For each of the two pairs of graphs below, determine whether they are isomorphic.
Pair 1:
Pair 2:
For convenience, let the graphs of the pairs, listed from left to right, be $G_1, G_2,H_1, $ and $H_2$, where $G$ and $H$ signify graphs of different pairs.
There are several conditions for isomorphic graphs: they must have the same degree sequence, number of edges, number of vertices, number of cycles, and vertices of the same degree must be mapped to each other.
However, these two pairs of graphs have vertices that are all the same degree. Also, from what I've learned, if I can carefully select which vertices to remove and obtain two nonisomorphic graphs by doing so, this'll suffice to show that the two graphs are not isomorphic.
I think that $G_1$ and $G_2$ are not isomorphic, but I'm not really sure how to show this. I think the possibilities for $3$ would be $c$ or $g$ since isomorphisms preserve adjacencies and two of the vertices that $3$ is adjacent to in $G_1$, specifically $5$ and $7$, are separated by a vertex. As well, $4$ must be mapped to $c$ or $g$. By a similar argument, $1$ can only be mapped to $a$ or $b$, and the other must be $2$. So we consider the cases where $1$ is mapped to $a$ or $b$ and where $3$ is mapped to $g$ or $h$. Consider when $3$ is mapped to $c$. Then $4$ is mapped to $g$. This is impossible because the vertices adjacent to $g$ and $c$ that are INSIDE the octagon in $G_2$ are not all adjacent whereas they are in $G_1$. We see a similar contradiction if $3$ is mapped to $g$. Now we consider when $1$ is mapped to $a$ or $b$. By symmetry, we may assume WLOG that $1$ is mapped to $a$. Then $f$ or $g$ must be $6$ or $5$. Also, $7$ and $8$ must be $e$ or $d$. But $8$ is not adjacent to $6$ or $5$, and so $8$ cannot be $e$ or $d$, a contradiction. Thus, $G_1$ and $G_2$ are not isomorphic.
Now we claim that $H_1$ and $H_2$ are not isomorphic. For a contradiction, suppose they are isomorphic. We see that $7$ and $8$ must be mapped to $A$ and $H$ respectively. By symmetry, we may assume WLOG that $7$ is mapped to $A$. Then $6$ and $5$ must be mapped to $B$ and $G$ respectively as those are adjacent to $7$. But since $B$ and $G$ are adjacent to $H$ and in $H_1$ $8$ is not adjacent to $5$, this is impossible.
I'm quite sure the graphs are not isomorphic, but I'm not sure if my justification is correct.