Graphs in groups or Groups in graphs : TIFR GS 2018

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.

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Hint 1

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.)

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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Vectors of prime length

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Given a prime number $p$ and let $\overline{v_1},\overline{v_2},\dotsc ,\overline{v_n}$ be $n$ distinct vectors of length $p$ with integer coordinates in an $\mathbb{R}^3$ Cartesian coordinate system. Suppose that for any $1\leqslant j<k\leqslant n$ , there exists an integer $0<\ell <p$ such that all three coordinates of $\overline{v_j} -\ell \cdot \overline{v_k}$ is divisible by $p$ . Prove that $n\leqslant 6$ .

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Kürschák Competition 2018[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.26.6" open="off"]Number Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.26.6" open="off"]Hard[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.22.7" open="off"][/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]

Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.26.6"]Prove that any two vectors are either perpendicular or colinear.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.26.6"]

Note that $latex p^2| |\overline{v_j}-\ell\cdot\overline{v_k}|^2$. This implies that $latex p^2|\langle \overline{v_j},\overline{v_k}\rangle$. Also, from the Cauchy-Schwarz inequality we get $latex |\overline{v_j}|\cdot |\overline{v_k}|\ge |\langle\overline{v_j}.\overline{v_k}\rangle|$ hence $latex -|\overline{v_j}|\cdot |\overline{v_k}|\le\langle\overline{v_j}.\overline{v_k}\rangle\le |\overline{v_j}|\cdot |\overline{v_k}|$. Also, $latex | \overline{v_j}|\cdot |\overline{v_k}|=p^2$. As the dot product is also divisible by $latex p^2$, it has to be equal to $latex \pm p^2$ or 0. It cannot be $latex p^2$ because the vectors are distinct, hence it is either $latex -p^2$ or 0. Thus the two vectors are either perpendicular or colinear (adding to 0).

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Let us plot the vectors in $latex \mathbb{R}^3$ and identify them with their tips (as the tails are at the origin). We join two tips by a straight line segment if they correspond to perpendicular vectors. Show that the number of straight lines is at least $latex \frac{n(n-2)}{2}$. Also prove that, there do not exist 4 vectors such that every possible pair of tips is joined by a line segment.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.26.6"]From the last assertion, show that the number of straight line segments in the figure is at most $latex \frac{n^2}{3}$ (this is a special case of a result called Turan's theorem). Thus, $latex \frac{n^2}{3}\ge \frac{n(n-2)}{2}$. Hence $latex n\le 6$. [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.26.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]

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Understand the problem

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Connected Program at Cheenta

Similar Problems