Question

Decide if the set {[0],[2],[4],[6],[8]} is in Z10 is a group under multiplication. If yes, is it abelian? If not, which conditions fail to hold?

Answer 1

\[{[0],[2],[4],[6],[8]} \subseteq Z _{10}\]

Answer 2

What exactly do the brackets mean?

Answer 3

Well, see
\[\huge \left\{0,2,4,6,8\right\}=\left<2\right>\subseteq \mathbb{Z}_{10}\]
Under addition, this is simply a cyclic subgroup, but under multiplication...there is no multiplicative identity, so no.

Answer 4

I am not sure about the brackets. That is how the example is written in my book.

Answer 5

How do you know that 2 is not in there?

Answer 6

Well, until further notice, I hope you don't mind that I left the brackets out...

Answer 7

How do I know that 2 is not in there? 2 is there, all right...

Answer 8

Yeah that is fine.

Answer 9

Are you sure the question asks if it's a group under *multiplication*?

Answer 10

Yes. It says decide if the set .... is a group under multiplication.

Answer 11

Okay... a group needs four things...

An associative binary operation
Closure
Existence of an identity
Existence of inverses....

Let's check these... shall we?

Answer 12

Is multiplication associative?

Answer 13

Yes. I understand that.

Answer 14

Yes multiplication is associative.

Answer 15

Okay, when you multiply two elements of the set {0,2,4,6,8} is the product still in the set?

Answer 16

Or we could just cut to the chase :)

Answer 17

No.

Answer 18

Yes, it is, actually closed under multiplication (in \(\large \mathbb{Z}_{10}\) )

Answer 19

But more importantly, the existence of an identity element.

Answer 20

\[4 * 6 = 24, \not \in (\mathbb{Z} _{10})\].
So how? I am not sure how to find the identity element

Answer 21

No, see (this must be one of my catchphrases by now, along with "catch me so far?" LOL)
Whenever you add or multiply in a set \(\large \mathbb{Z}_n\) you always take the answer in modulo n.
So actually
\[\huge 4\cdot 6 = 24(\mod 10 \ \ )=4\in\mathbb{Z}_{10}\]

Answer 22

Okay I got you on that

Answer 23

Which just made me realise... there IS an identity element after all!

Answer 24

How do I know if something is an identity element?

Answer 25

\[\huge 6\cdot 0 = 0\]\[\huge6\cdot2=12(\mod 10 \ \ )=2\]\[\huge 6\cdot 4=23(\mod 10 \ \ )=4\]\[\huge 6\cdot 6 = 36 \rightarrow 6\]\[\huge 6\cdot 8 =48\rightarrow 8\]

Answer 26

If there's an element, that, when multiplied (?) to any other element, the result is that element.

Answer 27

As you can see, 6 seems to satisfy that.

Answer 28

Okay I got you.

Answer 29

So far so good?

Answer 30

yes

Answer 31

So, apparently, it's the fourth one that presents a problem...
You agree that 6 is the identity element, right?

Answer 32

Yes I do.

Answer 33

Well, to complete the circle, every element in the group must have an element you can multiply to it so that you get the identity 6 (its inverse)
\[2\cdot 8 = 6\]\[4\cdot 4 = 6\]\[6\cdot 6 = 6\]\[8\cdot2=6\]

Answer 34

Did I leave anything out?

Answer 35

I have to go. I will be back later to get a better understanding of the inverse.

Answer 36

Well, anyway, there simply is nothing you can multiply to 0 so that you get 6, no matter what modulo you're using.
Hence, 0 has no inverse
Hence, this is no group.
Get it? Got it? Good :)
-------------------------------------------------
Terence out

Answer 37

So would this be abelian?

Answer 38

It is not even a group.