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Problem (Bonus) (The Point Group Of A Cube)
Consider a cube with vertices located at the 8 points $(\pm 1, 1,1)$, $(\pm 1, 1,-1)$, $(\pm 1, -1,1)$, $(\pm 1, -1,-1)$, and edges that connect each vertex to the three nearest points. For example, vertex $(1,1,1)$ has three edges, one connecting to each of $(1,1,-1)$, $(1,-1,1)$, and $(-1,1,1)$. We can view this as a graph whose vertices are these 8 points, along with the 12 edges. We will let $O_h$ represent the set of automorphisms of the graph of the cube.
- How many elements are in $O_h$? Explain.
- We will say that a set $S$ of permutations of $X$ is independent if for each $s\in S$, we have $\span (S\setminus\{s\})\neq \span (S)$. In other words, every element in $S$ is not in the span of the other permutations in $S$. Find an independent set $S$ of elements in $O_h$ so that $\span(S) = O_h$.
- Use sage to draw the Cayley graph of $O_h$ using S. In addition show the Cayley graph of $O_h$ using $S\setminus\{s\}$ for each $s\in S$. The code provided below can do this for you, by adding and/or removing elements from $S$.
- Challenge: Find another set $S_2$ with a different number of elements than $S$, so that $S_2$ is also an independent set whose span is $O_h$.
S = ["(1,2,3,4)(5,6,7,8)","(1,5,8,4)(2,6,7,3)"] g = PermutationGroup(S) d = g.cayley_graph() print("There are " + str(order(g)) + " elements in the graph below.") #To see a readable list of the elements, uncomment the line below. #print(g.list()) d.show(color_by_label=True, vertex_size=0.03, vertex_labels=True) #We now make a graph of what happens if we remove one element. for s in S: T=S[:]; #This makes a copy of S T.remove(s); #from which we remove s without affecting S. gT=PermutationGroup(T) print("Removing " + s + " yields " + str(order(gT)) + " elements.") gT.cayley_graph().show(color_by_label=True, vertex_size=0.03, vertex_labels=True) #The graph you create may be difficult to read in 2D. #Uncomment the next line to see a 3D Cayley graph. Note, it may cause your browser to hang for a moment. #d.show3d(color_by_label=True, vertex_size=0.03, vertex_labels=False)
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