Prove that |G| ≡ h mod 16.

**Turloughmack** · March 14th 2011, 02:54 AM

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.

**Drexel28** · March 14th 2011, 04:07 PM

Originally Posted by Turloughmack

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.

**TheArtofSymmetry** · March 15th 2011, 03:03 AM

Originally Posted by Drexel28

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.

**Drexel28** · March 15th 2011, 11:13 AM

Originally Posted by TheArtofSymmetry

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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