Question

Suppose that \({{\rm{G}}_1}\) and \({H_1}\) are isomorphic and that\({{\rm{G}}_2}\) and \({H_2}\) are isomorphic. Prove or disprove that \({{\rm{G}}_1} \cup {G_2}\) and \({H_1} \cup {H_2}\) are isomorphic.

Short answer

\({G_1} \cup {G_2}\)and\({H_1} \cup {H_2}\)are not isomorphic.

Step-by-step solution

Identification of given data

The given data is:

• The isomorphic sets are\({G_1},\;{H_1}.\)

Concept/Significance of isomorphism

An isomorphism describes the relationship between two graphs that have the same number of graph vertices linked in the same way. Two graphs G and H with graph vertices\({{\bf{V}}_{\bf{n}}} = {\bf{1,2,}}......{\bf{,n}}\)are said to be isomorphic if there is a permutation p of\({{\bf{V}}_{\bf{n}}}\)such that\({\bf{u,v}}\)is in the set of a graph’s edge\({\bf{E}}\left( {\bf{G}} \right)\)if\({\bf{p}}\left( {\bf{u}} \right){\bf{,}}\;{\bf{p}}\left( {\bf{v}} \right)\)are in the set of a graph’s edge\({\bf{E}}\left( {\bf{H}} \right)\).

Determination of the isomorphism of either sets

Assume that \({G_1} \cup {G_2}\)and \({H_1} \cup {H_2}\)are isomorphic by considering an elementary situation.

Let us take the graphs\({G_1},\)\({G_2},\)and\({H_1},\)where each has a single vertex, say x and we have another graph\({H_2},\)which has a single vertex, say y.

The isomorphic graphs are\({G_1}\),\({H_1}\),\({G_2}\)and\({H_2}.\)But, the union of\({G_1}\)and\({G_2}\)is a graph with a single vertex. Whereas, the union of\({H_1}\)and\({H_2}\)is a graph with two vertices, which is a contradiction to\({G_1} \cup {G_2}\)and\({H_1} \cup {H_2}.\)

Thus,\({G_1} \cup {G_2}\)and \({H_1} \cup {H_2}\)are not isomorphic.