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 Apr10 comment Counting the number of graphs with a certain property I found something in the text that says if a graph has an odd cycle, then $\chi (G) \geq 3$. But how do I check all simple graphs with odd cycles to see which follow the property above? Apr3 comment How to show a graph is not Hamiltonian? Thank you very much for the clear answer! I am curious. Does a proof exist that does not require referencing the Petersen graph? (Even if it is the only graph satisfying such requirements) That is, is there some general criteria that would force a graph to be non-Hamiltonian based solely on the number of vertices/girth/regularness? Apr3 comment How to show a graph is not Hamiltonian? $v_G$ is the number of vertices of $G$. I have edited my post to add this explanation! Apr3 comment How to show a graph is not Hamiltonian? Oops, upon reading my post again I realize I forgot to add the property that $G$ was 3-regular. My bad! But nice counterexample for the original post. Mar20 comment Proofs involving subtrees of a tree Sorry, I didn't notice the typo. Indeed it was a nonempty intersection, and I've corrected the post. Mar13 comment Constructing a 3-regular graph with no 3-cycles No, there cannot, but I'm having trouble writing this "formally". Do I simply write that this is because $k+1$, say, would only have edges between $k$, $k+2$, and $k+(n+1)$, and so $k+n$ is not included? Mar13 comment Constructing a 3-regular graph with no 3-cycles Oops, yes, my bad! Feb6 comment Finding the order of elements in a group? Alright, so I picked up where you left off, and I found the following. $[4]^2 = [16]$, so $n = 2$ does not work. However, $[4]^{11} = [1]$, so $n = 11$ and since $11$ is least, then the order of $[4]$ is 11. Am I correct? Feb6 comment Finding the order of elements in a group? Thank you so much for the clarification! I think I understand this problem now. Feb6 comment Finding the order of elements in a group? Yes, that is the exact question. Sorry for any confusion, I thought I might have had to clarify some of my notation. Feb6 comment Finding the order of elements in a group? The first part of my question IS the question I am asking. I simply do not know how to begin, and I am trying to seek help as well as provide background info on what we have learned prior to this. Feb6 comment Finding the order of elements in a group? $Ker(f) = \{(x_1, x_2) | f(x_1) = f(x_2)\}$. Feb6 comment Finding the order of elements in a group? Sorry, I was trying to define as much as I can while still keeping things terse. "For any sets $X, Y$ and any function $f: X \to Y$, there exists an equivalence relation defined on $X$, $Ker(f)$. Also, $X / Ker(f) = \{[x] | x \in X\}$ where for $x \in X$, $[x] = \{y \in X | f(y) = f(x)\}$". Dec5 comment Orders of a symmetric group I see. It is much clearer to me now! Likewise, the total number of elements of order 5 is $\frac{5 \times 4 \times 3 \times 2 \times 1}{5} = 24$, yes? Thank you for the help! Nov21 comment Finding all partial order relations on a set Thanks for the source! It looks rather useful. I'll definitely remember it! Nov21 comment Finding all partial order relations on a set The 3 element set is much clearer to me now. Thank you very much for the detailed answer! I tried finding all Hasse diagrams for the 4 element set, but I can only think of 12 types. (I drew them by hand, so it is difficult to display them on here.) Is there some list of diagrams I can compare mine to, to see which ones I missed? Nov21 comment Finding all partial order relations on a set My bad, I just selected finite small sets off the top of my head. I suppose we could just then consider the 3 and 4 element sets? (Since I know the 0, 1, and 2 element sets are trivial.) Oct31 comment Finding the equivalence class of a relation Is that it? Thank you for the help! The answer wasn't immediately visible to me, but now I see what to do. Oct31 comment Proving a relation is an equivalence relation Thank you for the hint and the advice! I will definitely keep this in mind for when I work through future problems like this. Oct10 comment Permutations of a sequence Thank you for the clarification! I'm new to this, so I'm still sorting out the terminology. Your explanation makes this seem much easier than I had first thought!