Need question 3 answered Answered must be correct Do not giv

Need question 3 answered. Answered must be correct. Do not give incorrect answer.

State the definition of a group. Check that SO_n(R^n) with matrix multiplication is indeed a group. Based on the definition, does there exist a group with 0 elements? Why? Consider the set S = Z/2Z times Z/2Z times 2/2Z with the following composition law: (a, b, c)*(a\', b\', c\') = (a + a\', b + b\', c + c\' + ab\'), where both addition and multiplication are taken mod 2. Prove that S with * is a noil-commutative group. Unimportant note. Can you see a nicer way to describe this group? Use the Euclidean algorithm to compute the greatest common divisor of 261 and 100. Use the Euclidean algorithm to find two integers a and b such that 11a+15b = 1. Draw two plane figures whose symmetry groups have order 10, but are not isomorphic. Decompose the following permutation of {1, 2, ...., 8} p(1 2 3 4 5 6 7 8 5 8 1 7 3 4 6 2) into disjoint cycles. Is it an even permutation or an odd permutation? What is its order? Is C_2 times C_6 isomorphic to C_3 times C_4? is C_10 times C_21 isomorphic to C_14 times C_15? Prove that C_a1 times C_a2 times .... times C_a0 is isomorphic to C_a1, a2, ...an if and only if a_1, a_2, ...a_n are pairwise. Let G be group with 25 elements, G a group with 21 elements and let phi be a group homomorphism from G to G. Prove that phi(x) = 1G for all x subset G. Here, I_G denotes the identity element of G. What are the possible values of a perfect square n^2 module 6? Prove that any 10-digit prime number has two equal digits. Prove that there exists a suejective group homomorphism phi: S_4 rightarrow S_2. How can you see this geometrically? How many injective group homomorphism phi:C_3 rightarrow S are there?

Solution

1) let a=164 ,b=100

*divide 164 by100 & get the result 1 with reminder 64,so 164=100(1)+64.

*divide 100 by64 & get the result 1 with reminder 36,so 100=64(1)+36.

*divide 64 by36 & get the result 1 with reminder 28,so 64=36(1)+28.

*divide 36 by28 & get the result 1 with reminder 8,so 36=28(1)+8.

*divide 28 by8 & get the result 3 with reminder 4,so 28=8(3)+4.

*divide 8 by4 & get the result 2 with reminder 0,so 8=4(2)+0.

The G.C.D is 4.

b) To find G.C.D of 11&15

15=11(1)+4

11=4(2)+3

4=3(1)+1

3=1(3)+0

so gcd is1.

1=4-1(3)

1=4-1(11-2*4)

1=4-11+2*4

1=3*4-11

1=3*(15-1*11)-11

1=3*15+(-4)11

so a=-4 & b=3