= Solution
{c}
\b[The exact orders printed in the question are inconsistent.] The source PDF says that the <orders of group elements> $a,b,c$ are $4,12,24$, respectively, and also imposes $3a+2b=0$. Multiplying this relation by $6$ and using $12b=0$ gives $18a=0$. Since $4a=0$ also holds, subtracting $4(4a)=0$ gives $2a=0$, contradicting the stated <order of a group element> $4$. Thus
$$
\boxed{\text{no group satisfies all the hypotheses literally as printed}}.
$$
The natural <abelian group presentation> interpretation is to impose the relations $4a=12b=24c=0$ and $3a+2b=0$, with no further relations, without requiring the orders of the images of the generators to stay $4,12,24$. Under this interpretation the calculation is as follows. Since $3a=-a$ follows from $4a=0$, the mixed relation gives $a=2b$. Consequently $8b=0$, and together with $12b=0$ this gives $4b=0$. Conversely, $a=2b$, $4b=0$, and $24c=0$ imply every original relation. Eliminating $a$ leaves two independent <cyclic groups>:
$$
\boxed{G_{\mathrm{pres}}\cong\mathbb Z_4\oplus\mathbb Z_{24},\qquad n=2,\quad d_1=4,\quad d_2=24}.
$$
To verify that these relations force no further identifications, map $a,b,c$ to $(2,0),(1,0),(0,1)$ in $\mathbb Z_4\oplus\mathbb Z_{24}$. All defining relations hold, and the inverse <group homomorphism> sends the two standard generators to $b,c$. These <group homomorphisms> are mutually inverse. The <divisibility> condition $4\mid24$ gives the requested <invariant factors of a finitely generated module> over $\mathbb Z$. In this presented <group>, $a,b,c$ actually have orders $2,4,24$.
Equivalently, the relation <matrix>
$$
\begin{pmatrix}4&0&0\\0&12&0\\0&0&24\\3&2&0\end{pmatrix}
$$
has nonzero <Smith normal form> entries $1,4,24$, with the trivial $\mathbb Z_1$ factor omitted. The displayed <direct sum of modules> is the answer to the presentation interpretation; the inconsistency of the literal exact-order formulation remains.
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