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## 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.
##### Source of the problem
TIFR GS 2018 Part A Problem 24
Abstract Algebra
Hard
##### Suggested Book
Abstract Algebra, Dummit and Foote

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.

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