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[ionic bone]

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[arctic carbon]

also got to say if they're cyclic groups

[bronze gust]

what does the * mean?

[arctic carbon]

Idk that either

[bronze gust]

Ok I'll ignore that for now

[arctic carbon]

I don’t get the _8 and _14 I’ve never seen that

[bronze gust]

the order of Zn is just n. The order or cardinality of a set is how many unique elements it has. For example, Z2 = {0, 1}, which has two elements, so order 2. Z100 is all integers from 0 to 99 inclusive, so order 100

oh

ok let me back up

so you know how Z is the set of integers right?

[arctic carbon]

Ye

[bronze gust]

when you have a little subscript like that in the picture, it means "mod _" (that number)

so Z8 is the integers mod 8

mod means you divide by 8 and take the remainder

for example, 5 mod 3 = 2, because 5/3 = 1 remainder 2

6 mod 3 = 0 because 3 divides 6 evenly

[arctic carbon]

Ok

Ty

Then what do I do for the order

[bronze gust]

So let's look at Z4 for example. 0 is in there, 1 is in there, 2 is in there, 3 is in there, but once you hit 4, 4 mod 4 = 0 (because they divide evenly). 5 mod 4 = 1, 6 mod 4 = 2, 7 mod 4 = 3, 8 mod 4 = 0... This pattern will repeat forever... 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3...

So Z4 = {0, 1, 2, 3}, which has 4 unique elements, so order 4

Note that this process removes the negative integers

So no Zn will have negatives, just the integers from 0 to (n - 1)

because once you hit n, you loop back around to 0 and start over

like a clock. You start at 00:00 hours, loop around to 12, and then start over again

So just looking at the Z4 example above, what would you guess Z8 to be?

[arctic carbon]

So it’s just 0-7

?

[bronze gust]

yep

[arctic carbon]

Oh ok

[bronze gust]

Zn = {0, 1, 2, ... n-1}

[arctic carbon]

And the order is 7

?

Or

[bronze gust]

don't forget 0

[arctic carbon]

Oh right ye

[timid vault]

The * denotes the set of invertible elements in the ring (Zn,+,x)

(Zn*,x) is a group

So it’s order is the number of invertible elements in Zn for x

[bronze gust]

oh

well that changes things

[timid vault]

(Which is exactly the number of elements that are coprime with n between 1 and n-1)

So you need to use euler’s function

[arctic carbon]

Never used that before

[bronze gust]

@arctic carbon so everything I said is true, but now that we know what * means, it changes what's in Z8, so it's order won't quite be 8

but it's still true that the order of Zn is n

without that *

[arctic carbon]

o ok

[bronze gust]

I have to go grade ochem labs now but I can come back later tonight if you still want to talk about this

[arctic carbon]

Ight ty

[timid vault]

You basically have to count the number of elements that are coprime with 8

Between 1 and 7

There is a general formula but if you don’t know it it’s not a big deal here considering we are dealing with small numbers so you can just count them manually

[arctic carbon]

Is it just 1 3 5 7?
Cause HCF of 2 4 and 6 with 8 is 2

[timid vault]

Yep

So $Z_8^*$ has 4 elements

[rough folio]

😑 ɿototoЯ | Rototor 😑

[arctic carbon]

So then its order is just 4?

[timid vault]

Yeah

[arctic carbon]

Ok ye ty

Then do the same with 14 but between 1 and 13

[timid vault]

Yep

[arctic carbon]

How do I show that they are/aren’t cyclic groups

[timid vault]

Take each of the elements of your group Z_8^* (except 1) and square them if it’s not 1 mod 8 then the group is cyclic, because the order of the element is 4 (the cardinal of the group)

It can’t be less because the smallest possible order before 4 is 2

In Z_14^* you need to square and cube each one

Also any abelian group that has order pq with p and q two distinct primes is cyclic but the proof requires the theorem of classification of finite abelian groups so it’s a lot more complicated

[arctic carbon]

Why do you square and cube Z_14 but not Z_8

[timid vault]

The orders of a non trivial element in Z_14* can be 2,3 or 6

In Z_8*, 2 or 4

If there is an element with order 6 then that means the group is cyclic and that element is a generator of the group