# Is union and intersection of two graphs preserve graph isomorphism?

Suppose $$G_1 = (V_1,E_1)$$ and $$G_2 = (V_2,E_2)$$ be two graphs, then union of two graphs is $$G_1\cup G_2 = (V_1\cup V_2, E_1\cup E_2)$$ and $$G_1\cap G_2 = (V_1\cap V_2, E_1\cap E_2)$$. Now, it is given that $$G_1$$ is isomorphic to $$G_2$$, denoted as $$G_1\simeq G_2$$.

My question is (very basic question, not found in any graph theory text book),

Suppose we consider two different graph $$G_1^{'}= (V_1^{'},E_1^{'})$$ and $$G_2^{'}= (V_2^{'},E_2^{'})$$ and $$G_1\simeq G_1^{'}$$ and $$G_2\simeq G_2^{'}$$. Then can I say that,

• $$G_1\cup G_2 \simeq G_1^{'}\cup G_2^{'}$$,
• $$G_1\cap G_2 \simeq G_1^{'}\cap G_2^{'}$$.

If not, is there any counter example.

Small Modification

I make a small assumption, that is at which position of the graph, common vertex appear that does not change for $$G_i^{‘}$$ for $$i=1,2$$ also. Does this assumption helps in preserving the set operation?

There's a basic answer to this basic question. For any algebraic structure (graphs, groups, fields, ...) any theorem you can state and prove that uses only the definitions of that structure (and things derived from those definitions) will be true when you replace any objects by isomorphic copies.

You can think of isomorphisms as simple renamings of elements that can have no consequences for their properties.

You should be able to write a proof for the particular questions you ask, starting from the definitions.

See What is an Homomorphism/Isomorphism "Saying"? (which is close to a duplicate).

• Thank you for the answer. But I unable to show that using the definition stated in en.wikipedia.org/wiki/Graph_isomorphism
– ann
Commented Jun 17 at 14:45
• Id $f_1$ maps $V(G_1)$ to $V(G_1')$ and $f_2$ does the same for $V(G_2)$ how will you define a map between the union of the vertex sets? Just think about them as sets. Then deal with the graph structure next. Commented Jun 17 at 15:18
• It is unclear what you are saying. You seem to be suggesting the answer to the question OP is asking is positive, but it is in fact not (unless you are assuming that the "union" of two graphs is always a disjoint union, and that the "intersection" is always empty). Commented Jun 18 at 13:37
• @xxxxxxxxx I think you are right. I think you should post a counterexample as an answer to the question. Commented Jun 18 at 14:04