I'm not really sure where to begin with this problem. Any insight as to where to begin or what to look out for would be much appreciated!
Orders of ##S_3##
##|e|=1##
##|f|=3##
##|f^2|=2##
##|g|=2##
##|gf|=2##
##|gf^2|=3##
Orders of ##Z_2##
##|0|=1##
##|1|=2##
Orders of ##S_3 x Z_2##...
Clearly e ∈ N. If a, b ∈ N, say ##a^k = b^l = e##, for some k,l ∈ N, then ##(ab)^{kl} = (a^k )^l (b^l )^k = e^l e^k = e##; thus, ab ∈ N. Also, ##|a|=|a^{−1}|##, so ##a^{−1}## ∈ N. Thus, N is a subgroup. As G is abelian, it is normal. Take any c ∈ G. If, for some n ∈ N, we have ##(cN)^n = eN##...
2u + 3v + 3w = 2
+ 5v + 7w = 2
6u + 9v + 8w = 5
##\begin{bmatrix}
2 & 3 & 3 & 2 \\
0 & 5 & 7 & 2 \\
6 & 9 & 8 & 5
\end{bmatrix}##
We have been asked to use Jordan exchange to solve the above equations. Can someone please explain how to determine the values for r, s for the equations...
Can someone please explain H-decomposable graphs. I understand that if all the subgraphs ##H_1, H_2, ..., H_k## are isomorphic to the same graph H, then G is H-decomposable. What I don't understand is, what is H? I'm reading this as there is a subgraph H, that contains a family of subgraphs...
1-factor is the same as a perfect matching-A graph G has a perfect matching iff for ##S \in V(G), k_o(G-S) \leq |S|##. If G has odd order, then G has no 1-factor. A 1-factor is a 1-regular sub-graph of G.
Also, every bridgeless cubic graph contains a 1-factor as well as every cubic graph with at...
When I asked my instructor about it she said to use partite sets. Hence, |U| = |W| = 6. One vertex from W can dominate 3 vertices in U and one vertex in U can dominate 3 vertices in W. This leaves 2 vertices in each set U and W. These must be dominated by two vertices. Hence, ##\gamma \neq 3##.
##\gamma(G) \leq 4##
##Claim: \gamma(G) = 4##
Since ##\Delta(G) = 3## a vertex can dominate at most four vertices. There are eight vertices equivalent to ##\Delta(G)##, so these are dominated by two vertices. The remaining four vertices are of degree two which require two vertices to dominate...