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Math Help - Prove that |G| ≡ h mod 16.

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    Prove that |G| ≡ h mod 16.

    Can anybody help me with these questions as I am completely lost and haven't done any character theory or rep theory for a long time.

    (1) Let G be a group of odd order (so |G| ≡ 1 mod 2) and (conjugacy) class number h. Prove that |G| ≡ h mod 16.
    Last edited by Ackbeet; March 14th 2011 at 05:42 AM. Reason: Split threads into two.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Turloughmack View Post
    Can anybody help me with these questions as I am completely lost and haven't done any character theory or rep theory for a long time.

    (1) Let G be a group of odd order (so |G| ≡ 1 mod 2) and (conjugacy) class number h. Prove that |G| ≡ h mod 16.
    I assume you're acquainted with the basics of rep. theory. One can prove that for non-trivial characters of G the characters of G come in pairs of equal degree, say \chi^{\text{triv}},\chi_1,\chi_1',\cdots,\chi_m,\c  hi_m' say with degress 1,d_1,\cdots,d_m. So, \displaystyle |G|=1+\sum_{j=1}^{m}2d_j^2. But, one has that if \kappa is the number of conjugacy classes then \kappa=2m+1. Now, since |G| is odd one has that each d_j is odd, say equal to 2\gamma_j+1. Then, plugging this into the equation \displaystyle |G|=1+\sum_{j=1}^{m}2d_j^2 will give the desired result.
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    Quote Originally Posted by Drexel28 View Post
    I assume you're acquainted with the basics of rep. theory. One can prove that for non-trivial characters of G the characters of G come in pairs of equal degree, say \chi^{\text{triv}},\chi_1,\chi_1',\cdots,\chi_m,\c  hi_m' say with degress 1,d_1,\cdots,d_m. So, \displaystyle |G|=1+\sum_{j=1}^{m}2d_j^2. But, one has that if \kappa is the number of conjugacy classes then \kappa=2m+1. Now, since |G| is odd one has that each d_j is odd, say equal to 2\gamma_j+1. Then, plugging this into the equation \displaystyle |G|=1+\sum_{j=1}^{m}2d_j^2 will give the desired result.
    It is not necessarily true that irreducible characters of finite group G come in pairs like
    \chi^{\text{triv}},\chi_1,\chi_1',\cdots,\chi_m,\c  hi_m'.

    For example, \mathbb{Z}_2 has two irreducible characters, one of which is the trivial character. The reason why irreducible characters \chi^{\text{triv}},\chi_1,\chi_1',\cdots,\chi_m,\c  hi_m' come in pairs is that the finite group G is of odd order by hypothesis. As my previous post showed, if G is of odd order, the trivial character is the only irreducible character that is real valued, and other irreducible characters are paired.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by TheArtofSymmetry View Post
    It is not necessarily true that irreducible characters of finite group G come in pairs like
    \chi^{\text{triv}},\chi_1,\chi_1',\cdots,\chi_m,\c  hi_m'.

    For example, \mathbb{Z}_2 has two irreducible characters, one of which is the trivial character. The reason why irreducible characters \chi^{\text{triv}},\chi_1,\chi_1',\cdots,\chi_m,\c  hi_m' come in pairs is that the finite group G is of odd order by hypothesis. As my previous post showed, if G is of odd order, the trivial character is the only irreducible character that is real valued, and other irreducible characters are paired.
    Right, which is what I was leaving out for him to conclude. Thus why I said "one can prove"...maybe I should have stated 'you can prove'.
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