Cheenta

Academy for Gifted Students

How Cheenta works to ensure student success?

Explore the Back-Storycan colour the elements of G with two colours in such a way that x and

gx have different colours for each x ∈ G.

One needs to know the basics of Graph Theory to understand the solution.

- As noted Colouring is a fundamental topic in Graph Theory,so we need to convert the problem into a graph theory problem.
- Consider the elements of the group G as the vertices and consider edges between two elements say x and y if x=gy or \(x=(g^{-1})x\). Call graph G*.
- Check that x---y---z (x is adjacent to y and y is adjacent to z) iff \(y = (g^{-1})x , z = (g^{-1})y = (g^{-2})x\) or \(y = x , z = (g)y = (g^2)x\).We will assume left multiplication by g ( the proof for \(g^{-1}\) is exactly the same.)

- Now observe that we need to answer whether this graph G* is 2-colourable.
- There is an elementary result in graph theory characterizing the 2-colourable graphs.

Theorem 1 : A graph is 2-colourable iff it is bipartite.

Theorem 2: A graph is bipartite iff it has no odd-cycle.

Hint 3 :
- Thus Theorem 1 and Theorem 2 ⇒ G* is 2 -colourable iff it has no odd cycle.
- Now what does odd cycle mean in here in terms of group.
- A path of odd length means that \(x---gx---(g^2)x---...---(g^k)x\) in odd number of steps i.e. k is odd.

- A cycle of odd length means that \((g^k)x=x ⇒ g^k=1\).
- We are given that g is even ordered so it can only happen if k is even.
- Hence an odd cycle cannot exist and we can colour G* with 2 colours.

The answer is therefore True.

can colour the elements of G with two colours in such a way that x and

gx have different colours for each x ∈ G.

One needs to know the basics of Graph Theory to understand the solution.

- As noted Colouring is a fundamental topic in Graph Theory,so we need to convert the problem into a graph theory problem.
- Consider the elements of the group G as the vertices and consider edges between two elements say x and y if x=gy or \(x=(g^{-1})x\). Call graph G*.
- Check that x---y---z (x is adjacent to y and y is adjacent to z) iff \(y = (g^{-1})x , z = (g^{-1})y = (g^{-2})x\) or \(y = x , z = (g)y = (g^2)x\).We will assume left multiplication by g ( the proof for \(g^{-1}\) is exactly the same.)

- Now observe that we need to answer whether this graph G* is 2-colourable.
- There is an elementary result in graph theory characterizing the 2-colourable graphs.

Theorem 1 : A graph is 2-colourable iff it is bipartite.

Theorem 2: A graph is bipartite iff it has no odd-cycle.

Hint 3 :
- Thus Theorem 1 and Theorem 2 ⇒ G* is 2 -colourable iff it has no odd cycle.
- Now what does odd cycle mean in here in terms of group.
- A path of odd length means that \(x---gx---(g^2)x---...---(g^k)x\) in odd number of steps i.e. k is odd.

- A cycle of odd length means that \((g^k)x=x ⇒ g^k=1\).
- We are given that g is even ordered so it can only happen if k is even.
- Hence an odd cycle cannot exist and we can colour G* with 2 colours.

The answer is therefore True.

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIALAcademic Programs

Free Resources

Why Cheenta?

Online Live Classroom Programs

Online Self Paced Programs [*New]

Past Papers

More