= Solution
For $X\in\mathfrak g=T_eG$, let $X^L$ be the <left-invariant vector field> with value $X$ at the identity. Its integral curve through the identity is the unique <one-parameter subgroup> $\gamma_X$ with $\gamma_X'(0)=X$. It exists for all real time: local existence followed by translation and the identity $\gamma_X(t+s)=\gamma_X(t)\gamma_X(s)$ extends it indefinitely. The <Exponential map of a Lie group> is
$$
\boxed{\exp X=\gamma_X(1),\qquad \exp(tX)=\gamma_X(t).}
$$
It is smooth and $d_0\exp=\operatorname{id}_{\mathfrak g}$, because differentiating $t\mapsto\exp(tX)$ at zero gives $X$. The <inverse function theorem> therefore makes it a local <diffeomorphism> at zero.
If $G$ is abelian, the product $t\mapsto\exp(tX)\exp(tY)$ is a <one-parameter subgroup> with initial tangent $X+Y$. Uniqueness gives $\exp(t(X+Y))=\exp(tX)\exp(tY)$, hence
$$
\boxed{\exp(X+Y)=\exp X\exp Y.}
$$
Conversely, if this additive <homomorphism> identity holds, all $\exp(tX)$ and $\exp(sY)$ commute. Differentiating their adjoint action shows $[X,Y]=0$. Alternatively, their commuting images contain an identity neighbourhood; that neighbourhood generates the <identity component of a Lie group>, so the <identity component> is abelian. Conversely, if the <Lie algebra> is abelian, its <left-invariant vector fields> commute, so their flows commute. Hence all <one-parameter subgroups> commute; their product with initial tangent $X+Y$ is the <one-parameter subgroup> for $X+Y$. The <Lie exponential map> is therefore additive, and its image generates an abelian identity component. Thus \b[the stated equivalence holds for connected $G$]. Without connectedness the exact assertion is that the <Lie exponential map> is a <homomorphism> precisely when $G^\circ$ is abelian. The converse implication to all of $G$ is false: in $O(2)$ the <Lie algebra> consists of $tJ$, where
$$
J=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad
\exp(tJ)=R_t.
$$
Rotations satisfy $R_{t+s}=R_tR_s$, but a reflection $S$ satisfies $SR_tS^{-1}=R_{-t}$, so the full <orthogonal group> is not abelian.
Now take a connected <abelian Lie group> of dimension $n$. Its <Lie exponential map> is a <homomorphism> whose image contains an identity neighbourhood, hence an open <subgroup>. A connected <group> has no proper open <subgroup>, so this map is onto. Its kernel $\Lambda$ is discrete, since it is injective on a neighbourhood of zero. The induced map is a bijective <local diffeomorphism> and therefore a <Lie group isomorphism>
$$
G\cong\mathbb R^n/\Lambda.
$$
Here is the needed classification of $\Lambda$. Let $r$ be the dimension of its real span. Choose $\lambda_1,\ldots,\lambda_r\in\Lambda$ that form a real basis of that span. Their <integer> span $\Lambda_0$ has finite index in $\Lambda$: subtract <integer> multiples of the $\lambda_i$ to put each <coset> representative in a compact parallelepiped. Only finitely many points of $\Lambda$ lie there, since discreteness at zero gives a uniform positive separation between distinct points. Thus $\Lambda$ is finitely generated and torsion-free, and the <structure theorem for finitely generated abelian groups> makes it free abelian of rank $r$. Its <integer> basis is a real basis of the span. A linear change of coordinates consequently sends $\Lambda$ to $\mathbb Z^r\times\{0\}$ in $\mathbb R^n$.
This proves the <classification of connected abelian Lie groups>:
$$
\boxed{G\cong(\mathbb R/\mathbb Z)^r\times\mathbb R^{n-r}
=\mathbb T^r\times\mathbb R^s,\qquad r+s=n.}
$$
The isomorphism depends on the choice of lattice basis and complementary <vector space>. Compactness is equivalent to $s=0$.
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