Understand the problem
Let G be a finite group and g ∈ G an element of even order. Then we
can colour the elements of G with two colours in such a way that x and
gx have different colours for each x ∈ G.Start with hints
Do you really need a hint? Try it first! Hint 1One 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.)
Hint 2- 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.
Hint 4 :- 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.
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Connected Program at Cheenta
College Mathematics Program
the higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.Similar Problems
Related
Understand the problem
Let G be a finite group and g ∈ G an element of even order. Then we
can colour the elements of G with two colours in such a way that x and
gx have different colours for each x ∈ G.Start with hints
Do you really need a hint? Try it first! Hint 1One 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.)
Hint 2- 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.
Hint 4 :- 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.
Watch the video
Connected Program at Cheenta
College Mathematics Program
the higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.Similar Problems
Related
