Proving that the set of units of a ring is a cyclic group of order 4 The set of units of \mat

kadetskihykw 2022-04-30 Answered

Proving that the set of units of a ring is a cyclic group of order 4
The set of units of Z10Z is {1,3,7,9}, how can I show that this group is cyclic?
My guess is that we need to show that the group can be generated by some element in the set, do I need to show that powers of some element can generate all elements in the other congruence classes?
For example 72=499±mod10, i.e. using 7 we can generate an element in the congruence class of 9, but can not generate 29 for example from any power of 7, so is it sufficent to say that an element is a generator if it generates at least one element in all other congruence classes?

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Answers (2)

belamontern9i
Answered 2022-05-01 Author has 19 answers
You had just the right idea, then got sidetracked. What you've seen is that 72=9. From this, it does, indeed, follow that 72=29, since we're thinking modulo 10.
From there, 7397=633,
so that 73=3, and similarly, 74=1.
As an alternative, we could show that 32=9,33=7, and that 34=1, which is arguably even simpler, since 32=9,33=27, and 34=81 are math facts you probably know.
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Dexter Conner
Answered 2022-05-02 Author has 15 answers

Step 1
My guess is that we need to show that the group can be generated by some element in the set, do I need to show that powers of some element can generate all elements in the other congruence classes?
It is enough if you find one element such that its powers cover exactly the group. I would give 3 a try.
For example 72=499mod10, i.e. using 7 we can generate an element in the congruence class of 9, but can not generate 29 for example from any power of 7, so is it sufficent to say that an element is a generator if it generates at least one element in all other congruence classes?
Step 2
Edited following Cameron Buie's comment: 7 actually works, as it generates 9,3 and 1, in that order.
Let's see what happens with 3:
31=3
32=9
33=27=7
34=81=1
This is enough. On the one hand, we have covered all the elements in the subgroup. On the other hand, we are now satisfied that 34 is the neutral element of the group, so we are positive that for any natural number n, 3n=3nmod4.

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