= Solution
For a <finite group> $G$ and a <subgroup> $H$, <Lagrange's theorem> states
$$
|G|=[G:H]|H|.
$$
In particular $|H|$ divides $|G|$: the left <cosets> partition $G$, and multiplication by a representative is a <bijection> from $H$ to each <coset>. Apply this to the <cyclic subgroup> $\langle g\rangle$. Its size is the <order of a group element> $g$, so \b[every element order divides $|G|$].
Under the square condition, every <group> element is its own <inverse element>. Consequently
$$
gh=(gh)^{-1}=h^{-1}g^{-1}=hg
$$
for all $g,h$. This proves that a <group of exponent two is abelian>, including the trivial <group>.
For the fourth-power condition a counterexample to commutativity is the <quaternion group>
$$
\boxed{Q_8=\{\pm1,\pm i,\pm j,\pm k\}.}
$$
Here $i^2=j^2=k^2=-1$ and $ij=k=-ji$. Thus its six elements outside $\{\pm1\}$ have <order of a group element> four, while $1$ and $-1$ have orders one and two. All fourth powers are the identity, yet $ij\ne ji$.
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