= Solution
Fix the <Riemann curvature tensor> convention
$$
R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z.
$$
With this convention the round sphere has positive <sectional curvature>. Write $R(X,Y,Z,W)=g(R(X,Y)Z,W)$. We use skew-symmetry in the first and last pairs, pair interchange $R(X,Y,Z,W)=R(Z,W,X,Y)$, and the first Bianchi identity. These are the standard curvature symmetries of the <Levi-Civita connection>.
For an <orthonormal basis> $e_1,\ldots,e_n$ of $T_pM$, the <Ricci curvature> is the trace
$$
\operatorname{Ric}_p(X,Y)=\sum_i g(R(e_i,X)Y,e_i).
$$
For a plane spanned by linearly independent vectors $U,V$, its <sectional curvature> is
$$
K(U,V)=\frac{g(R(U,V)V,U)}{g(U,U)g(V,V)-g(U,V)^2}.
$$
Both definitions are independent of the chosen basis. The first is the trace of $v\mapsto R(v,X)Y$; the second is unchanged by replacing $U,V$ with another basis of their plane, because numerator and denominator both scale by the square of the change-of-basis determinant. For $U=e_i,V=e_j$, $i\ne j$, the denominator is one. The terms with $i=j$ vanish, so
$$
\boxed{\operatorname{Scal}_p=\sum_j\operatorname{Ric}(e_j,e_j)=\sum_{i\ne j}K(e_i,e_j)=2\sum_{i<j}K(e_i,e_j)}.
$$
The ordered-pair sum counts every unoriented coordinate plane twice.
For $A_X(v)=R(X,v)X$, pair interchange immediately gives
$$
g(A_Xv,w)=R(X,v,X,w)=R(X,w,X,v)=g(v,A_Xw).
$$
Thus \b[$A_X$ is self-adjoint], including when $X=0$. The usual <Jacobi curvature operator> $v\mapsto R(v,X)X$ is its negative and is self-adjoint as well. Keeping these two signs distinct will also fix the <Jacobi field> equation below.
The <Bonnet-Myers theorem> states that a connected, complete $n$-dimensional <Riemannian manifold>, $n\ge2$, with $\operatorname{Ric}\ge(n-1)\kappa g$ for a constant $\kappa>0$ has diameter at most $\pi/\sqrt\kappa$ and is compact. Completeness is required; a pointwise positive Ricci tensor without a uniform positive lower bound would not suffice for this stated conclusion.
A <left-invariant metric> on a <Lie group> is a <Riemannian metric> for which every left translation $L_a$ is an <isometry>, equivalently $L_a^*g=g$. It is determined by an inner product on the <Lie algebra> $\mathfrak g=T_eG$. If the metric is also right-invariant, this inner product is invariant under the adjoint action; differentiating yields
$$
\langle[X,Y],Z\rangle+\langle Y,[X,Z]\rangle=0.
$$
Use left-invariant vector fields and the permitted formula $\nabla_XY=\tfrac12[X,Y]$. The Jacobi identity gives
$$
R(X,Y)Z=\frac14\bigl([X,[Y,Z]]-[Y,[X,Z]]\bigr)-\frac12[[X,Y],Z]
=-\frac14[[X,Y],Z].
$$
Adjoint skew-symmetry then implies
$$
g(R(X,Y)Y,X)=\frac14\|[X,Y]\|^2,\qquad
\operatorname{Ric}(X,X)=\frac14\sum_i\|[X,e_i]\|^2.
$$
This is the <Ricci curvature of a bi-invariant Riemannian metric>. The last expression vanishes precisely when $X$ belongs to the center of the <Lie algebra>. If that center is zero, it is positive for every nonzero $X$. Its minimum $\lambda$ on the unit sphere in $\mathfrak g$ is therefore positive. Left invariance transfers this bound to every point: $\operatorname{Ric}\ge\lambda g$.
Work in the identity component $G^0$, since $\pi_1(G,e)=\pi_1(G^0,e)$. It is compact and hence complete. Give its <universal cover> $\widetilde G^0$ the lifted metric. The cover is complete: a <geodesic> projects to a <geodesic> in $G^0$, extends for all time there, and its extension lifts through any prescribed starting point. The <local isometry> also preserves <Ricci curvature>, so the same uniform bound holds on the cover. Applying Bonnet-Myers with $\kappa=\lambda/(n-1)$ makes $\widetilde G^0$ compact. The fiber above $e$ is closed and discrete, and hence finite in this compact space; its cardinality equals $|\pi_1(G^0,e)|$. Consequently \b[$\pi_1(G,e)$ is finite]. This is the <finite fundamental group from a uniform positive Ricci bound>. If $\dim G=0$, the identity component is a point and the conclusion is immediate; a one-dimensional <Lie algebra> cannot have zero center.
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