In short, if we consider the group of symmetries of a regular octahedron, we see (or at least, the author of "Groups, Graphs and Trees" saw...) that the group is isomoprhic to(adsbygoogle = window.adsbygoogle || []).push({}); Z[tex]\otimes[/tex]_{2}Z[tex]\otimes[/tex]_{2}Z[tex]\otimes[/tex]_{2}S- particularly since if we break up the vertices into 3 groups of front-back, top-bottom and left-right we get the first three factors and the second factor is obtained by further permutation. But my question is how does an element of a direct product act on an element of a graph? If we take a vertex v in the graph, since all symmetries commute here, if we apply a symmetry h in the direct product to v, we are thus applying 4 symmetries each in one of the factors of the direct product. So is the vertex taken to "coordinates" of whatever the direct product indicates similar to the way we consider 3 dimensional coordinates as in if we have the coordinates (1,2,3) we could move 2 in the y direction then 1 in the x and 3 in the z or 3 in the z direction THEN 2 in the y then 1 in the x etc. and thus the same process for the location of the vertex v after h is applied. Thanks in advance for your help - I know I probably rambled a bit..._{3}

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# Validity of Direct Product Structure of Symmetry Group

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