Example 4.1.4.

Consider the graphs:

The graph G1 consisting of six vertices arranged in a triangle.  Starting with the bottom left vertex and traveling around the triangle clockwise, the vertices are labeled a, d, f, e, c, b.  Edges connect vertices at the corners of the triangle to vertices along their edges of the triangle.  These center vertices are also connected to each of the other center vertices.
The graph G2 of four vertices: a horizontal row of three, labeled a, b, c, and a vertex labeled d centered between and above a and b.  Edges connect a, b, and d (forming a triangle) and a fourth edge connects b to c.
The graph G3 of five vertices.  Vertices a, b, and c are aligned in a horizontal row.  Vertices d and f form a line with a in a line that slants up and to the right.  There are edges between a and d, between d and f, between d and b, and between b and c.
The graph G4 with five vertices: a, b, and c form a horizontal row, d and f line up with a in a line slanting up and the to the right.  The vertices form a triangle, and edges are arranged to fill in the outside border of the triangle.  Vertices d and b are also connected by and edge.

Here both \(G_2\) and \(G_3\) are subgraphs of \(G_1\text{.}\) But only \(G_2\) is an induced subgraph. Every edge in \(G_1\) that connects vertices in \(G_2\) is also an edge in \(G_2\text{.}\) In \(G_3\text{,}\) the edge \(\{a,b\}\) is in \(E_1\) but not \(E_3\text{,}\) even though vertices \(a\) and \(b\) are in \(V_3\text{.}\)

The graph \(G_4\) is NOT a subgraph of \(G_1\text{,}\) even though it looks like all we did is remove vertex \(e\text{.}\) The reason is that in \(E_4\) we have the edge \(\{c,f\}\) but this is not an element of \(E_1\text{,}\) so we don't have the required \(E_4 \subseteq E_1\text{.}\)

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