# Group Table

• Aug 26th 2010, 04:14 PM
novice
Group Table
I don't know whether or not I am right about $\displaystyle 2^n$ and $\displaystyle n \in G$, where $\displaystyle G = \{-2,-1,0,1,2\}$ being an operation on $\displaystyle (G,*)$. Please tell me.

Suppose that I am correct, then I may say: Since it is a Group, it is a binary operation as well.

Suppose that the group $\displaystyle (G,*)$ is under multiplication.
Then $\displaystyle (n\cdot n)=2^n$ on $\displaystyle (G,\cdot)$. I constructed the group table below. Please verify whether it's correct.

$\displaystyle \begin{array}{c|c|c|c|c|c}\cdot&-2&-1&0&1&2\\ \hline -2&2^4&2^2&2^0&2^1&2^2\\ \hline -1&2^2&2^1&2^0&2^1&2^2\\ \hline 0&2^0&2^0&2^0&2^0&2^0\\ \hline 1&2^{-2}&2^{-1}&2^0&2^1&2^2\\ \hline 2&2^{-4}&2^{-2}&2^0&2^1&2^2\end{array}$

Since $\displaystyle (1*n)=2^n$, it follows that $\displaystyle 1$ is the identity of $\displaystyle (G,\cdot)$.

Yes?
• Aug 26th 2010, 04:43 PM
Defunkt
I can't quite understand what is the set of elements you're defining your operation on, or how it is defined...
• Aug 26th 2010, 04:59 PM
novice
I don't quite understand the question. Do you mean --uh--I don't even know how to put it into words?

Well, I think $\displaystyle 2^n$ is under multiplication on $\displaystyle (G,\cdot)$ where $\displaystyle G=\{-2,-1,0,1,2\}$

I made up this question myself after I saw

$\displaystyle (a*b)=a^b$ on $\displaystyle \mathbb{N}$, and I want to know what $\displaystyle 2^n , n\in \mathbb{Z}$ is, a group or not a group?

I really don't know what I am doing because I don't have enough reading material.
• Aug 26th 2010, 05:10 PM
Defunkt
So you're defining your group over the set $\displaystyle \{\pm2, \pm1, 0\}$ with the operation $\displaystyle (\cdot )$. How do you define $\displaystyle a \cdot b$?
• Aug 26th 2010, 05:20 PM
novice
Quote:

Originally Posted by Defunkt
So you're defining your group over the set $\displaystyle \{\pm2, \pm1, 0\}$ with the operation $\displaystyle (\cdot )$. How do you define $\displaystyle a \cdot b$?

Oh, I see.

Does $\displaystyle (2\cdot n) = 2^n$ on $\displaystyle \{\pm2, \pm1, 0\}$ look right?

Will the group table look like this?

$\displaystyle \begin{array}{c|c|c|c|c|c} \cdot&-2&-1&0&1&2\\ \hline 2&2^{-2}&2^{-1}&2^0&2^1&2^2\end{array}$

Then what is the identity? is it $\displaystyle 0$?
• Aug 26th 2010, 05:29 PM
Defunkt
This is not good. You need the operation to be defined for every two elements of your set. Also note that by your current definition, $\displaystyle 2\cdot 2 = 2^2=4 \notin \{\pm 2, \pm 1, 0\}$ so this can't be a group.
• Aug 26th 2010, 05:35 PM
novice
I see.

Well, at any rate, I want to thank you for your time. You have been a very nice friend, and I want you to stay alive on the line of duty. I hope when you see a terrorist in Israel, you would shoot first before questioning.
• Aug 26th 2010, 05:37 PM
Defunkt
You're very welcome!
• Aug 26th 2010, 05:38 PM
novice
You have given me an idea, and I am going to change it to $\displaystyle (2 \cdot n)$ on $\displaystyle \mathbb{Z}$.

Will it work this time?
• Aug 27th 2010, 03:45 AM
Defunkt
Your operation has to be defined for every $\displaystyle a,b \in \mathbb{Z}$. What you're suggesting is not a binary operation - rather a function $\displaystyle f:\mathbb{Z} \to \mathbb{Z}$, $\displaystyle f(n) = 2^n$.

A binary operation can be looked at as a function
$\displaystyle \cdot : G \times G \to G$.

For example, the binary operation in the example you saw, $\displaystyle a * b = a^b$ can be looked at as $\displaystyle * : \mathbb{N} \times \mathbb{N} \to \mathbb{N}, ~ *(a,b) = a^b$

Another example for a binary operation: Take $\displaystyle \mathbb{Z}_n = \{0,1,...,n-1 \}$ and define $\displaystyle a * b = a + b ~ (mod ~ n)$.