problem, it
may be more convenient to use one series $\arctan T$ rather than two series $\arcsin$ and $\arccos$. So: Given $X,T$ with $X^2+Y^2=1$ we claim there is $A$ with $\sin A = X, \cos A = Y$. We should take either $A = \arctan Y/X$ or that plus $\pi$, depending on the signs of $X$ and $Y$. 


This is getting to be too long for an answer...


