MMT-003 Solved Assignment 2023

IGNOU MMT-003 Solved Assignment 2023 | M.Sc. MACS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU MMT-003 Assignment Question Paper 2023

mmt-003-2023-ext-aa2d2d76-613f-4e79-9c79-c97616952506
  1. Which of the following statements are true? Give reasons for your answers. Marks will only be given for valid justification of your answers.
i) If G G GGG is a finite abelian group and p p ppp is a prime factor of o ( G ) o ( G ) o(G)o(G)o(G), then the number of Sylow p-subgroups of G G G\mathrm{G}G is a prime.
ii) The minimum polynomial of 5 1 / 3 5 1 / 3 5^(1//3)5^{1 / 3}51/3 over Q Q Q\mathbb{Q}Q is x 1 / 3 x 1 / 3 x^(1//3)x^{1 / 3}x1/3.
iii) Z m n Z m × Z h m , n Z Z m n Z m × Z h m , n Z Z_(mn)≃Z_(m)xxZ_(h)AAm,ninZ\mathbb{Z}_{\mathrm{mn}} \simeq \mathbb{Z}_{\mathrm{m}} \times \mathbb{Z}_{\mathrm{h}} \forall \mathrm{m}, \mathrm{n} \in \mathbb{Z}ZmnZm×Zhm,nZ
iv) If G G GGG is a finite group and m o ( G ) , m N m o ( G ) , m N m∣o(G),m inNm \mid o(G), m \in \mathbb{N}mo(G),mN, then G G GGG has a subgroup of order m m mmm.
v) If β 1 β 1 beta_(1)\beta_{1}β1 and β 2 β 2 beta_(2)\beta_{2}β2 are two 15 th 15 th  15^(“th “)15^{\text {th }}15th  roots of unity, then Q ( β 1 ) = Q ( β 2 ) Q β 1 = Q β 2 Q(beta_(1))=Q(beta_(2))\mathbb{Q}\left(\beta_{1}\right)=\mathbb{Q}\left(\beta_{2}\right)Q(β1)=Q(β2).
vi) There exists an extension field of Z 3 Z 3 Z_(3)\mathbb{Z}_{3}Z3 of order 25 .
vii) Every group of order 18 has a normal subgroup of order 2.
viii) If I I I\mathrm{I}I and J J J\mathrm{J}J are ideals of a ring R R R\mathrm{R}R, then I J = I J I J = I J IJ=InnJ\mathrm{IJ}=\mathrm{I} \cap \mathrm{J}IJ=IJ.
ix) If f : R S f : R S f:RrarrS\mathrm{f}: \mathrm{R} \rightarrow \mathrm{S}f:RS is a ring homomorphism and I I I\mathrm{I}I is an ideal of R R R\mathrm{R}R, then f ( I ) f ( I ) f(I)\mathrm{f}(\mathrm{I})f(I) is an ideal of S.
x x x\mathrm{x}x Every prime ideal of an integral domain is a maximal ideal.
  1. a) Let G G GGG be a group and let H G , K G , o ( H ) = o ( K ) = p H G , K G , o ( H ) = o ( K ) = p H <= G,K <= G,o(H)=o(K)=pH \leq G, K \leq G, o(H)=o(K)=pHG,KG,o(H)=o(K)=p, a prime. Show that either H K = { e } H K = { e } HnnK={e}\mathrm{H} \cap \mathrm{K}=\{\mathrm{e}\}HK={e} or H = K H = K H=K\mathrm{H}=\mathrm{K}H=K. Is this result still true if p p p\mathrm{p}p is not a prime? Justify your answer.
b) Let G G G\mathrm{G}G be the group of all rigid motions of a plane and S S S\mathrm{S}S be the set of all rectangles in the plane. Show that G G GGG acts on S . Also obtain the orbit and stabiliser of a square under this action.
c) Let G G GGG be a finite group and H H HHH be a normal subgroup of G G GGG. Prove that H = C x H = C x H=uuC_(x)H=\cup C_{x}H=Cx where the C x C x C_(x)C_{x}Cx are all the distinct conjugacy classes of G G GGG such that H C x H C x H nnC_(x)!=O/H \cap C_{x} \neq \varnothingHCx.
  1. a) Find the number of Sylow 5-subgroups, Sylow 7-subgroups and Sylow 2-subgroups A 5 A 5 A_(5)A_{5}A5 has.
b) Let G 1 G 1 G_(1)G_{1}G1 and G 2 G 2 G_(2)G_{2}G2 be finite groups such that p p ppp divides | G 1 | G 1 |G_(1)|\left|G_{1}\right||G1| and | G 2 | G 2 |G_(2)|\left|G_{2}\right||G2|. Prove that the Sylow p-subgroups of G 1 × G 2 G 1 × G 2 G_(1)xxG_(2)G_{1} \times G_{2}G1×G2 are precisely of the form P 1 × P 2 P 1 × P 2 P_(1)xxP_(2)P_{1} \times P_{2}P1×P2, where P 1 P 1 P_(1)P_{1}P1 and P 2 P 2 P_(2)P_{2}P2 are Sylow p-subgroups of G 1 G 1 G_(1)G_{1}G1 and G 2 G 2 G_(2)G_{2}G2, respectively.
  1. a) Write [ 1 1 0 2 0 3 5 5 1 ] 1 1 0 2 0 3 5 5 1 [[1,-1,0],[2,0,3],[5,5,1]]\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 0 & 3 \\ 5 & 5 & 1\end{array}\right][110203551] as a product of O ( 3 ) O ( 3 ) O(3)\mathrm{O}(3)O(3) and an element of B 3 ( R ) B 3 ( R ) B_(3)(R)B_{3}(\mathbb{R})B3(R).
b) Show that S U ( 2 ) S U ( 2 ) SU(2)\mathrm{SU}(2)SU(2) and S 3 S 3 S^(3)\mathrm{S}^{3}S3 are structurally the same.
  1. a) Find all the possible abelian groups, up to isomorphism, of order 900 .
b) Construct the free group on the set { α , β , γ } { α , β , γ } {alpha,beta,gamma}\{\alpha, \beta, \gamma\}{α,β,γ}. Further, check if it is a free abelian group or not.
  1. a) Check whether or not the ring R = Z 3 [ x ] / < x 6 1 > R = Z 3 [ x ] / < x 6 1 > R=Z_(3)[x]// < x^(6)-1 >\mathrm{R}=\mathbb{Z}_{3}[\mathrm{x}] /<\mathrm{x}^{6}-1>R=Z3[x]/<x61>
    i) is finite;
    ii) has zero divisors;
    iii) has nilpotent elements.
b) Give two distinct rings whose quotient field is { a + i b a , b R } { a + i b a , b R } {a+ib∣a,b inR}\{a+i b \mid a, b \in \mathbb{R}\}{a+iba,bR}. Justify your answer.
c) Evaluate the following Legendre Symbols:
i) ( 139 431 ) 139 431 ((139)/(431))\left(\frac{139}{431}\right)(139431)
ii) ( 149 439 ) 149 439 ((149)/(439))\left(\frac{149}{439}\right)(149439)
d) Check whether 9782957210008 is a valid ISBN number.
  1. a) Check whether or not Z [ 7 ] Z [ 7 ] Z[sqrt(-7)]\mathbb{Z}[\sqrt{-7}]Z[7] is a Euclidean domain.
b) Use the division algorithm to find the inverse of 18 ¯ 18 ¯ bar(18)\overline{18}18¯ in Z 35 Z 35 Z_(35)\mathbb{Z}_{35}Z35.
  1. i) Let G G GGG be a group of automorphisms of a field K K KKK. Is the fixed field K G K G K^(G)K^{G}KG a subfield of K K K\mathrm{K}K ? Why, or why not?
ii) Find K G K G K^(G)K^{G}KG, where K = Q ( i , 3 ) , G = G ( K / Q ) K = Q ( i , 3 ) , G = G ( K / Q ) K=Q(i,sqrt3),G=G(K//Q)K=\mathbb{Q}(i, \sqrt{3}), G=G(K / \mathbb{Q})K=Q(i,3),G=G(K/Q).
\(sin^2\left(\frac{\theta }{2}\right)=\frac{1-cos\:\theta }{2}\)

MMT-003 Sample Solution 2023

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MMT-003 Solved Assignment 2023

Question:-01

  1. Which of the following statements are true? Give reasons for your answers. Marks will only be given for valid justification of your answers.
    i) If G G G\mathrm{G}G is a finite abelian group and p p p\mathrm{p}p is a prime factor of o ( G ) o ( G ) o(G)\mathrm{o}(\mathrm{G})o(G), then the number of Sylow p-subgroups of G G G\mathrm{G}G is a prime.
Answer:
The statement “If G G GGG is a finite abelian group and p p ppp is a prime factor of o ( G ) o ( G ) o(G)o(G)o(G), then the number of Sylow p p ppp-subgroups of G G GGG is a prime” is false.

Justification:

In a finite abelian group G G GGG, every Sylow p p ppp-subgroup is normal. Therefore, all Sylow p p ppp-subgroups are conjugate to each other, and they are all isomorphic. In an abelian group, the number of Sylow p p ppp-subgroups is given by n p n p n_(p)n_pnp, and according to Sylow’s Third Theorem, n p n p n_(p)n_pnp divides | G | | G | |G||G||G| and n p 1 mod p n p 1 mod p n_(p)-=1modpn_p \equiv 1 \mod pnp1modp.
However, there is no requirement that n p n p n_(p)n_pnp must be a prime number.

Counterexample:

Consider the abelian group G = Z 4 × Z 2 G = Z 4 × Z 2 G=Z_(4)xxZ_(2)G = \mathbb{Z}_4 \times \mathbb{Z}_2G=Z4×Z2, where Z n Z n Z_(n)\mathbb{Z}_nZn denotes the integers modulo n n nnn. The order of G G GGG is o ( G ) = 4 × 2 = 8 o ( G ) = 4 × 2 = 8 o(G)=4xx2=8o(G) = 4 \times 2 = 8o(G)=4×2=8.
The prime factors of o ( G ) o ( G ) o(G)o(G)o(G) are 2 2 222 and 2 2 222 (i.e., 8 = 2 3 8 = 2 3 8=2^(3)8 = 2^38=23).
The Sylow 2 2 222-subgroups of G G GGG are generated by ( 2 , 0 ) ( 2 , 0 ) (2,0)(2, 0)(2,0), ( 0 , 1 ) ( 0 , 1 ) (0,1)(0, 1)(0,1), and ( 2 , 1 ) ( 2 , 1 ) (2,1)(2, 1)(2,1). So, there are 3 3 333 Sylow 2 2 222-subgroups, and 3 3 333 is not a prime divisor of 8 8 888.
Thus, the statement is false, as demonstrated by this counterexample.

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ii) The minimum polynomial of 5 1 / 3 5 1 / 3 5^(1//3)5^{1 / 3}51/3 over Q Q Q\mathbb{Q}Q is x 1 / 3 x 1 / 3 x^(1//3)x^{1 / 3}x1/3.
Answer:
The statement “The minimum polynomial of 5 1 / 3 5 1 / 3 5^(1//3)5^{1/3}51/3 over Q Q Q\mathbb{Q}Q is x 1 / 3 x 1 / 3 x^(1//3)x^{1/3}x1/3” is false.

Justification:

The minimum polynomial of an algebraic number α α alpha\alphaα over a field F F FFF is the unique monic polynomial f ( x ) f ( x ) f(x)f(x)f(x) of least degree such that f ( α ) = 0 f ( α ) = 0 f(alpha)=0f(\alpha) = 0f(α)=0 and f ( x ) f ( x ) f(x)f(x)f(x) has coefficients in F F FFF.
For 5 1 / 3 5 1 / 3 5^(1//3)5^{1/3}51/3, the minimum polynomial over Q Q Q\mathbb{Q}Q is x 3 5 x 3 5 x^(3)-5x^3 – 5x35, not x 1 / 3 x 1 / 3 x^(1//3)x^{1/3}x1/3.
  1. x 3 5 x 3 5 x^(3)-5x^3 – 5x35 is a polynomial with coefficients in Q Q Q\mathbb{Q}Q.
  2. ( 5 1 / 3 ) 3 5 = 5 5 = 0 ( 5 1 / 3 ) 3 5 = 5 5 = 0 (5^(1//3))^(3)-5=5-5=0(5^{1/3})^3 – 5 = 5 – 5 = 0(51/3)35=55=0, so 5 1 / 3 5 1 / 3 5^(1//3)5^{1/3}51/3 is a root of x 3 5 x 3 5 x^(3)-5x^3 – 5x35.
  3. x 3 5 x 3 5 x^(3)-5x^3 – 5x35 is irreducible over Q Q Q\mathbb{Q}Q, which means it has the least degree among all polynomials that have 5 1 / 3 5 1 / 3 5^(1//3)5^{1/3}51/3 as a root and have coefficients in Q Q Q\mathbb{Q}Q.
Moreover, x 1 / 3 x 1 / 3 x^(1//3)x^{1/3}x1/3 is not even a polynomial over Q Q Q\mathbb{Q}Q because its exponent 1 / 3 1 / 3 1//31/31/3 is not a non-negative integer.
Thus, the statement is false.

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iii) Z m n Z m × Z n m , n Z Z m n Z m × Z n m , n Z Z_(mn)≃Z_(m)xxZ_(n)AAm,ninZ\mathbb{Z}_{\mathrm{mn}} \simeq \mathbb{Z}_{\mathrm{m}} \times \mathbb{Z}_{\mathrm{n}} \forall \mathrm{m}, \mathrm{n} \in \mathbb{Z}ZmnZm×Znm,nZ
Answer:
The statement “ Z m n Z m × Z n m , n Z Z m n Z m × Z n m , n Z Z_(mn)≃Z_(m)xxZ_(n)AA m,n inZ\mathbb{Z}_{mn} \simeq \mathbb{Z}_{m} \times \mathbb{Z}_{n} \forall m, n \in \mathbb{Z}ZmnZm×Znm,nZ” is false.

Justification:

The isomorphism Z m n Z m × Z n Z m n Z m × Z n Z_(mn)≃Z_(m)xxZ_(n)\mathbb{Z}_{mn} \simeq \mathbb{Z}_{m} \times \mathbb{Z}_{n}ZmnZm×Zn holds if and only if m m mmm and n n nnn are coprime (i.e., gcd ( m , n ) = 1 gcd ( m , n ) = 1 gcd(m,n)=1\gcd(m, n) = 1gcd(m,n)=1).

Counterexample:

Let m = 4 m = 4 m=4m = 4m=4 and n = 2 n = 2 n=2n = 2n=2. Then Z 4 × 2 = Z 8 Z 4 × 2 = Z 8 Z_(4xx2)=Z_(8)\mathbb{Z}_{4 \times 2} = \mathbb{Z}_8Z4×2=Z8.
However, Z 4 × Z 2 Z 4 × Z 2 Z_(4)xxZ_(2)\mathbb{Z}_4 \times \mathbb{Z}_2Z4×Z2 is not isomorphic to Z 8 Z 8 Z_(8)\mathbb{Z}_8Z8.
  1. Z 8 Z 8 Z_(8)\mathbb{Z}_8Z8 has elements [ 0 ] , [ 1 ] , [ 2 ] , [ 3 ] , [ 4 ] , [ 5 ] , [ 6 ] , [ 7 ] [ 0 ] , [ 1 ] , [ 2 ] , [ 3 ] , [ 4 ] , [ 5 ] , [ 6 ] , [ 7 ] [0],[1],[2],[3],[4],[5],[6],[7][0], [1], [2], [3], [4], [5], [6], [7][0],[1],[2],[3],[4],[5],[6],[7] and is cyclic, generated by [ 1 ] [ 1 ] [1][1][1].
  2. Z 4 × Z 2 Z 4 × Z 2 Z_(4)xxZ_(2)\mathbb{Z}_4 \times \mathbb{Z}_2Z4×Z2 has elements ( 0 , 0 ) , ( 0 , 1 ) , ( 1 , 0 ) , ( 1 , 1 ) , ( 2 , 0 ) , ( 2 , 1 ) , ( 3 , 0 ) , ( 3 , 1 ) ( 0 , 0 ) , ( 0 , 1 ) , ( 1 , 0 ) , ( 1 , 1 ) , ( 2 , 0 ) , ( 2 , 1 ) , ( 3 , 0 ) , ( 3 , 1 ) (0,0),(0,1),(1,0),(1,1),(2,0),(2,1),(3,0),(3,1)(0,0), (0,1), (1,0), (1,1), (2,0), (2,1), (3,0), (3,1)(0,0),(0,1),(1,0),(1,1),(2,0),(2,1),(3,0),(3,1) and is not cyclic.
Since Z 8 Z 8 Z_(8)\mathbb{Z}_8Z8 is cyclic and Z 4 × Z 2 Z 4 × Z 2 Z_(4)xxZ_(2)\mathbb{Z}_4 \times \mathbb{Z}_2Z4×Z2 is not, they cannot be isomorphic.
Thus, the statement is false, as demonstrated by this counterexample.

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iv) If G G G\mathrm{G}G is a finite group and m o ( G ) , m N m o ( G ) , m N m∣o(G),minN\mathrm{m} \mid \mathrm{o}(\mathrm{G}), \mathrm{m} \in \mathbb{N}mo(G),mN, then G G G\mathrm{G}G has a subgroup of order m m m\mathrm{m}m.
Answer:
The statement “If G G GGG is a finite group and m o ( G ) , m N m o ( G ) , m N m∣o(G),m inNm \mid o(G), m \in \mathbb{N}mo(G),mN, then G G GGG has a subgroup of order m m mmm” is false.

Justification:

The statement would be true if G G GGG were a cyclic group, as every divisor of the order of a cyclic group corresponds to a unique subgroup. However, the statement is not generally true for all finite groups.

Counterexample:

Consider the symmetric group S 3 S 3 S_(3)S_3S3 with o ( S 3 ) = 6 o ( S 3 ) = 6 o(S_(3))=6o(S_3) = 6o(S3)=6. The divisors of 6 are 1 , 2 , 3 , 6 1 , 2 , 3 , 6 1,2,3,61, 2, 3, 61,2,3,6.
While S 3 S 3 S_(3)S_3S3 does have subgroups of orders 1, 2, 3, and 6, it does not have a subgroup of order 4, even though 4 is a natural number ( m N m N m inNm \in \mathbb{N}mN).
Thus, the statement is false, as demonstrated by this counterexample.

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v) If β 1 β 1 beta_(1)\beta_1β1 and β 2 β 2 beta_(2)\beta_2β2 are two 15 th 15 th  15^(“th “)15^{\text {th }}15th  roots of unity, then Q ( β 1 ) = Q ( β 2 ) Q β 1 = Q β 2 Q(beta_(1))=Q(beta_(2))\mathbb{Q}\left(\beta_1\right)=\mathbb{Q}\left(\beta_2\right)Q(β1)=Q(β2).
Answer:
The statement “If β 1 β 1 beta_(1)\beta_1β1 and β 2 β 2 beta_(2)\beta_2β2 are two 15 th 15 th 15^(“th”)15^{\text{th}}15th roots of unity, then Q ( β 1 ) = Q ( β 2 ) Q ( β 1 ) = Q ( β 2 ) Q(beta_(1))=Q(beta_(2))\mathbb{Q}(\beta_1) = \mathbb{Q}(\beta_2)Q(β1)=Q(β2)” is true.

Justification:

The 15 th 15 th 15^(“th”)15^{\text{th}}15th roots of unity are the complex numbers of the form
β k = cos ( 2 π k 15 ) + i sin ( 2 π k 15 ) , β k = cos 2 π k 15 + i sin 2 π k 15 , beta _(k)=cos((2pi k)/(15))+i sin((2pi k)/(15)),\beta_k = \cos\left(\frac{2\pi k}{15}\right) + i \sin\left(\frac{2\pi k}{15}\right),βk=cos(2πk15)+isin(2πk15),
where i i iii is the imaginary unit and k = 0 , 1 , 2 , , 14 k = 0 , 1 , 2 , , 14 k=0,1,2,dots,14k = 0, 1, 2, \ldots, 14k=0,1,2,,14.
These roots of unity are all solutions to the equation x 15 1 = 0 x 15 1 = 0 x^(15)-1=0x^{15} – 1 = 0x151=0, and they generate a cyclotomic field Q ( ζ 15 ) Q ( ζ 15 ) Q(zeta_(15))\mathbb{Q}(\zeta_{15})Q(ζ15), where ζ 15 = e 2 π i / 15 ζ 15 = e 2 π i / 15 zeta_(15)=e^(2pi i//15)\zeta_{15} = e^{2\pi i / 15}ζ15=e2πi/15.
Now, any 15 th 15 th 15^(“th”)15^{\text{th}}15th root of unity can be expressed as a power of ζ 15 ζ 15 zeta_(15)\zeta_{15}ζ15. Specifically, β 1 = ζ 15 k 1 β 1 = ζ 15 k 1 beta_(1)=zeta_(15)^(k_(1))\beta_1 = \zeta_{15}^{k_1}β1=ζ15k1 and β 2 = ζ 15 k 2 β 2 = ζ 15 k 2 beta_(2)=zeta_(15)^(k_(2))\beta_2 = \zeta_{15}^{k_2}β2=ζ15k2 for some k 1 , k 2 { 0 , 1 , 2 , , 14 } k 1 , k 2 { 0 , 1 , 2 , , 14 } k_(1),k_(2)in{0,1,2,dots,14}k_1, k_2 \in \{0, 1, 2, \ldots, 14\}k1,k2{0,1,2,,14}.
Since both β 1 β 1 beta_(1)\beta_1β1 and β 2 β 2 beta_(2)\beta_2β2 can be expressed using powers of ζ 15 ζ 15 zeta_(15)\zeta_{15}ζ15, it follows that Q ( β 1 ) Q ( β 1 ) Q(beta_(1))\mathbb{Q}(\beta_1)Q(β1) and Q ( β 2 ) Q ( β 2 ) Q(beta_(2))\mathbb{Q}(\beta_2)Q(β2) are both subfields of Q ( ζ 15 ) Q ( ζ 15 ) Q(zeta_(15))\mathbb{Q}(\zeta_{15})Q(ζ15).
Moreover, β 1 β 1 beta_(1)\beta_1β1 and β 2 β 2 beta_(2)\beta_2β2 generate the same field as ζ 15 ζ 15 zeta_(15)\zeta_{15}ζ15 because they are powers of ζ 15 ζ 15 zeta_(15)\zeta_{15}ζ15. Therefore, Q ( β 1 ) = Q ( ζ 15 ) Q ( β 1 ) = Q ( ζ 15 ) Q(beta_(1))=Q(zeta_(15))\mathbb{Q}(\beta_1) = \mathbb{Q}(\zeta_{15})Q(β1)=Q(ζ15) and Q ( β 2 ) = Q ( ζ 15 ) Q ( β 2 ) = Q ( ζ 15 ) Q(beta_(2))=Q(zeta_(15))\mathbb{Q}(\beta_2) = \mathbb{Q}(\zeta_{15})Q(β2)=Q(ζ15), which implies Q ( β 1 ) = Q ( β 2 ) Q ( β 1 ) = Q ( β 2 ) Q(beta_(1))=Q(beta_(2))\mathbb{Q}(\beta_1) = \mathbb{Q}(\beta_2)Q(β1)=Q(β2).
Thus, the statement is true.

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vi) There exists an extension field of Z 3 Z 3 Z_(3)\mathbb{Z}_3Z3 of order 25 .
Answer:
The statement “There exists an extension field of Z 3 Z 3 Z_(3)\mathbb{Z}_3Z3 of order 25″ is false.

Justification:

An extension field K K KKK of a finite field F F FFF of order p n p n p^(n)p^npn (where p p ppp is a prime and n n nnn is a positive integer) must have order p m p m p^(m)p^mpm for some integer m > n m > n m > nm > nm>n. This is because the extension field K K KKK can be thought of as a vector space over F F FFF, and the size of K K KKK must be p m p m p^(m)p^mpm to be compatible with the vector space structure.
In the given statement, Z 3 Z 3 Z_(3)\mathbb{Z}_3Z3 is a field of order 3 3 333, and we are asked about an extension field of order 25 25 252525. The number 25 = 5 2 25 = 5 2 25=5^(2)25 = 5^2