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Math Help - Group Theory(1):Center of groups and other problems

  1. #1
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    Group Theory(1):Center of groups and other problems

    Hello MHF,

    Now I finished reading Artin, Chapter 2 and was attempting exercises. Few problems look really tricky. I will be grateful if anybody could help me out with these:

    1) Prove that if a group contains exactly one element of order 2, then that element is in the center of the group.
    2) Prove that a group whose order is a power of prime p,contains an element of order p.


    Something that looked like a promising direction:If g_i denotes a general element in G, I proceeded like this:

    Lets say there does not exist an element of order prime. Consider the set S = \{g_1 ^p, g_2 ^p,.....g_{p^k - 1}^p\} where I have not considered the identity. Then it forces S to be a permutation of G \ {e}. But I could not proceed further...
    3) What is the center of the group of all unit upper triangular matrices?
    Clearly it has the set of all scalar matrices(matrices of form cI) as its subgroup... But do you know any other matrix that commutes with all unit upper triangular matrices?


    I would appreciate hints compared to solutions.

    Thanks,
    Srikanth
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    when you put your question in "Quote", we can't quote you anymore.


    Question 1: suppose a is the unique element of order 2 and b is any element of the group. then bab^{-1} has order 2 and thus bab^{-1}=a. \ \ \Box


    Question 2: proof by induction on |G|. if |G|=p, then every non-identity element of G has order p and we're done. let |G|=p^n, and suppose that the claim is true for any group of order

    p^m, \ 1 \leq m < n. if G is cyclic, then there's nothing to prove. if it's not cyclic, then choose 1 \neq x \in G. now apply induction on <x>. \ \ \Box


    Question 3: the center has only one element, i.e. the identity matrix. suppose G is the group of all n \times n unit upper triangular matrices over a field or a unitary commutative ring. as usual

    let e_{ij} be the matrix with 1 in its (i,j)-entry and 0 elsewhere. let g \in Z(G). we have g=I + \sum_{1 \leq i < j \leq n} a_{ij}e_{ij}, for some scalars a_{ij}. now let 1 \leq k < \ell \leq n. then I + e_{k \ell} \in G and so we must

    have (I+e_{k \ell})g=g(I+e_{k \ell}), which gives us: \sum_{1 \leq i < j \leq n} a_{ij}e_{k \ell}e_{ij}=\sum_{1 \leq i < j \leq n} a_{ij}e_{ij}e_{k \ell}. using the fact that e_{rs}e_{tu}=\delta_{st}e_{ru}, we'll have: \sum_{\ell < j \leq n} a_{\ell j}e_{kj}=\sum_{1 \leq i < k} a_{ik}e_{i \ell}, and this gives us:

    a_{ik}=a_{\ell j} = 0, \ \ \forall 1 \leq i < k < \ell < j \leq n. therefore g=I, and thus: Z(G)=\{I \}. \ \ \Box
    Last edited by NonCommAlg; December 28th 2008 at 01:23 PM.
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    Quote Originally Posted by NonCommAlg View Post
    when you put your question in "Quote", we can't quote you anymore.
    Oops! I will keep that in mind.

    Quote Originally Posted by NonCommAlg View Post
    Question 1: suppose a is the unique element of order 2 and b is any element of the group. then bab^{-1} has order 2 and thus bab^{-1}=a. \ \ \Box
    Thanks. Looks like we can replace 2 by any prime.

    Quote Originally Posted by NonCommAlg View Post
    Question 2: proof by induction on |G|. if |G|=p, then every non-identity element of G has order p and we're done. let |G|=p^n, and suppose that the claim is true for any group of order

    p^m, \ 1 \leq m < n. if G is cyclic, then there's nothing to prove. if it's not cyclic, then choose 1 \neq x \in G. now apply induction on <x>. \ \ \Box
    This technique is nice. Thank you.I never thought of using induction.

    Quote Originally Posted by NonCommAlg View Post
    Question 3: the center has only one element, i.e. the identity matrix. suppose G is the group of all n \times n unit upper triangular matrices over a field or a unitary commutative ring. as usual

    let e_{ij} be the element of G with 1 in its (i,j)-entry and 0 elsewhere. let g \in Z(G). we have g=I + \sum_{1 \leq i < j \leq n} a_{ij}e_{ij}, for some scalars a_{ij}. now let 1 \leq k < \ell \leq n. then I + e_{k \ell} \in G and thus

    we must have (I+e_{k \ell})g=g(I+e_{k \ell}), which gives us: \sum_{1 \leq i < j \leq n} a_{ij}e_{k \ell}e_{ij}=\sum_{1 \leq i < j \leq n} a_{ij}e_{ij}e_{k \ell}. using the fact that e_{rs}e_{tu}=\delta_{st}e_{ru}, we'll have: \sum_{\ell < j \leq n} a_{\ell j}e_{kj}=\sum_{1 \leq i < k} a_{ik}e_{i \ell}, and this gives us:

    a_{ik}=a_{\ell j} = 0, \ \ \forall 1 \leq i < k < \ell < j \leq n. therefore g=I, and thus: Z(G)=\{I \}. \ \ \Box
    I got confused with the definition of Z(G) and thats why I thought cI will work. Clearly for c \neq 1, cI does not belong to that group. Thanks again
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