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Math Help - Inverse of an element in a group

  1. #1
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    Inverse of an element in a group

    Let G be a group. Then

    (a) G has a unique identity, and
    (b) each element in G has a unique inverse.

    The table below shows an abelian group with 3 elements:

    \begin{tabular}{lccr}<br />
*&a&b&c\\<br />
 \cline{2-4}a&a&b&c\\<br />
b&b&c&c\\<br />
c&c&a&b<br />
\end{tabular}

    I can see the identity is show on the first row and column, but I don't see the inverse of each elements. Does it mean that the inverses are not in the group?
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  2. #2
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    Quote Originally Posted by novice View Post
    (b) each element in G has a unique inverse.
    You are correct, each element does not have an inverse, therefore G is not a group.
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    Quote Originally Posted by novice View Post

    The table below shows an abelian group with 3 elements:

    \begin{tabular}{lccr}<br />
*&a&b&c\\<br />
 \cline{2-4}a&a&b&c\\<br />
b&b&c&c\\<br />
c&c&a&b<br />
\end{tabular}
    Also bc \neq cb \implies G does not commute.
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  4. #4
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    Let G be group. Would it look like this:

    G=\{e, 1/a, a, 1/b, b,...\} provided that G is associative and commutative?
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    Quote Originally Posted by novice View Post
    Let G

    G=\{e, 1/a, a, 1/b, b,...\} provided that G is associative and commutative?
    I don't think so as a=e and the inverses a^{-1},b^{-1}, c^{-1} are not listed in the make up. You would need a rule linking the other elements.
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    Quote Originally Posted by novice View Post
    Let G be group. Would it look like this:

    G=\{e, 1/a, a, 1/b, b,...\} provided that G is associative and commutative?
    A group (G, *) is always associative by definition, but not always commutative. It is commutative if and only if it is abelian, by definition.

    In general the inverse of \displaystyle a is denoted \displaystyle a^{-1}. Also keep in mind that it's possible to have \displaystyle a=a^{-1}, so your enumeration of the set above could have duplicates. There would also be duplicates if you listed a pair of inverses twice, for example if \displaystyle a = b^{-1}.
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    I see. No wonder a*b=b.
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    Quote Originally Posted by undefined View Post
    In general the inverse of \displaystyle a is denoted \displaystyle a^{-1}. Also keep in mind that it's possible to have \displaystyle a=a^{-1}, so your enumeration of the set above could have duplicates. There would also be duplicates if you listed a pair of inverses twice, for=e, where e is an indentity of (G,*) example if \displaystyle a = b^{-1}.
    So if G is a group, the inverse of each element is in (G,*). Yah?
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  9. #9
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    Yes, by definition of a group.
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  10. #10
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by novice View Post
    Let G be a group. Then

    (a) G has a unique identity, and
    (b) each element in G has a unique inverse.

    The table below shows an abelian group with 3 elements:

    \begin{tabular}{lccr}<br />
*&a&b&c\\<br />
 \cline{2-4}a&a&b&c\\<br />
b&b&c&c\\<br />
c&c&a&b<br />
\end{tabular}

    I can see the identity is show on the first row and column, but I don't see the inverse of each elements. Does it mean that the inverses are not in the group?
    A quick check to see that this is not a group is to note that in each row and column of the Cayley table every element must occur. This reflects the fact that one can get from each element of a group to every other, for example one can get to c from a by (post-)multiplying a by a^{-1}c.
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