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ECE-C490 Security in Computing
Winter term 2005
Homework #8
Due in the TA’s mailbox by 5pm on Thursday, March 10th
1. Consider the group J24+, +24.
a. Identify a subgroup of 8 elements.
The easiest subgroup to choose is the multiples of 3: {0, 3, 6, 9, 12, 15, 18,
21}. It’s easy to see that since every element is a multiple of 3 (i.e. 3x or 3y,
where x, y are elements of the set {0, 1, …7}) adding any two elements will
also result in a multiple of 3. ex: 6 = 3*2; 15 = 3*5. 3*2 + 3*5 = 3(5+2) =
21 = 3*7. So closure holds. Similarly, the group identity (0) is present, and we
can recover the identity from any member of the subgroup by adding a
different subgroup member mod 24. Thus there is a unique inverse for each
element. It is a valid subgroup.
b. Find the factor structure and create the Cayley table for it.
Cosets of the subgroup above: [0], [1], [2]. Factor structure’s Cayley table:
[0]
[1]
[2]
[0]
[0]
[1]
[2]
[1]
[1]
[2]
[0]
[2]
[2]
[0]
[1]
2. Given the positive integers and 0 as a group (over addition), show that the multiples
of 6 constitute a subgroup. Then find the cosets of this subgroup.
This is not, in fact, a group. If all integers were part of the set, not just the positive
integers, it would be a group either in the case of addition or multiplication, and the
multiples of 6 would be a subgroup for either. As it is, there’s no identity for the set.
3. Consider J18+, +18, *18.
a. Is it a ring? Why or why not?
Yes, this is a ring. All group properties hold for addition; multiplication is
associative and closed. These are the required ring properties.
b. A commutative ring? Why or why not?
Yes; multiplying any two elements modular 18 will yield the same result no
matter what order they are multiplied in.
c. If it is a ring, what is its identity?
The ring’s identity is 0. (Remember, no identity for multiplication here.)
d. Is it a field?
This is not a field. Some elements have no multiplicative inverse; for
example, 4 cannot be multiplied mod 18 by anything in the ring to recover 0.