Groups of order \(p^3, pq\)

Group and Representation Theory
Alessandro Avellino
a.avellino2@studenti.unipi.it

Today we will discuss groups of order \(p^3\) and \(pq\), where \(p\) and \(q\) are distinct primes. We will explore the structure of these groups, their representations, and we will try to generalize.

Groups of Order \(pq\)

Let's find a non abelian group of order \(pq\). Clearly the groups we are talking about are non abelians, since in that case the irreducible representations would be trivials. So we are going to consider the case in which $q \mid p-1$. In that case. Let's consider the following group:

\[ G=\left\{ \bigl(\begin{smallmatrix} 1 & x \\ 0 & y \\ \end{smallmatrix}\bigr)\mid y \in \left(\mathbb{Z}/(p)\right)^* , x \in \mathbb{Z}/(p)\right\} \]

Since $q\mid p-1$ e $\left(\mathbb{Z}/(p)\right)^*$ is cyclic of order $p-1$, there exists an element $u$ of order $q$ in $\left(\mathbb{Z}/(p)\right)^*$. Thus, if we let:

\[ \sigma = \bigl(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \\ \end{smallmatrix}\bigr), \quad \tau = \bigl(\begin{smallmatrix} 1 & 0 \\ 0 & u \\ \end{smallmatrix}\bigr) \]

It's easy to see that $\tau^{-1}\sigma \tau=\sigma^u$, this implies that $\tau \in N_G(\langle \sigma\rangle )$. Noticing that $\sigma$ is of order $p$ and $\tau$ is of order $q$, this let us conclude that, if $\mathfrak{U}=\langle \sigma, \tau\rangle$, then $\mf=\langle \sigma, \tau \mid \sigma^p=\tau^q=\tau^{-1}\sigma \tau\sigma^{-u}=I\rangle$.
which implies that $\langle \tau, \sigma \rangle$ is a group where all the elements are of the form $\sigma^i\tau^j$
Let $H=\langle \sigma\rangle$ and $K=\langle \tau\rangle$, then $H$ is a normal subgroup of $\mf$ and $K$ is a subgroup of index $p$ in $\mf$.

Conjugacy classes: now we want to find the conjugacy classes of $\mf$. First we take, in $(\mathbb{Z}/(p))^*$, the quotient by $S=\langle u \rangle$, let's call $V=\left\{ v_1,v_2,\ldots,v_r\right\}$, where $r=\frac{p-1}{q}$.
Now let's check the conjugacy classes of $\mf$. First we notice that $Z_{\mf}(\sigma)$ as order multiple of $p$, since $\sigma$ is of order $p$, also it's not the total groups since $\tau \notin Z_{\mf}(\sigma)$, actually non of the element of the form $\sigma ^i \tau^j$ commutates with it, unless $j=0$. So we can conclude that $Z_{\mf}(\sigma)=\langle \sigma \rangle$.
This let us conclude that the conjugacy classes of $\sigma$ is a set of $q$ elements, similarly for $\tau$, and all its powers, its conjugacy class is composed by $p$ elements. So far we said that there are the following conjugacy classes:
for each non trivial power of $\tau$ we got $p$ elements in their conjugacy class, that is $\left\{ \sigma^i\tau^j \mid 0\leq i \leq p-1 \right\}$. Also, the relation between $\sigma$ and $\tau$ gives us that $\tau^{-j}\sigma \tau^j=\sigma^{u^j}$ tells us that $\langle \sigma\rangle $ is partitioned in $U_i=\left\{ \sigma^{v_iu^j}\mid 0\leq j \leq q-1\right\}$.
Now we now the number of conjugacy classes of $\mf$ is $r+q$. Let's take as reppresentatives for the conjugacy classes: $\tau^j, \sigma^{v_i}, 0\leq i \leq r-1, 0\leq j \leq q-1$.

Linear representations: Let's count the number of one-dimensional irreducible representations of $\mf$. As usual, those are in one-to-one correspondence with the irreducible representations of the quotient gorup $\mf/[\mf, \mf]$. It's easy to see that $[\mf, \mf]=\langle \sigma\rangle $. In fact since the quotient is abelian, the commutator must lie under $H$, but it hasn't any proper subgroup, so it is itself.
This quotient is $\mathbb{Z}/(q)$, so there are exactly $q$ one-dimensional irreducible representations of $\mf$.
Those are the ones that maps $\tau H$ to a root of unity, so the ones of $\mf$ are:

\begin{align*} \theta_n: \sigma^i\tau^j \mapsto \zeta_p^{nj} \end{align*}

where $0\leq n \leq q-1$.

Induced representations: Now we want to prove the other irreducible representations of $\mf$ are the ones induced by $H$. Let's call $\zeta=\zeta_p$ to simplify notation.
Let's consider the representations $\omega_k$ which send $\sigma^j \mapsto \zeta^{jk}$. Let's call $\beta_k$ the induced representations and $\chi_{\beta_k}$ the characters of them.
Noticing that $\tau^j$ are representatives for cosets of $H$, we want to show that those are irreducible representations of $\mf$. Let's remind that the induced characters are definied as follows:

\[ \chi_{\beta_k}(\tau^j)=\sum_{h\in \mf/H}\omega_k(h^{-1}\tau^j h)=\sum_{h\leq q-1} \omega_k(\tau^h\tau^j\tau^{-h})=\sum_{h\leq q-1} \omega_k(\tau^j) \]

And this is equalt to $0$. Now if we consider the conjugacy class of $\sigma^{v_i}$:

\[ \chi_{\beta_k}(\sigma^{v_i})=\sum_{h\in \mf/H}\omega_k(h^{-1}\sigma^{v_i} h)=\sum_{h\leq q-1} \omega_k(\tau^h \sigma^{v_i} \tau^{-h})=\sum_{h\leq q-1} \omega_k(\sigma^{v_i\cdot u^{-h}}) \]

Which lead as to:

\[ \chi_{\beta_k}(\sigma^{v_i})=\sum_{h\leq q-1} \zeta^{v_i\cdot u^{-h}\cdot k}=\sum_{h\leq q-1} (\zeta^{v_i\cdot k})^{u^{-h}}=\sum_{h\leq q-1} (\zeta^{v_i\cdot k})^{u^h} \]

We notice that if $k=0$, $\omega_0$ is the trivial representation, and the induced does $q$ on every power of $\sigma$. And if we do $\langle \chi_{\beta_0}, \chi_{\theta_n} \rangle_{\mf}=1 $ which implies that it's not irreducible. Moreover notice that is $k'=k\cdot u^l$, since it is just a permutation of the sum, then the representations are the same, in particular they have the same character.
So we are interested in the case $k \in \left\{v_1,\ldots, v_r\right\}$.
Now we want to show that the representations $\beta_k$ are irreducible.
Using Frobenius reciprocity, we have:

\[ \langle Res_{\mf}^H(\beta_{v_j}), \omega_{v_j s}\rangle_H=\langle \beta_{v_j}, Ind_H^{\mf}(\omega_{v_j s})\rangle_{\mf}=\langle \beta_{v_j}, \beta_{v_j}\rangle_{\mf} \]

Then $Res_{\mf}^H(\beta_{v_j})=\sum_{s \in S} \langle \beta_{v_j}, \beta_{v_j}\rangle_{\mf} \omega_{v_j s} +\chi$
where $\chi$ is some other character of $H$ possibly nothing.
Considering the dimensions, we have:

\[ q=\langle \beta_{v_j}, \beta_{v_j}\rangle_{\mf} \cdot q + \chi(1) \]

This implies that $\langle \beta_{v_j}, \beta_{v_j}\rangle_{\mf} = 1$, so the representations are irreducible.
Now we have $q$ one-dimensional irreducible representations and $r$ irreducible representations of dimension $p$. we reached the number of the conjugacy classes, which is $r+q$, so we found all the irreducible representations.
For the sake of completeness notice that the sum of the squares of the dimensions of the irreducible representations is equal to the order of the group, that is $pq$.

Explicit Characters: The characters of the irreducible representations are:

\begin{align*} \chi_{\theta_n}(\sigma^i\tau^j) &= \zeta_p^{nj} \quad \text{for } 0\leq n \leq q-1, 0\leq i \leq p-1 \end{align*}

Where $\theta_n$ are the one-dimensional irreducible representations.
The characters of the irreducible representations induced by $H$ are:

\[ \chi_{\beta_k}(\sigma^l \tau^j)= \left\{\begin{matrix} \sum_{i\leq q-1} \zeta_p^{lu^iv_k} & j=0 \\ 0 & otherwise \\ \end{matrix}\right. \]

Can we generalize:
The only thing we used about $q$ being prime is that the non-abelian groups of order $pq$ is unique up to isomorphism. But if we want to find the representations of the following group:

\[ G_{p,q}=\langle a,b \mid a^p, b^q, bab^{-1}=a^u\rangle \]

then we can reproduce the same arguments, obtaining though the irreducible representations.

Some Examples:
If $q=2$ is the already studied dihedral group.
IndeEd another nice example is when $q=p-1$. In this case $r=\frac{p-1}{q}=1$, so there are $p$ irreducible representations.
For example, by substiting $p$ with $5$, we get the following character table:

\[ \begin{array}{c|ccccc} \text{Classi} & C_1 = \{ id \} & C_2 = \{\sigma\} & C_3 = \{ \tau \} & C_4 = \{ \tau^2 \} & C_5 = \{ \tau^3\} \\ \text{Card.} & 1 & 4 & 5 & 5 & 5\\ \hline triv & 1 & 1 & 1 &1 & 1\\ \chi_1 & 1 & 1 & i & -1 & -i\\ \chi_2 & 1 & 1 & -1 & 1 & -1\\ \chi_3 & 1 & 1 & -i & -1 & i\\ \phi & 4 & -1 & 0 & 0 & 0 \end{array} \]
Table 1: \( G_{5,4}\)

Some Application:
We want to show that $\mathbb{Q}(\zeta_p)$ contains $\sqrt{\delta_p p}$, where

\[ \delta_p=\left\{\begin{matrix} 1 & p\equiv 1 \mod 4 \\ -1 & p\equiv 3 \mod 4 \\ \end{matrix}\right. \]

let $q=\frac{p-1}{2}$. Let $u$ having order $q$ in $(\mathbb{Z}/(p))^*$. By cardinality, the only elements of order $q$ are the square of the element of order $p-1$.
This implies that $-1$ is a power of $u$ if and only if $-1$ is a square modulo $p$, if and only if $p \equiv 1 \mod 4$.
Considering $G_{p,q}$ as before, $a$ is conjugate to all the elements of the form $a^{u^k}$. So $a$ and $a^{-1}$ are conjugate if and only if $p \equiv 1 \mod 4$.
Since $(p-1)/q=2$, there are $2$ non linear characters, let's call them $\phi_1$ and $\phi_2$. For what we said before, we know that for all the linear characters, $\chi_i(a)=1$.
We also know the following:

  1. $\sum_{\chi irr} \chi(1)\chi(g) = 0$, in fact this sum is equal to $\chi_{\mathbb{C}G_{p,q}}(g)=0$ for $g \neq id$
  2. $\sum_{\chi irr} \chi(g)\overline{\chi(g)} = \mid G_{p,q}/Cl_{G_{p,q}}(g) \mid$, from column orthogonality

Applying this to $a$, we get that $\phi_1(a)+\phi_2(a)=-1$ and that $\phi_1(a)\overline{\phi_1(a)}+\phi_2(a)\overline{\phi_2(a)}=\frac{p+1}{2}$
If $a$ and $a^{-1}$ are conjugates, this implies that $\chi(a)$ is real for all characters, in particular we get that $2\phi_1(a)=-1+\sqrt{p}$ and $2\phi_2(a)=-1-\sqrt{p}$
In a similar way if they are not conjugates, then $\phi_1(a)$ and $\phi_2(a)$ are both non real or both real, since they have the same imaginary part. But they cannot be both real, otherwise we would have that for each irreducible representation, $\chi(a)=\chi(a^{-1})$, remembering that the irreducible representations generate the class function,
let's say $f:G \mapsto \mathbb{C}$ which sends every conjugacy class into 0 except for the one of $a$. This is clearly a class function, which implies that is generated by the irreducible characters, though this implies that $f(a)=f(a^{-1})$ but it's a contraddiction since as hypothesis we said that they aren't in the same conjugacy class.
So they are both complexes, and $\phi_1(a)=\overline{\phi_2(a)}$ which implies that $\phi_1(a)=\frac{-1+\sqrt{-p}}{2}$
Finally, as we said before, the induced representations $\phi_1(a)=\sum_{l\leq q} \zeta_p^{u^l}=\sum_{k \in Sq} \zeta_p^{k}$, where $Sq$ is the set of square residues modulo $(\mathbb{Z}/(p))^*$. This let us conclude what we were seeking for.

Groups of Order \(p^3\)

As before, we eclude the trivial cases of abelian groups, which are $C_{p^3}, (C_p)^3, C_p\times (C_p)^2$.

The existence of H:
First of all, we claim that every group of order \(p^3\) has an abelian subgroup of order $p^2$. Since it's not abelian, the centre is not the whole group, as it's not trivial, and moreover it hasn't order $p^2$, otherwise the quotient group would be cyclic.
So it has order $p$ and the quotient is $C_p\times C_p$. In the quotient we consider one of the subgroups of order $p$, rising it to the preimage thanks to correspondenc theorem we are sure that the subgroup is of order $p^2$. Moreover is abelian since it has order $p^2$.
Now, let's call the centre $Z$ and the subgroup of order $p^2$ $H$. Since the quotient of $Z$ is abelian, and it has no non-trivial subgroups, we can conclude that $Z$ is the commutator.

The linear representations of the groups: As before, the linear representations of the groups are the ones of the abelianized group, so they are $p^2$ and they are one-dimensional.
Now we want to show that the other irreducible representations of the group are the ones induced by $H$.

The induced representations:
We know the following:

Moreover they cannot be one-dimensional since we already counted them, but they can't even have dimension $p^2$, since the sum of the squares of even one addend equal $p^4$ would exceed the order of the group, so the only possible dimension is $p$.
The group $H$ is abelian, so it's irreducible representations are one-dimensional, so the induced representations are of dimension $p$.

We want to prove that the induced representations are irreducible. We notice that if $\psi$ is a representation of dimension $p$ in $G$, then it is either irreducible or sum of linear representations. But this would implies that $Z The one-dimensional representations of $H$ are in total $p^2$, but from those we have to exclude the ones that are trivial on $Z$, which are $p$ in total (since they are in correspondence with the ones of $H/Z$), so there are $p^2-p$ one-dimensional representations of $H$. But not all of them are different when induced.
Actually, let $\psi_1, \psi_2$ be two one-dimensional representations of $H$, such that the induced representations are equal. By Frobenius reciprocity, we have:

\[ 1=\langle Ind_H^G(\psi_1), Ind_H^G(\psi_2)\rangle_G=\langle Res_G^H(Ind_H^G(\psi_1)), \psi_2\rangle_H \]

So there can be utmost $p$ different irreducible representations of $H$ that induce the same representations for each one. Which implies that the number of different irreducible representations of $G$ that we are seeking are at least $p-1$, but this is also an upperbound, since all of them is p-dimensional, and as we said before, the remaining space for them is $p^3-p^2=p^2(p-1)$, which gives as the limit we were looking for.

Explicit Characters:
The linears characters induced by the quotient for the centre, since it is $C_p\times C_p=\langle aZ, bZ \rangle $, are

\begin{align*} \chi_{ij}(a^nb^mz^l)=\zeta_p^{in}\zeta_p^{jm} \end{align*}

To calculate the characters of the remaining representations, we write $H=\langle a,z\rangle$. Clearly this gruop is an abelian group of order $p^2$, if it were the whole group it would have implied the abelianity of the whole group itself.
Now as we said before we do not want that those representations are trivial on $Z$, so let's consider $\psi_u(z^t)=\zeta_p^{ut}$. For every power of $a$ that doesn't lie on the centre, $Cl(a^r)=\left\{z^ta^r \mid 0\leq t < p \right\}$
So $Ind(\psi_u)(a^r)=\psi_u(a^r)+\psi_u(za^r)+\dots+\psi_u(z^{p-1}a^r)=\psi_u(a^r)\cdot \sum \psi_u(z^t)=\psi_u(a^r)\cdot \sum \zeta_p^{ut}=0$, in fact $u While $Ind(\psi_u)(z^t)=p\zeta_p^{ut}$ and $Ind(\psi_u)(g)=0$ for each element not in $H$. Therefore the characters we got are the following:

\[ \psi_u(a^nb^mz^l)= \left\{\begin{matrix} p\zeta_p^{ul} & n=m=0 \\ 0 & otherwise \\ \end{matrix}\right. \]

Can we generalize this?
Yes, we can generalize this to groups of order \(p^n\), but we need a condition, the existence of an abelian subgroup of index \(p\) in $G$.
In fact, let $H$ be such subgroup, in this case it's easy to prove the existence of a subgroup of order $p$ which is containted in $K=H\cap Z \cap [G,G]$.
Now $K$ takes the role of $Z$ in the previous case. We lift the irriducible representations of $G/K$, which correspond to the ones which have $K$ in the kernel.
And the sum of the squares of the dimensions of the irreducible representations lifted like this is equal to $\mid G/K \mid = p^{n-1}$. The remaining space as before is $p^n-p^{n-1}=p^2(p^{n-2}-p^{n-3})$. Again the claim is that the other irreducible representations are the ones induced by the subgroup of order $p^{n-1}=H$, which is abelian, so they are all linear, from which we exclude the ones that have $K$ in the kernel, which are $p^{n-2}$. After induced will become of dimension $p$ and will be partiioned as subset of p as before
givinig a lower bound which coincides with the upperbound we gave before.