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Math Help - Prove <a> is subgroup of C(a)

  1. #1
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    Prove <a> is subgroup of C(a)

    For any element a in G, a group, prove that <a> is a subgroup of C(a).

    Proof:

    Now <a> = { e, a, a^2, ... , a^{n-1}} if |a| = n. So e would be in <a> for the least possible n, implies that <a> is nonempty. If <a> = {e}, then <a> is already a subgroup of C(a), so I assume <a> do not equal to {e}.

    Let a^i and a^j be in <a>, and consider C(a) = \{g \in G : ga = ag\ \forall a\}

    Now (a^i)(a^j) = a^{i+j} , which is in <a>, and (a^i)^{-1} = a^{-i}, which is in <a> as well.

    Furthermore, (a^i)(a) = a^{i+1} = a^{1+i} = (a)(a^{i}), so a^{i} \in C(a) , thus proves <a> is a subgroup of C(a).

    Q.E.D.

    Is that right?
    Last edited by CaptainBlack; September 7th 2007 at 11:52 PM. Reason: fix LaTeX
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    For any element a in G, a group, prove that <a> is a subgroup of C(a).

    Proof:

    Now <a> = { e, a, a^2, ... , a^{n-1}} if |a| = n. So e would be in <a> for the least possible n, implies that <a> is nonempty. If <a> = {e}, then <a> is already a subgroup of C(a), so I assume <a> do not equal to {e}.

    Let a^i and a^j be in <a>, and consider C(a) = \{g \in G : ga = ag\ \forall a\}

    Now (a^i)(a^j) = a^{i+j} , which is in <a>, and (a^i)^{-1} = a^{-i}, which is in <a> as well.

    Furthermore, (a^i)(a) = a^{i+1} = a^{1+i} = (a)(a^{i}), so a^{i} \in C(a) , thus proves <a> is a subgroup of C(a).

    Q.E.D.

    Is that right?
    Where did you show that for any i>0 that there exits a j \ge 0 such that a^{-i}=a^j

    RonL
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  3. #3
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    The centralizer C(a) = \{ g\in G | ga = ag \} contains all \left< a \right>. Because if g \in \left< a \right> i.e. g=a^k then ga = a^k a = aa^k = ag implies g=a^k \in C(a).
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  4. #4
    Senior Member sfspitfire23's Avatar
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    then would a not be an element of the center because a does not commute with everything? how would you show this?
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  5. #5
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    Quote Originally Posted by sfspitfire23 View Post
    then would a not be an element of the center because a does not commute with everything? how would you show this?
    You are talking to a vacum, all the participants in this thread have moved on.

    Also quote the post you are refering to so we know exactly what you are refering to.

    CB
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  6. #6
    Senior Member sfspitfire23's Avatar
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    Ah, sorry,

    So, if <a>=C(a) of a non abelian group, G, then would the element a be in the center? that is, would a\in Z(G). This is similar to a homework Q i have here at university in Canada
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  7. #7
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    Quote Originally Posted by sfspitfire23 View Post
    Ah, sorry,

    So, if <a>=C(a) of a non abelian group, G, then would the element a be in the center? that is, would a\in Z(G). This is similar to a homework Q i have here at university in Canada
    Not necessarily.

    For instance, let's take an example for a symmetric group G=S_4.
    (1,2,3,4) generates a subgroup <(1,2,3,4)> of S_4, which equals to C_G((1,2,3,4)). But the center of S_4 is {e}. Actually, a symmetric group S_n of n>=3 has a trivial center {e}.
    Last edited by aliceinwonderland; October 22nd 2009 at 05:08 PM. Reason: clarification
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  8. #8
    Senior Member sfspitfire23's Avatar
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    would there be a non-symmetric group this could work for? A permutation?
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  9. #9
    Senior Member sfspitfire23's Avatar
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    ah, possibly if the group was non-abelian, the identity would be the only thing to commute with everything and thus an element a would not commute with everything and thus would not be in Z(G).

    thoughts?
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