Cyclic Groups & Subgroups : IIT 2018 Problem 1

Understand the problem

Which one of the following is TRUE? (A) $\Bbb Z_n$ is cyclic if and only if n is prime
(B) Every proper subgroup of $\Bbb Z_n$ is cyclic
(C) Every proper subgroup of $S_4$ is cyclic
(D) If every proper subgroup of a group is cyclic, then the group is cyclic.

Start with hints

Hint 1:

We will solve this question by the method of elimination. Observe that if n is prime then $\mathbb{Z}_n$ is obviously cyclic as any of the subgroup <a> has order either 1 or n by Lagrange's theorem. Now if the order is 1 then a=id. So choose a($\neq$e) $\in \mathbb{Z}_n$ then |<a>|=n and <a> $\subseteq$ $\mathbb{Z}_n$ $\Rightarrow$ <a>= $\mathbb{Z}_n$. The problem will occur with the converse see $\mathbb{Z}_6$ is cyclic but 6 is not prime. In general $\mathbb{Z}_n$ = <$\overline{1}$> is always cyclic no matter what n is!! so option (A) is false. Can you rule out option (C)

Hint 2:

Consider option (C) every proper subgroup of $S_4$ is cyclic. Consider { e , (12)(34) , (13)(24) , (14)(23) } = G Observe that this is a subgroup and |G|=4. Moreover o(g)=2 $\forall$ g($\neq$e) $\in$ G So G is not cyclic. Hence option (C) is not correct. Can you rule out option (D)?

Hint 3:

Consider $\mathbb{Z}_2$*$\mathbb{Z}_2$ which is also known as Klein's 4 group then it is not cyclic but all of it's proper subgroups are {0}*$\mathbb{Z}_2$ , $\mathbb{Z}_2$*{0} and {0}*{0} which are cyclic. Hence we can rule out option (D) as well.

Hint 4:

So option (B) is correct. Now let prove that H $\leq$ $\mathbb{Z}_n$ = {$\overline{0}$,$\overline{1}$,.....,$\overline{n-1}$}. By well ordering principle H has a minimal non zero element 'm'. Claim: H=<m> clearly <m> $\subset$ H. For any r $\in$ H by Euclid's algorithm we have r=km+d where 0 $\leq$ d < m which $\Rightarrow$ d=r-km $\in$ H If d $\neq$ 0 then d<m which is a contradiction So, d=0 $\Rightarrow$ r=km $\Rightarrow$ H=<m> and we are done

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Order of rings: TIFR GS 2018 Part B Problem 12

Understand the problem

The number of rings of order 4, up to isomorphism, is:
(a) 1
(b) 2
(c) 3
(d) 4.

Start with hints

Hint 1:

First, ask yourself how many groups are there of order 4.
the answer is simple => Z/4Z and Klein’s four group (K).

Hint 2 :

So intuitively there should be two rings with 4 elements.

Okay, I will give you two rings or same order=> (R,+,.), (R,+,*) . see that order of the rings are same but I have changed the multiplication, and

I define a*b=0 for all a,b in R. [(R,+,*) is called zero ring]

Hint 3:

Question: Prove that (R,+,.) and (R,+,*) are not isomorphic! (easy)
So, we had (Z/4Z,+,.) and (K,+,.) as our answers. But if you change the multiplication 
to “*” then there will be 4 different rings 
*upto isomorphism* right?
Hence the answer is 4.

Hint 4:

Bonus Problem:
Prove that there are only two non-ismorphic p-rings(ring with p elements) upto isomorphism.

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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.

Start with hints

Do you really need a hint? Try it first!

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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Are juniors countable if seniors are?: TIFR GS 2018 Part A Problem 21

Understand the problem

A countable group can have only countably many distinct subgroups.

Start with hints

Hint 1

This is a really artistic problem.

Pre-Solution Thoughts:

Lemma: A group is finite iff the number of subgroups of a group is finite.(Check!)

The lemma is a way to understand that an infinite group will have infinite number of subgroups.

Hint 2

We are considering here set of all subsets of a countable group G, which is a subset of the Power Set of G.

Given G is countable the Power Set of G is uncountable.

Now we know that $2^d$, where d is the set of natural numbers denotes the cardinality of the Power Set of G which is an uncountable set as it is bijective to the Real Numbers [Here d is not finite by the way, d is countably infinite].

So it is kinda intuitive that it may be uncountable.

First I took the group (Z,+), but we all know that the subgroups of Z are nZ only. This doesn’t solve our purpose.

So naturally the next choice was the group (Q.+) whose subgroup is (Z,+).

While understanding the subgroups of (Q.+), the question is solved.

Hint 3

We need to understand the subgroups of (Q.+).

Consider any rational number q and the subgroup qZ of (Q.+) generated by q.

What if we take two rational numbers?

For simplicity check the subgroup generated by {1/2 , 1/3}.

Prove that the subgroup generated by {1/2 , 1/3} is (1/2.3)Z.{Observe that co-prime property of 2 and 3 is playing an important role}.

Hint 4

Now what if we take three mutually coprime natural numbers say a,b,c and see the subgroup generated by { 1/a , 1/b ,1/c}. In fact for simplicity take a,b,c to be primes.

Observe that it is of the form (1/a.b.c)Z.

Hence for every finite subset of primes, we generate a distinct subgroup of (Q,+).

Hence the total number of subgroups contained all the subsets of primes, which has a bijection with the Real Numbers as mentioned above.

So the answer is False.

Exercise

Find all the subgroups of (Q,+) with proof.

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Diagonilazibility in triangular matrix: TIFR GS 2018 Part A Problem 20

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" hover_enabled="0" _i="1" _address="0.0.0.1"]Let A ∈ $M_n(\Bbb R)$ be upper triangular with all diagonal entries 1 such that $ A \neq I$. Then A is not diagonalizable. [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27.3" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="3.27.3" hover_enabled="0" _i="0" _address="0.1.0.0.0"]

TIFR GS 2018 Part A Problem 20[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27.3" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Linear Algebra[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27.3" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Hard[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27.3" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="on"]Linear Algebra; Hoffman and Kunze[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" 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" _i="1" _address="0.1.0.1"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" _i="1" _address="0.1.0.2.1" hover_enabled="0"]

Let us start by assuming A is diagonalizable. Let’s see what it means to say A is diagonalizable.

A is diagonalizable if there exists a basis of $\Bbb R^n$ consisting of eigenvalues of A.
Now as A is upper triangular the diagonal elements of A are the eigenvalues of A. So the only eigenvalue of A is 1.
Now what we can say about a linear operator A with a basis of eigenvectors and all the eigenvectors have eigenvalue 1?

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" _i="2" _address="0.1.0.2.2" hover_enabled="0"]

Now write any vector as a linear combination of eigenvectors of A.
Prove every vector under A is an eigenvector corresponding to eigenvalue 1.
Doesn’t it seem geometrically that A is identity i.e. keeps every vector the same?

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" _i="3" _address="0.1.0.2.3" hover_enabled="0"]

Yes, indeed. Prove it mathematically by taking the Euclidean Basis vectors $e_1,e_2,..,e_n$
Hence the only such matrix is I, so no non-identity upper triangular matrix fits this category.
So the answer is True.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" _i="4" _address="0.1.0.2.4" hover_enabled="0"]

Food for Thought:

What would have happened if the diagonal entries are not all equal?
If it is hard to see try to play with 2x2 matrices!
Prove that it can be diagonalized.

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The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/collegeprogram/" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark" _i="7" _address="0.1.0.7"][/et_pb_button][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="50px||50px" custom_padding="20px|20px|20px|20px" _i="8" _address="0.1.0.8"]

Multiplicative group from fields: TIFR GS 2018 Part A Problem 17

Understand the problem

The multiplicative group $F^*_7$ is isomorphic to a subgroup of the multiplicative group $F^*_{31}$.

Start with hints

Hint 1

We will write them as (Z/7Z)* and (Z/31Z)* respectively instead of the notations used.

Observe that (Z/7Z)* has order 6 and (Z/31Z)* has order 30.So there is a possibility that (Z/7Z)* is a subgroup of (Z/31Z)* by Lagrange’s Theorem.
So we need to go into the structure of the groups to solve this problem.Hence we proceed!
Let us investigate the group (Z/7Z)*.It consists of {1,2,3,4,5,6 mod 7}.Observe that 3 mod 7 generates the group.
So naturally the next question is that whether (Z/31Z)* rather is there any general result?

Hint 2

In fact the following theorem is true and describes the cyclicity (Z/nZ)* to some extent.
Theorem: If p is a prime then (Z/pZ)* is cyclic. (Check!) {Check the bonus question for the complete characterization of cyclicity of (Z/nZ)* done by Gauss.}

Hint 3

So (Z/7Z)* and (Z/31Z)* are cyclic groups of order 6 and 30 respectively with generators say A and B respectively.
Now take the element $B^5$.The following Lemma describes its order.
Lemma: If g is the generator of the cyclic group of order n. Then \(g^k\_ has order n/gcd(n,k).(Check !)
So $B^5$ has order 6 and hence it is isomorphic to (Z/7Z)*.
Hence the answer is True.

Hint 4

Bonus Problem:

Theorem: The group (Z/nZ)* is cyclic if and only if n is $1, 2, 4,p^k$ or 2.$p^k$, where p is an odd prime and k > 0. This was first proved by Gauss. (Wow!)

Solve and Salvage if Possible.

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Group with Quotient : TIFR GS 2018 Part A Problem 16

Understand the problem

Let G be a finite group with a normal subgroup H such that G/H has
order 7. Then $G \cong$ H × G/H.

Start with hints

Hint 1

This is also an interesting question. First of all we need to understand something in general.

If G is a finite group and H Δ G. So Consider the quotient group G/H.

Observe the following!

Lemma: If G ≈ H x G/H , then G/H is isomorphic to a normal subgroup of G. [Consider the projection homomorphism of G to (H,1) which contains G/H as the kernel.]
But in general G/H is not even a subgroup of G.

We will illustrate this by giving a simple example.

Hint 2

Naturally we took the group (Z,+) and we know all the subgroups of Z are nZ ,which are normal subgroups as Z is an abelian group.
Consider the quotient group Z/nZ.We know that this is not even isomorphic to a subgroup of Z.
Hence comes our counter-example.
G=Z, H=7Z. G/H =Z/7Z. But G is not isomorphic to 7Z x Z/7Z as Z/7Z is not isomorphic to a subgroup of Z.
Hence the answer is False.

Hint 3

But we will give an example where the given statement is also False.

Consider the Dihedral group on n elements $D_n$ as a subgroup of O(2) [The orthogonal group in $R^2$.] There is a homomorphism (Determinant) from O(2) → {-1,1},whose kernel is SO(2).
Hence consider the homomorphism from $D_n$ → {-1,1} formed by the composition of inclusion homomorphism and the determinant homomorphism.
Observe that the Kernel of the above defined homomorphism is the Rotation Group of angle 2 π /n and the Quotient Group is the Reflection Group around a specific line(?)[Which is essentially Z/2Z.]
But observe that Dn is not isomorphic to Rotation Group of angle 2 π /n x Z/2Z .[As there is an interaction between rotation and reflection. $ref.rot.ref=rot^{-1}$ .]

Hint 4

#Bonus

Prove that the finite subgroups of the group of rigid body motion are only

Rotation Group of Angle 2 π /n for all n in N.
Dihedral Group $D_n$

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Direct product of groups: TIFR 2018 Part A, Problem 9

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]Let G, H be finite groups. Then any subgroup of G × H is equal to A × B for some subgroups A<G and B<H[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]TATA INSTITUTE OF FUNDAMENTAL RESEARCH -GS-2018 (Mathematics) -Part A -Question 9[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Group Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Medium [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Abstract Algebra Dummit and Foote[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" 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" _i="1" _address="0.1.0.1"]

Start with hints

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]Consider K $\leq$ G x H . Now if ( a , b ) $\to$ K then $ a^{-1} , b^{-1}$ $\to$ K . Also if ( a , b) , ( p,q ) $\in$ K then ( ap , bq ) , ( pa , qb ) $\in$ K So , G $|_k$ = { g $\in$ G | ( g , h ) $\in$ K for some h $\in$ H } is a subgroup of G . Similarly, H $|_k$ $\leq$ H . [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]We have arrived to the conclusion that G$|_k$ $\leq$ G & H$|_k$ $\leq$ H . Now use this as a fact to guess the answer . Are you sure that G$|_k$ x H$|_k$ = K ? I mean that who confirms that K can be within as A X B [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]Yes this is true that for every g $\in$ G$|_K$ $\exists$ k s.t ( g , h ) $\in$ K . But G$|_K$ x H$|_K$ contain ( g , h ) $\forall h$ $\in$ H$|_K$ . This is certainly having a bigger expectation . [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]So , here is a nice counter example . Take G = $\mathbb{Z_2}$ & H = $\mathbb{Z_4}$ Consider a cyclic subgroup of G x H ; < ( 1 , 1 ) > = { ( 1 , 1 ) , ( 0, 2 ) , ( 1, 3) , (0,0) } Observe that G$|_k$ = G & H$|_k$ = H But < ( 1 , 1 ) > $\neq$ G x H Hence , the answer is false.[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.23.3" 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" _i="3" _address="0.1.0.3"]

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Real Symmetric matrix: TIFR 2018 Part A, Problem 6

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][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_padding="20px|20px|20px|20px"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.26.6" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]True or false: Let $latex A$ be a $latex 3 \times 3$ real symmetric matix s.t $latex A^6=I$. Then $latex A^2=I$[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.26.8" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.26.8"]TIFR 2018 Part A, Problem 6[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.26.8" open="off"]Linear Algebra[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.26.8" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.26.8" open="off"]Linear Algebra, Hoffman and Kunze[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" 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

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.26.8" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.26.8"]First investigate the properties of Real Symmetric Matrices.

The most important property of a real symmetric matrix A is encoded in “Spectral Decomposition” of A.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.26.8"]

If matrix $latex A$ then there exists $latex Q$ with $latex Q'Q=I$ such that $latex A = Q'BQ$,where $latex B$ is a diagonal matrix with diagonal entries being the eigenvalues of A which are real numbers.

Here assume that the eigen values are $latex a,b,c$.

We will use this spectral decomposition of A.

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.26.8"]

Observe that $latex A^6 = I \Rightarrow Q'(B^6)Q = I$.(Check!)

$latex Q'(B^6)Q=I \Rightarrow Q[Q'(B^6)Q]Q'= QQ'= I \Rightarrow B^6=I$ . [as Q is orthogonal]

$latex B^6$ is a diagonal matrix with diagonal entries real numbers raised to the power 6 i.e $latex a^6,b^6$ and $latex c^6$.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.26.8"]

Now hint 3 implies $latex a^6=1,b^6=1$ and $latex c^6=1$.

$latex a^2=1,b^2=1$ and $latex c^2=1$ as $latex a,b$ and $latex c$ are real numbers. This means $latex B^2=I$.

$latex A^2 = Q'(B^2)Q = Q'Q = I$.

Hence the given statement is TRUE.

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Connected Program at Cheenta

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