## 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.

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

gx have different colours for each x ∈ G.

##### Source of the problem

TIFR GS 2018 Part A Problem 24

##### Topic

Abstract Algebra

##### Difficulty Level

Hard

##### Suggested Book

Abstract Algebra, Dummit and Foote

## Start with hints

Do you really need a hint? Try it first!

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.

- 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.

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