How Cheenta works to ensure student success?

Explore the Back-Story

TIFR 2014 Problem 20 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.

The image is a front cover of a book named Abstract Algebra by David S. Dummit and Richard M. Foote. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program

Let (C) denote the cube ([-1,1]^3\subset \mathbb{R}^3 ). How many rotations are there in (\mathbb{R}^3) which take (C) to itself?

Let us label the six faces of the cube by (F_1,F_2,...,F_6).

Let (G) be the set consisting of all the rotations of (\mathbb{R}^3) which take (C) to itself. This set forms a group under the binary operation composition. In fact, (G) is a subgroup of the group of isometries of (\mathbb{R}^3). The identity is the identity transformation (rotation by 0 degree).

This group (G) acts on the set {(F_1,...F_6)}. The action is the most obvious one. A face (F_i) goes to face (F_j) under any rotation that preserves (C). That is, every rotation is a permutation of the faces (F_1,...,F_6). Said in other words: (\sigma . F_i = F_j) is the action where the face (F_i) goes to (F_j) under the rotation (\sigma).

We recall the **Orbit-Stabilizer theorem**:

( |G|=|orbit(x)||stab(x)| ). Where (G) acts on the set (X) and (x\in X).

Where (orbit(x)={ g.x | g\in G}) and (stab(x)={g\in G | g.x=x } ) is the stabilizer of (x). Here "." denotes the action.

In this context, we have our group (G) as described above. And (X={F_1,...,F_6}).

In order to apply the orbit-stabilizer theorem, we look at (orbit(F_1)) and (stab(F_1)).

We can always rotate the cube in order that the face (F_1) goes to the face (F_2,..,F_6) and also the zero-rotation takes (F_1) to (F_1). Therefore (orbit(F_1)={F_1,...,F_6}).

What is the stabilizer of (F_1)? Well, the only rotations which fix the face (F_1) are the (0,\pi/2,2\pi/2,3\pi/2) rotations in the plane of (F_1) (i.e, having axis perpendicular to (F_1). Hence (|stab(F_1)|=4).

By orbit-stabilizer theorem we now have (|G|=6\times 4=24 ).

**What is this topic:**Abstract Algebra**What are some of the associated concept:**Orbit-Stabilizer Theorem**Book Suggestions:**Abstract Algebra by David S. Dummit and Richard M. Foote.

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