Math 3124 Monday, November 3
Seventh Homework Solutions
1. 22*28 = 616.
(a)
616 =9∗65 +31,
65 =2∗31 +3,
31 =10 ∗3+1
so working from bottom to top,
1=31 −10 ∗3=31 −10 ∗(65 −2∗31) = 21 ∗31 −10 ∗65
=21 ∗(616 −9∗65)−10 ∗65 =21 ∗616 −199 ∗65
and we deduce that 417 ∗65 ≡1 mod 616. Therefore Peter Linnell’s public key is
(417,667).
Another way to do this is with Mathematica with the command
PowerMod[65,-1,616]
(b) We need to calculate 123417 mod 667. We have 12316 =393 mod 667. Therefore
123416 =39316 =285 mod 667. Thus 123417 =123 ∗285 =371 mod 667 and we
conclude that the number 371 has to be sent.
(c) We need to calculate 12365 mod 667. We have 12313 =547 mod 667. Therefore
12365 =5475=49 mod 667 and hence the number 49 was originally sent.
2. S5×S4has elements of order 15, for example ((1 2 3 4 5), (1 2 3)). On the other
there is no element of order 15 in S6(the only way to have a permutation of order 15
is if there is a 15 cycle, or a product of a 5 cycle and a 3 cycle (which are disjoint),
which would require at least S8). Since the elements of Z4have order 1, 2 or 4 and
o(x,y) = [o(x),o(y)], we see that S6×Z4can have an element of order 15 only if
S6has an element of order 15. Therefore S6×Z4has no element of order 15 and it
follows that S5×S4S6×Z4.
3. Problem 21.6(a) on page 109. Show that α:Z3→Z6by α([a]3) = [a]6is not well-
defined.
[0]3= [3]3, so for αto be well defined, we must have α[0]3=α[3]3, that is [0]6= [3]6.
But [0]66= [3]6and it follows that αis not well-defined.
4. (a) We will use subscripts to indicate which Znwe are in; thus [3]7is an element of
Z7. Suppose we have [x]35 = [y]35. Then [x−y]35 = [0]35 and hence 35|x−y.
Therefore 5,7|x−y, consequently [x−y]5= [0]5and [x−y]7= [0]7. It follows
that θis well defined.
