Recent content by Robb

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##...

Much appreciated! I think I get it.

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...

THanks

Thanks, I don't suppose you would have an example of the graphs G and H, and possibly the partitions?

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...

I understand how to show a given graph does/does not contain a 1-factor but I'm not sure how to show existence (or the lack thereof). Please advise.

I thought it was ok. Thanks for the help..., again!

Right, I get what you're saying. Going to revisit.

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##.

FYI, attached is another option.

##\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...

This is perfect! Thanks so much. Makes sense. I didn't consider sending y to infinity. Good day, my friend!