# Prove that |G| ≡ h mod 16.

• Mar 14th 2011, 02:54 AM
Turloughmack
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.
• Mar 14th 2011, 04:07 PM
Drexel28
Quote:

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 $\displaystyle G$ the characters of $\displaystyle G$ come in pairs of equal degree, say $\displaystyle \chi^{\text{triv}},\chi_1,\chi_1',\cdots,\chi_m,\c hi_m'$ say with degress $\displaystyle 1,d_1,\cdots,d_m$. So, $\displaystyle \displaystyle |G|=1+\sum_{j=1}^{m}2d_j^2$. But, one has that if $\displaystyle \kappa$ is the number of conjugacy classes then $\displaystyle \kappa=2m+1$. Now, since $\displaystyle |G|$ is odd one has that each $\displaystyle d_j$ is odd, say equal to $\displaystyle 2\gamma_j+1$. Then, plugging this into the equation $\displaystyle \displaystyle |G|=1+\sum_{j=1}^{m}2d_j^2$ will give the desired result.
• Mar 15th 2011, 03:03 AM
TheArtofSymmetry
Quote:

Originally Posted by Drexel28
I assume you're acquainted with the basics of rep. theory. One can prove that for non-trivial characters of $\displaystyle G$ the characters of $\displaystyle G$ come in pairs of equal degree, say $\displaystyle \chi^{\text{triv}},\chi_1,\chi_1',\cdots,\chi_m,\c hi_m'$ say with degress $\displaystyle 1,d_1,\cdots,d_m$. So, $\displaystyle \displaystyle |G|=1+\sum_{j=1}^{m}2d_j^2$. But, one has that if $\displaystyle \kappa$ is the number of conjugacy classes then $\displaystyle \kappa=2m+1$. Now, since $\displaystyle |G|$ is odd one has that each $\displaystyle d_j$ is odd, say equal to $\displaystyle 2\gamma_j+1$. Then, plugging this into the equation $\displaystyle \displaystyle |G|=1+\sum_{j=1}^{m}2d_j^2$ will give the desired result.

It is not necessarily true that irreducible characters of finite group G come in pairs like
$\displaystyle \chi^{\text{triv}},\chi_1,\chi_1',\cdots,\chi_m,\c hi_m'$.

For example, $\displaystyle \mathbb{Z}_2$ has two irreducible characters, one of which is the trivial character. The reason why irreducible characters $\displaystyle \chi^{\text{triv}},\chi_1,\chi_1',\cdots,\chi_m,\c hi_m'$ 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.
• Mar 15th 2011, 11:13 AM
Drexel28
Quote:

Originally Posted by TheArtofSymmetry
It is not necessarily true that irreducible characters of finite group G come in pairs like
$\displaystyle \chi^{\text{triv}},\chi_1,\chi_1',\cdots,\chi_m,\c hi_m'$.

For example, $\displaystyle \mathbb{Z}_2$ has two irreducible characters, one of which is the trivial character. The reason why irreducible characters $\displaystyle \chi^{\text{triv}},\chi_1,\chi_1',\cdots,\chi_m,\c hi_m'$ 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'.