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## TIFR 2014 Problem 23 Solution – Homomorphisms from $S_n$

TIFR 2014 Problem 23 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.
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## Problem:True/False?

There exists an onto group homomorphism

A. from (S_5) to (S_4)

B. from (S_4) to (S_2)

C. from (S_5) to (\mathbb{Z}_5)

D. from (S_4) to (\mathbb{Z}_4)

## Discussion:

(S_2) is the permutation group on 2 letters. It has order 2, so it is isomorphic to (\mathbb{Z}_2).

And we know a group homomorphism from (S_n) to  (\mathbb{Z}_2), namely the signature map.

(\sigma \to 0 ) if ( \sigma ) is even

(\sigma \to 1 ) if ( \sigma ) is odd.

This map is onto. And it is a homomorphism. This is fairly well known fact so let us not prove it here.

So we know for sure that B is true.

How do we know that C is not true?

For that, suppose (\phi:S_5 \to \mathbb{Z}_5 ) be an onto group homomorphism. Apply first isomorphism theorem for groups, then (S_5/ker{\phi} ) is isomorphic to (\mathbb{Z}_5). That implies their orders are same. And (|G/H|=|G|/|H| ) so we get (|ker{\phi}| = 4! =24 ).

Now remember that (ker{\phi} ) is a normal subgroup of (S_5). Also, notice that any 4-cycle in (S_5) must go to the identity of (\mathbb{Z}_5) since the order of image must be divisible by the order of domain which in the case of 4-cycles are 4. So the image of any 4-cycle must have order 1,2 or 4 in (\mathbb{Z}_5). But out of these, only order 1 is possible in (\mathbb{Z}_5). Hence the image of any 4-cycle is the identity. So, the kernel must contain all the 4-cycles in (S_5). Now we count the number of 5-cycles in (S_5).

There are ({{5}\choose{4}} ) ways to choose 4 elements out of 1,2,3,4,5. Next, given any 4 elements, we can always write them in increasing order. Let’s say we chose 2,4,5,3. Then we write this as 2,3,4,5. Now to count the number of possible 4-cycles (distinct) we fix 2 in the first position and permute the rest. Each of these permutations will give a different 4-cycle. There are (3!) such permutations. So in total, there are ({{5}\choose{4}}\times 3! =30 ) 4-cycles in (S_5). This is more than our cardinality of (ker{\phi}). This is a contradiction.

So option C is false.

The same kind of argument will apply to option D as well. Here one can consider the 3 cycles in (S_4) and get a contradiction as above.

For option A, consider the 5-cycles in (S_5). They must map to identity (because of order reasons as above). And there are (4!=24) 5-cycles (start the cycles with 1, permute rest). But first isomorphism theorem will give the cardinality of the kernel as (\frac{5!}{4!}=5), a contradiction. This disproves A.

## Helpdesk

• What is this topic:Abstract Algebra
• What are some of the associated concept: Permutation Group, Homeomorphism
• Book Suggestions: Contemporary Abstract Algebra by Joseph A. Gallian
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## TIFR 2014 Problem 30 Solution – Number of Maps

TIFR 2014 Problem 30 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 Topics in Algebra by I.N.Herstein. This book is very useful for the preparation of TIFR Entrance.

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## PROBLEM:

How many maps (\phi: \mathbb{N} \cup  {0} \to \mathbb{N} \cup  {0}) are there satisfying (\phi(ab)=\phi(a)+\phi(b)) , for all (a,b\in \mathbb{N} \cup  {0}) ?

## Discussion:

Take (n\in \mathbb{N} \cup  {0} ).

By the given equation (\phi(n\times 0)=\phi(n)+\phi(0)).

This means (\phi(0)=\phi(n)+\phi(0)).

Oh! This means (\phi(n)=0). (n\in \mathbb{N} \cup  {0}) was taken arbitrarily. So…

(\phi(n)=0) for all (n\in \mathbb{N} \cup  {0} ).

There is only one such map.

## HELPDESK

• What is this topic:Algebra
• What are some of the associated concept: Number of Function
• Book Suggestions: Topics in Algebra by I.N.Herstein
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## TIFR 2014 Problem 28 Solution – Continuous Functions from Discrete Space

TIFR 2014 Problem 28 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 Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem:

Let (X) be a topological space such that every function $f: X \to \mathbb{R}$ is continuous. Then

A. (X) has the discrete topology.

B. (X) has the indiscrete topology.

C. (X) is compact.

D. (X) is not connected.

Discussion:

We know that if (Y) is a discrete space then any function (g: Y \to Z ) is continuous.

Option A asks whether the converse to this is true in the case that (Z= \mathbb{R} ).

To prove/disprove whether (X) has the discrete topology or not it is enough to prove whether every singleton set is open or not.

If we can show that for every (x\in X) there exists a function (f_x :X \to \mathbb{R}) such that (f_x^{-1} (-1,1) = {x} ) then we are done. Because we are given that (f_x) if exists must be continuous, and since ((-1,1)) is open in (\mathbb{R}) we will have the inverse image of it open in (X), so ({x} ) will be open in (X).

Now, this target is easy to handle. We define for each (x\in X)

(f_x (x) = 0 ) and (f_x (y) =2) for (y \neq x ).

This (f_x) satisfies our desired property. So (X) is discrete.

Taking (X= \mathbb{Z}) (for example) shows that (X) does not need to be indiscrete nor does it have to be compact.

Taking (X= {0} ) shows that (X) may be connected. Ofcourse if (X) has cardinality more than 1, it is not connected.

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Continuity, Discrete Space
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert
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## TIFR 2014 Problem 27 Solution -Homeomorphism of product space with mother space

TIFR 2014 Problem 27 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 Topology by James.R. Munkres. This book is very useful for the preparation of TIFR Entrance.

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## Problem:

(X) is a topological space with infinite cardinality which is homeomorphic to (X \times X). Then

A. (X) is not connected.

B. (X) is not compact.

C. (X) is not homeomorphic to a subset of (\mathbb{R})

D. none of the above

## Discussion:

Let us take (X=\mathbb{Z}). The topology is the subspace topology obtained from (\mathbb{R}). Then, in fact, (X) has the discrete topology. This is because for every (n\in \mathbb{Z}), ((n-1,n+1)) is an open subset of (\mathbb{R}) and hence ((n-1,n+1)\cap \mathbb{Z}={n} ) is open in (\mathbb{Z}).

We have a bijection from (\mathbb{Z} \times \mathbb{Z}) to (\mathbb{Z}). The product of two discrete spaces is again a discrete space because we have all sets of the form ({n} \times {m} ={(n,m)} ) as open sets. Now, any map from a discrete space is continuous. Because the inverse image of any open set is a subset of the space and every subset is open in that space. So in fact, the bijection that we know of from (\mathbb{Z} \times \mathbb{Z}) to (\mathbb{Z}) is continuous in both ways. Therefore in this setup (X \times X) is homeomorphic to (X). And also (X=\mathbb{Z}) is a subset of (\mathbb{R}). So this provides us with a counterexample for option C.

Next, we will take (X=\mathbb{Z}) to disprove A and B together. But now we will consider a different topology for (\mathbb{Z}). We will consider the indiscrete topology. That is, the only sets which are open in (X) are (X) and the empty set. Note firstly, that since we only have finitely many open sets to start with, (X) is compact. (X) is also connected because the only non-empty set which is both open and closed is (X) itself. We use the same bijection that we know of between (\mathbb{Z}) and (\mathbb{Z}\times \mathbb{Z}). Note that if a map has its co-domain set as an indiscrete space then that map must be continuous. This is because the only open sets are empty and whole set, and their inverse images are empty and the whole domain set respectively which are always open. Also, note that if (X) has the indiscrete topology then so does (X \times X). Because the open sets in (X \times X) are (\phi \times \phi = \phi , \phi \times X = \phi , X \times \phi = \phi, X \times X ). So both the spaces in consideration have indiscrete topology. So the bijection that we have is continuous in both ways. This, therefore, gives the counterexample for A and B.

## Helpdesk

• What is this topic: Topology
• What are some of the associated concept: Homeomorphism, indiscrete topology
• Book Suggestions: Topology by James.R. Munkres
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## TIFR 2014 Problem 26 Solution -Number of irreducible Polynomials

TIFR 2014 Problem 26 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 Linear Algebra Done Right by Sheldon Axler. This book is very useful for the preparation of TIFR Entrance.

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## Problem

The number of irreducible polynomials of the form (x^2+ax+b) , with (a,b) in the field (\mathbb{F}_7) of 7 elements is:

A. 7

B. 21

C. 35

D. 49

## Discussion:

First, what is the number of polynomials of the form (x^2+ax+b) in (\mathbb{F}_7) ? (a) has 7 choices and (b) has 7 choices. So we have a total of (7 \times 7 =49) monic degree 2 polynomials in (\mathbb{F}_7).

How many of these are reducible? If a monic polynomial of degree 2 is reducible then it must break into two factors in the form (p(x)=(x-\alpha)(x-\beta) ).

We want to count the number of polynomials of the form (p(x)=(x-\alpha)(x-\beta)). This will give us the number of reducible polynomials and we will subtract it from the total number to get the number of irreducible.

For the counting purposes, we must be careful not to have repetitions. For (\alpha=0) we can allow (\beta=0,1,..,6). For (\alpha=1) we can allow (\beta=1,…,6) (NOT 0 because that will again give ((x-0)(x-0)) which we have already counted in the (\alpha=0) situation ).

For (\alpha=2) we have (\beta=2,…,6) and so on… for (\alpha=6) we have (\beta=6) only.

So in total, we have (7+6+…+1=\frac{7\times 8}{2} = 28 ) polynomials which are reducible, has degree 2 and is monic.

Therefore, the number of irreducible polynomials are (49-28=21).

## HELPDESK

• What is this topic: Linear Algebra
• What are some of the associated concept: Monic polynomial
• Book Suggestions: Linear Algebra done Right by Sheldon Axler
Categories

## TIFR 2014 Problem 24 Solution – Order of Generated Subgroup

TIFR 2014 Problem 24 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 Contemporary Abstract Algebra by Joseph A. Gallian. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program

## Question:

Let (H_1, H_2) be two distinct subgroups of a finite group G, each of order 2. Let (H) be the smallest subgroup containing (H_1) and (H_2). Then order of (H) is

A. always 2

B. always 4

C. always 8

D. none of the above

## Discussion:

Let’s check out the simplest example where we suspect things might get wrong.

Take (S_3) as the group. It has 3 order 2 subgroups. If we take any two of those then the subgroup generated by (i.e the smallest subgroup containing) should have order greater than 2; so it must have order 3 or 6 and that itself leads us to conclude that the answer is none of the above.

For example take (<(1 2)>=H_1,<(13)>=H_2). Then (H) contains ( (1 2), (1 3), (1 2)(1 3)=(1 3 2), (1 3)(1 2)=(1 2 3)). Therefore (H) has atleast 4 elements, so it must have 6 elements i.e, (H=S_3).

## Helpdesk

• What is this topic:Abstract Algebra
• What are some of the associated concept: Order of a Subgroup
• Book Suggestions: Contemporary Abstract Algebra by Joseph A. Gallian
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## TIFR 2014 Problem 25 Solution – Identifying Isomorphic Groups

TIFR 2014 Problem 25 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 Topics in Algebra by I.N.Herstein. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program

## PROBLEM:TRUE/FALSE?

Which of the following groups are isomorphic?

A. (\mathbb{R}) and (\mathbb{C})

B. (\mathbb{R}^* ) and   (\mathbb{C}^*)

C. (S_3 \times \mathbb{Z}_4) and (S_4)

D. (\mathbb{Z}_2 \times \mathbb{Z}_2 ) and ( \mathbb{Z}_4 )

## DISCUSSION:

(\mathbb{R}) and (\mathbb{C}) are isomorphic:

Both these spaces are vector spaces ove (\mathbb{Q}).

We state a result without proof.

Theorem: Let (V) be an infinite dimensional vector space over a countable field (F). Then the dimension of (V) over (F) is (|V|) (the cardinality of (V) ).

This immediately tells us that (\mathbb{R}) has dimension (|R| = 2^{\aleph _0 } ).

Now the dimension of (\mathbb{C}) over (\mathbb{Q}) is (dim(\mathbb{R}^2) = 2 \times 2^{\aleph _0 } =2^{\aleph _0 } ).

What did we just show? We showed that (dim(\mathbb{C})=dim(\mathbb{R}) ) over (\mathbb{Q}).

Therefore they are isomorphic as vector spaces. So they are isomorphic as groups with respect to addition.

(\mathbb{R}^* ) and   (\mathbb{C}^*) are not isomorphic:

(\mathbb{C}^) has an element of order 4, namely (i) has order 4. If there was an isomorphism then the corresponding element in (\mathbb{R}^) will also have order 4. But there is no element of order 4 in (\mathbb{R}^*). Hence we conclude that these two groups are not isomorphic.

(S_3 \times \mathbb{Z}_4) and (S_4) are not isomorphic:

The order of (((1 2 3), [1] ) \in S_3 \times \mathbb{Z}_4) is 12. (S_4) does not contain an element of order 12.

(\mathbb{Z}_2 \times \mathbb{Z}_2 ) and ( \mathbb{Z}_4 ) are not isomorphic:

(\mathbb{Z}_2 \times \mathbb{Z}_2 ) is not cyclic and ( \mathbb{Z}_4 ) is cyclic. So they can not be isomorphic.

## HELPDESK

• What is this topic:Modern Algebra
• What are some of the associated concept: Isomorphism
• Book Suggestions: Topics in Algebra by I.N.Herstein
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## TIFR 2014 Problem 22 Solution -An application of Intermediate Value Theorem

TIFR 2015 Problem 7 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 Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem:

Let (f:\mathbb{R}^2 \to \mathbb{R} ) be a continuous map such that (f(x)=0 ) for only finitely many values of (x). Which of the following is true?

A. either (f(x) \le 0 )for all (x) or (f(x) \ge 0 ) for all (x).

B. the map (f) is onto

C. the map (f) is one-to-one

D. none of the above

## Discussion:

Let (f) be the map (f(x_1,x_2)=x_1^2 +x_2^2 ). Then (f) is zero at only (0,0). (f) is continuous because (x=(x_1,x_2) \to x_1 \to x_1^2 ) is continuous from (\mathbb{R}^2 \to \mathbb{R} \to \mathbb{R} ). i.e, each arrow is continuous. The first arrow is the projection map, and such maps are always continuous, and the second arrow is just squaring, which is continuous. And composition of continuous functions are continuous, so (x \to x_1^2 ) is continuous function from (\mathbb{R}^2 \to \mathbb{R}). Where here and henceforth (x=(x_1,x_2)\in \mathbb{R}^2 ).

Similar reasoning will show that (x \to x_2^2 ) is continuous function from (\mathbb{R}^2 \to \mathbb{R}).

Sum of continuous functions is continuous, so the map (x \to x_1^2+ x_2^2 ) is continuous function from (\mathbb{R}^2 \to \mathbb{R}).

This function (f) is not one-one since (f(1,0)=f(0,1)=1) and it is not onto since it only takes values in ([0,\infty ) ).

So we now are sure that B,C are false options.

We will prove A now.

Let (x=(x_1,x_2)) and (y=(y_1,y_2)) be two points in (\mathbb{R}^2) such that (f(x)>0) and (f(y)<0).

We will prove that this will imply infinitely many zeros in between (x) and (y). But wait a second… what does between mean in this context? For that we consider the paths between (x) and (y). Note that there are infinitely many paths between any two points in (\mathbb{R}^2). Further, we can in fact have infinitely many paths completely disjoint except for the initial and final points. We show that corresponding to each path (\alpha: [0,1] \to \mathbb{R}^2) which connects (x) and (y) we have a zero in the path. Since there are infinitely many disjoint paths, we get infinitely many distinct zeros for (f).

Now, (\alpha: [0,1] \to \mathbb{R}^2) is a continuous function , ( \alpha(0)=x ), (\alpha(1)=y).

Consider the composition (g=f o \alpha : [0,1] \to \mathbb{R}). (g) is continuous. g(0)=f(x)>0 and g(1)=f(y)<0.

Therefore by the intermediate value theorem, (g(c)=0) for some (c\in [0,1]).

That means, (f(\alpha(c))=0). And using the discussion above we get a contradiction.

This proved the option A.

## Chatuspathi

• What is this topic: Real Analysis
• What are some of the associated concept: Continuity,Intermediate Value Theorem
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert
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## TIFR 2014 Problem 21 Solution – Determining function

TIFR 2014 Problem 21 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 Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem:

Let (f:[0,1]\to [0,\infty ) ) be continuous. Suppose ( \int_{0}^{x} f(t)dt \ge f(x) ) for all (x\in [0,1]).

Then

A. no such function exists

B. there are infinitely many such functions

C. there is only one such function

D. there are exactly two such functions

Discussion:

Basically, the question is to find out how many such functions can exist.

Let (F(x)=\int_{0}^{x} f(t)dt ). That is, (F(x)) is the indefinite integral of (f).

We know from the fundamental theorem of calculus that:

(F'(x)=f(x)) .

So we have (F'(x) \ge F(x) ) for all (x\in [0,1] ).

We will manage this equation as we do in case of differential equations. Except that, we have an inequality here.

(F'(x)-F(x) \ge 0) for all (x\in [0,1] ).

Multiplying both sides by (e^{-x}) the inequality remain unchanged. This is because (e^{-x} >0 ).

(e^{-x}F'(x)-e^{-x}F(x) \ge 0) for all (x\in [0,1] ).

Now, ( (e^{-x}F(x))’ = e^{-x}F(x) -e^{-x}F'(x) ).

So we have (  (e^{-x}F(x))’ \le 0 ) for all (x\in [0,1] ).

Therefore, by taking integral from o to y and by using fundamental theorem of calculus:

(  e^{-y}F(y)-e^{0}F(0) \le 0 ) for all (y\in [0,1] ).

i.e, (  e^{-x}F(x) \le F(0) ) for all (x\in [0,1] ).

Also, note that by definition of (F), (F(0)=0). So we have

(  e^{-x}F(x) \le 0 ) for all (x\in [0,1] ).

But, since (e^{-x} >0 ) for all (x\in [0,1] ), we have (  F'(x) \le 0 )  for all (x\in [0,1] ).

So far, we have not used the fact that (f) is a non-negative function. Now we use it. Since (f(x) \ge 0) for all (x\in [0,1] ), therefore by monotonicity of the integral, (F) is an increasing function. This means (F'(x) \ge 0 ) for all (x\in [0,1] ).

By the two inequalities obtained above, we get (  F'(x) = 0 )  for all (x\in [0,1] ).

By the fundamental theorem (again!) we get (f(x)=F'(x)=0)  for all (x\in [0,1] ).

So there is only one such (f) namely the constant function (f=0).

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Fundamental theorem of calculus, Increasing Function
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert
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## TIFR 2014 Problem 20 Solution – Symmetries of Cube

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

## Problem

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?

## Discussion:

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 ).

## Helpdesk

• 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.