The set {1,2,4,8,13,16,26,32} forms a group R under multiplication modulo 5. Show that R is cyclic?

If we're working modulo 5 then our set is really {1, 2, 3, 4} because:

13 = 3 (5)

32 = 2 (5)

16, 26 = 1 (5)

Then we observe that in modulo 5:

2 = 2

2² = 4

2³ = 8 = 3

2⁴ = 16 = 1

So every element in our set is a power of 2. You can check it's a group rather easily. So it's cyclic.

b) {1} s a subgroup trivially.

You can check {1, x} is not a group for any x = 2, 3, 4

{1, 4} is a subgroup

Again for anything else it's simple to show no other combinations apart from the actual group itself forms a group, so we're done.

I might go to the definition of a gaggle (a multiplicative group in the event you insist). I might then show that this collection is a bunch. That it meets the entire necessities for a multiplicative team (for instance, that it's closed underneath multiplication -- meaning: multiply any two factors, modulo 40, and the reply might be an detail of the group). Show that there's an identity aspect (what we name a "1"). An identity element is one who leaves the other aspect unchanged in a multiplication. --- 5 * 15 = 35 (modulo 40) This suggests that neither 5 nor 15 are the identity 25*35 = 35 So, might be 25 is the identification: 5 * 25 = 5 15 * 25 = 15 and so on (you need to verify them all and you ought to assess the opposite direction 25 * 5 = ? (nevertheless, integer multiplication is abelian, so you don't relatively ought to fear) to make sure that 25 rather behaves like an identity.