= Solution
In this setting a <formal group law> means a one-dimensional commutative series $F(X,Y)\in R[[X,Y]]$ with identity zero and
$$
F(X,0)=X,\quad F(0,Y)=Y,\quad F(X,Y)=F(Y,X),\quad F(F(X,Y),Z)=F(X,F(Y,Z)).
$$
It starts with $X+Y$; there is a unique inverse series $\iota(T)=-T+\cdots$ satisfying $F(T,\iota(T))=0$. For coefficients in $\mathbb Z$, the series and its inverse converge on $p\mathbb Z_p$, making this ideal a topological <abelian group> under $F$.
Set $a(T)=\partial_YF(T,0)\in1+T\mathbb Z[[T]]$, and write $a(T)^{-1}=\sum_{j\ge0}b_jT^j$. The <formal logarithm> is
$$
L(T)=\int_0^T\frac{dU}{a(U)}=T+\sum_{j\ge1}\frac{b_j}{j+1}T^{j+1}\in\mathbb Q[[T]].
$$
Differentiate associativity in the third variable at zero. It gives $a(F(X,Y))=\partial_YF(X,Y)a(Y)$, so $\partial_YL(F(X,Y))=L'(Y)$. Subtracting $L(Y)$ leaves a series independent of $Y$, whose value at zero is $L(X)$. Thus \b[$L(F(X,Y))=L(X)+L(Y)$].
For odd $p$, the logarithm converges on $I=p\mathbb Z_p$: the valuation of its degree-$n$ term is at least $n-v_p(n)$, which tends to infinity. Put $u(T)=L(T)-T$. For $x,y\in I$, factor $x^n-y^n$ to obtain
$$
v_p\left(\frac{b_{n-1}}n(x^n-y^n)\right)\ge v_p(x-y)+n-1-v_p(n)\ge v_p(x-y)+1
$$
for every $n\ge2$ and odd $p$. Hence $u$ is a contraction with constant at most $1/p$ and maps $I$ into $pI$. The equation $L(x)=z$ is equivalent to $x=z-u(x)$; the <contraction mapping theorem> gives a unique solution in $I$ for every $z\in I$. The same estimate shows $|L(x)-L(y)|_p=|x-y|_p$, so the logarithm is a topological group isomorphism. Dividing its values by $p$ gives
$$
\boxed{F(p\mathbb Z_p)\cong(p\mathbb Z_p,+)\cong(\mathbb Z_p,+)\qquad(p\text{ odd}).}
$$
This is the <deep logarithm subgroup of a formal group> argument, here with depth one.
For $p=2$, the degree-two estimate at depth one need not be strict. At depth two, however,
$$
2(n-1)-v_2(n)\ge1\qquad(n\ge2),
$$
so the same proof gives \b[$F(4\mathbb Z_2)\cong(4\mathbb Z_2,+)\cong\mathbb Z_2$]. This subgroup has index two in $F(2\mathbb Z_2)$, because $F(x,y)\equiv x+y\pmod4$ when $x,y\in2\mathbb Z_2$.
Indeed the <topological structure of a formal group on twice the 2-adic integers> has just two possibilities. Let $G=F(2\mathbb Z_2)$, let its index-two subgroup $H$ have topological generator $h$, and choose $v\notin H$. Write $2v=c h$, with $c\in\mathbb Z_2$, using group notation. If $c$ is odd, $2v$ generates $H$ and $v$ generates $G$, giving $G\cong\mathbb Z_2$. If $c$ is even, $v-(c/2)h$ has order two and lies outside $H$, giving $G\cong H\times\mathbb Z/2\mathbb Z$. The $\mathbb Z_2$ multiples are defined by continuity in this compact pro-two group.
For explicit contrasting examples, the <formal additive group> $F_a(X,Y)=X+Y$ gives $F_a(2\mathbb Z_2)\cong\mathbb Z_2$, which is a <torsion-free group>. The <formal multiplicative group> $F_m(X,Y)=X+Y+XY$ is identified by $x\mapsto1+x$ with $1+2\mathbb Z_2$. Its element $x=-2$ corresponds to $-1$ and has order two. Moreover
$$
1+2\mathbb Z_2=\{\pm1\}\times(1+4\mathbb Z_2),
$$
and the logarithm identifies the second factor with $4\mathbb Z_2$. \b[The two resulting groups are $\mathbb Z_2$ and $\mathbb Z/2\mathbb Z\times\mathbb Z_2$, so they are not isomorphic.]
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