= Solution
Use additive notation for this <finitely generated abelian group>. The change of generators $A=a+3d$, $e=c-b$, together with $b,d$, is invertible over the integers: $a=A-3d$ and $c=b+e$. The second relation gives $2e=0$. Subtracting it from the first gives $4A=0$, while the third becomes $4b+2e=0$, hence $4b=0$. The last relation becomes $8A+4b+4e=0$ and is consequently redundant. Conversely $4A=4b=2e=0$ imply every original relation, so no further relation has been introduced or lost. Thus the <cyclic group> decomposition is
$$
\boxed{G\cong\mathbb Z\times(\mathbb Z/4\mathbb Z)\times(\mathbb Z/4\mathbb Z)\times(\mathbb Z/2\mathbb Z).}
$$
The infinite <cyclic group> factor contributes only zero to an element killed by two. Each finite factor has two such elements, so the kernel of multiplication by two has $2^3=8$ elements. Removing the identity leaves
$$
\boxed{7\text{ elements of order }2.}
$$
The invertible generator change also supplies a direct presentation proof of the decomposition, without relying solely on a <Smith normal form> computation.
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