= Solution
{c}
The <Abelian group criterion for a product of two prime squares> applies because $11<19$ and $11\nmid19^2-1=360$. Here its <Sylow theorems> proof is especially short:
$$
n_{19}\mid121,\quad n_{19}\equiv1\pmod{19};\qquad n_{11}\mid361,\quad n_{11}\equiv1\pmod{11}.
$$
The candidates for $n_{19}$ are $1,11,121$, of which only $1$ has the required remainder. The candidates for $n_{11}$ are $1,19,361$, again leaving only $1$. Thus both <Sylow subgroups> $P,Q$ are <normal subgroups>. Their <group commutators> lie in $P\cap Q=\{1\}$, and $PQ=G$, so $G\cong P\times Q$. Both factors are <abelian groups of prime-square order> by (a).
The <Fundamental theorem of finitely generated abelian groups> gives precisely
$$
\boxed{\begin{gathered}
C_{121}\times C_{361},\qquad C_{121}\times C_{19}\times C_{19},\\
C_{11}\times C_{11}\times C_{361},\qquad C_{11}\times C_{11}\times C_{19}\times C_{19}.
\end{gathered}}
$$
These four <groups> are pairwise nonisomorphic: their characteristic <Sylow subgroups> differ in whether they contain an element of order $11^2$ or $19^2$. The first is also the <cyclic group> $C_{43681}$.
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