 Solution

In agglomerative hierarchical clustering, start with eight singleton groups. Single-linkage clustering uses the smallest distance between cross-pairs; complete-linkage clustering uses the largest. Updating these cluster distances after each merge, one valid order of the tied merges givesThe merges at height $0.6$ may be reordered; they give the same requested three-cluster cuts. For example, the single-link distance from $AC$ to $BD$ is $\min (1.2,1.4,0.6,0.8)=0.6$, while its complete-link distance is their maximum $1.4$. After forming $EG$, its single-link distance to $H$ is $\min (0.9,1.2)=0.9$. The complete-link distance between $EG$ and $FH$ is $\max (1.2,0.9,1.5,1.2)=1.5$. Finally the closest cross-pair between $ABCD$ and $EFGH$ is $CG$ at $1.9$, while the farthest is $BF$ at $3.8$.

At three groups the partitions are thereforeTwo groups, $\{A,B,C,D\}$ and $\{E,F,G,H\}$, are more strongly supported by the merge heights. For single linkage the three-to-two merge is at $1.0$, close to the preceding height $0.9$, whereas the two-to-one merge is delayed until $1.9$. For complete linkage the three-to-two merge at $1.5$ is close to $1.4$, whereas the final merge jumps to $3.8$. Both methods agree on the two-group partition but split its second group differently when forced to return three groups. A three-group choice is possible if required substantively, but is not particularly compelling from these dissimilarities alone.