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Math Help - Group Table

  1. #1
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    Group Table

    I don't know whether or not I am right about 2^n and n \in G, where G = \{-2,-1,0,1,2\} being an operation on (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 (G,*) is under multiplication.
    Then (n\cdot n)=2^n on (G,\cdot). I constructed the group table below. Please verify whether it's correct.

    \begin{array}{c|c|c|c|c|c}\cdot&-2&-1&0&1&2\\<br />
\hline<br />
-2&2^4&2^2&2^0&2^1&2^2\\<br />
\hline<br />
-1&2^2&2^1&2^0&2^1&2^2\\<br />
\hline<br />
0&2^0&2^0&2^0&2^0&2^0\\<br />
\hline<br />
1&2^{-2}&2^{-1}&2^0&2^1&2^2\\<br />
\hline<br />
2&2^{-4}&2^{-2}&2^0&2^1&2^2\end{array}

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

    Yes?
    Last edited by novice; August 26th 2010 at 05:45 PM.
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  2. #2
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    I can't quite understand what is the set of elements you're defining your operation on, or how it is defined...
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  3. #3
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    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 2^n is under multiplication on (G,\cdot) where G=\{-2,-1,0,1,2\}

    I made up this question myself after I saw

    (a*b)=a^b on \mathbb{N}, and I want to know what 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.
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  4. #4
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    So you're defining your group over the set \{\pm2, \pm1, 0\} with the operation (\cdot ). How do you define a \cdot b?
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  5. #5
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    Quote Originally Posted by Defunkt View Post
    So you're defining your group over the set \{\pm2, \pm1, 0\} with the operation (\cdot ). How do you define a \cdot b?
    Oh, I see.

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

    Will the group table look like this?

    \begin{array}{c|c|c|c|c|c}<br />
\cdot&-2&-1&0&1&2\\<br />
\hline<br />
2&2^{-2}&2^{-1}&2^0&2^1&2^2\end{array}

    Then what is the identity? is it 0?
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  6. #6
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    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, 2\cdot 2 = 2^2=4 \notin \{\pm 2, \pm 1, 0\} so this can't be a group.
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  7. #7
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    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.
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  8. #8
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    You're very welcome!
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  9. #9
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    You have given me an idea, and I am going to change it to (2 \cdot n) on \mathbb{Z}.

    Will it work this time?
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  10. #10
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    Your operation has to be defined for every a,b \in \mathbb{Z}. What you're suggesting is not a binary operation - rather a function f:\mathbb{Z} \to \mathbb{Z}, f(n) = 2^n.

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

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

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