Isomorphism of Groups and Automorphisms

Ans 1.
Solution (a) Every element of Z2 x Z2 x Z2 has order 1 or 2. For example, ([1], [0], [1])2 ([1] e [1] , [0] e
[0], [1] e [1] ) Go] , [0], [0] ) is the identity element of Z2 x 7Z2 x Z2, SO ([1], [0], [1]) has order 2. There's
an element of Z2 x Z4 of order 4, namely ([0]2, [1]4), but there's no element of order 8 in Z2 X Z4. The
element [1]8 of Z8 has order 8. Hence the three groups 7G2 x 7G2 x Z2, 7G2 x Z4 and Z8 are not
isomorphic, by Theorem 41(d). On the other hand, it's easy to show that G x H is always isomorphic to H
x G, since the mapping (g, h) (h, g) is an isomorphism. [Check this!] So Z2 x Z4 c 7Z4 x Z2. And Z4 x Z2 is
not isomorphic to any other group in the list, since if it were then Z2 x Z4 would be too (by transitivity of
Ras), but we've already shown that this is not true.
Ans 2.
Consider Z12. We have 4)(n) = 4 since 1, 5, 7 and 11 are the only integers r such that 1 r < 12 and (r, 12) =
1. Therefore, there are exactly four auto-morphisms of Z12 and are given below. = I, the identity map of
76122 rs : Z12 -' Z12 defined by f5(m) = 5m (mod 12), ri : Z12 -> Z12 defined by /7(m) = 7m (mod 12), fii :
Z12 -> 712 defined by fit(m) = 1 lm (mod 12).
The following table gives a complete description of all the four automor-phisms of 7612.
0 1 2 3 4 5 6 7 8 9 10 11
F1 0 1 2 3 4 5 6 7 8 9 10 11
F2 0 5 10 3 8 1 6 11 4 9 2 7
F3 0 7 2 9 4 11 6 1 8 3 10 5
F4 0 11 10 9 8 7 6 5 4 3 2 1
Ans 3. 1. dmin(c)=5
2. C can detect 3 error at max
3. C can correct 2 error at max
4. Decoded word is 1111011001011
Ans 4. S2(c): {1010,0010,1110,1000,1011,1001,1111,0101,0110,0011,1100}
Ans 5 . S3(c): ∑
k=0
12
Ck
n
2. If c can correct 3 error so maximum number of codeword in c is 6.
Ans 6. is a finite, non-empty set of states;
 is a finite, non-empty set of tape alphabet symbols;
 is the blank symbol (the only symbol allowed to occur on the tape infinitely often at any step
during the computation);