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Math Help - Abstract Algebra: Lagrange's Theorem 2

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    is up to his old tricks again! Jhevon's Avatar
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    Abstract Algebra: Lagrange's Theorem 2

    Hail Mathematicians,

    My last question for the day... I think. Check my solution please.


    Problem:

    The exponent of a group G is the smallest positive integer n such that a^n = e for each a \in G ( e is the identity element), if such an integer n exists. Prove that every finite group has an exponent, and that this exponent divides the order of the group. [Suggestion: Consider the LCM of \{ o(a)~:~a \in G \}.]


    Solution:

    We need to show: (1) If G is finite, then G has an exponent; and (2) we need to find such an exponent.

    Proof:

    (1) By the contrapositive, assume that G does not have an exponent. Then there exists some a \in G such that a^n \ne e for all n \in \mathbb{Z}^+. But that would mean a is of infinite order (by the definition of o(a)), and so, G would have to be infinite.

    (2) Consider n = \mbox{LCM}( \{ o(a)~:~a \in G \}). Now, for any a \in G, we have a^k = e for some k \in \mathbb{Z}^+ by (1). But since k \mid n, we can write n = mk for some k \in \mathbb{Z}. Thus, a^n = a^{mk} = (a^k)^m = e^m = e. And since n is the LCM of the orders of all a \in G, it is the smallest integer for which this is true. Moreover, n \mid |G| as a corollary of Lagrange's Theorem.

    QED


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    If a has infinite order than a,a^2,a^3,... must all be distinct. Because otherwise if not then a^i = a^j \implies a^{i-j}=e which is not possible. But then \{a,a^2,a^3,..\} implies G has infinitely many elements, a contradiction.

    If a\in G for finite group construct \left< a\right> then this is a subgroup and its order is the order of a. Since |\left< a \right>| | |G| it means the order of a divides |G|.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    If a has infinite order than a,a^2,a^3,... must all be distinct. Because otherwise if not then a^i = a^j \implies a^{i-j}=e which is not possible. But then \{a,a^2,a^3,..\} implies G has infinitely many elements, a contradiction.

    If a\in G for finite group construct \left< a\right> then this is a subgroup and its order is the order of a. Since |\left< a \right>| | |G| it means the order of a divides |G|.
    i thought of doing the proof you gave, pretty much exactly, but (1) i didn't want to single out <a>, i tend to have a problem choosing specific examples when i am trying to prove something in general. and (2) the question suggested using the LCM of the orders, so i wanted to do it that way. does my proof do the job with those two in mind?

    going back to what i said in (1) above: are there subgroups of G that are not generated? that is, they are not of the form <a> for any a in G?
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