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

1. ## 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.

2. 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.

3. 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.

4. 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'.