I would like the write down the argument of $\S 8.2$ of Serre’s book Linear Representations of Finite Groups.

Let $A$ and $H$ be two subgroups of finite group $G$, with $A$ normal and abelian. $G$ is the semidirect product of $H$ by $A$ (i.e. $G=AH$ and $A\cap H={1}$, or in other words, each element of $G$ can be written uniquely as product of $ah$). Irreducible representations of $G$ can then be constructed from those of certain subgroups of $H$, as follow:

Since $A$ is abelian, its irreducible characters are of degree $1$ (i.e. dimension of the space is $1$) and form a group $X=\text{Hom }(A,\mathbf{C}^*)$. The group $G$ acts on $X$ by

\[s_{\chi}(a)=\chi(s^{-1}as), s\in G,\chi\in X, a\in A.\]

Let $(\chi_i)_{i\in X/H}$ be a system of representatives of for the orbits of $H$ in $X$. For each $i\in X/H$, let $H_i$ be the subgroup of $H$ consisting of those elements $h$ such that $h\chi_i=\chi_i$, and let $G_i=A\cdot H_i$ the corresponding subgroup of $G$. Extend the function $\chi_i$ to $G_i$ by setting

\[\chi_i(ah)=\chi_i(a) \; \text{for }a\in A,h\in H_i.\]

Using the fact that $h\chi_i=\chi_i$ for $h\in H_i$, we see that $\chi_i$ is a character of degree $1$ of $G_i$.

Proof. It suffices to show $\chi_i:G_i\to \mathbf{C}^*$ is a group homomorphism. Note since $A$ normal so $t^{-1}at\in A$ for all $t\in G,a\in A$. We have for $ah,a’h’\in G_i$ then

\[\begin{aligned} \chi_i((ah)(a'h')) & = \chi_i(aha'h^{-1}hh'), \\ & = \chi_i (aha'h^{-1}), \\ & = \chi_i(a)\chi_i(ha'h^{-1}), \\ & = \chi_i(a)(h\chi_i)(a'), \\ & = \chi_i(a)\chi_i(a'), \; (h\in H_i) \\ & = \chi_i(ah)\chi_i(a'h'). \end{aligned}\]

Now let $\rho$ be an irreducible representation of $H_i$. By composing $\rho$ with the canonical projection $G_i\to H_i$ we obtain an irreducible representation $\tilde{\rho}$ of $G_i$. Finally, by making the tensor product of $\chi_i$ and $\tilde{\rho}$ we obtain an irreducible representation $\chi_i\otimes \tilde{\rho}$ of $G_i$. Note since $\chi_i:G_i\to \mathbf{C}^{\times}$ so one can view $\chi_i\otimes \tilde{\rho}$ as $g \mapsto\chi_i(g)\tilde{\rho}(g)$ (where $\chi_i(g)\in\mathbf{C}^{\times}$ and $\tilde{\rho}(g):V\to V$ where $V$ representation space of $\rho$). Let $\theta_{i,\rho}$ be the corresponding induced representation of $G$.

Proposition 25.

  1. $\theta_{i,\rho}$ is irreducible.
  2. If $\theta_{i,\rho}$ and $\theta_{i’,\rho’}$ are isomorphic, then $i=i’$ and $\rho$ is isomorphic to $\rho’$.
  3. Every irreducible representation of $G$ is isomorphic to one of $\theta_{i,\rho}$.

(1) is proved using Mackey’s criterion ($\S 7.4$ of Serre’s book), as follow: Let $s\not\in G_i$ and let $K_s=G_i\cap sG_is^{-1}$. We have to show that if we compose the representation $\chi_i\otimes \tilde{\rho}$ of $G_i$ with the two injections $K_s\to G_i$ defined by $x\mapsto x$ and $x\mapsto s^{-1}xs$, we obtain two disjoint representations of $K_s$ (i.e. they have no common subrepresentations). To do this, it is enough to check that the restrictions of these representations to subgroup $A$ of $K_s$ are disjoint (since common subrepresentation of $K_s$ implies common subrepresentation of $A$).

But the first restricts to a multiple of $\chi_i$ and the second to a multiple of $s\chi_i$; since $s\not\in G_i=A\cdot H_i$ we have $s\chi_i\ne \chi_i$ and so the two representations in question are indeed disjoint.

Proof. Consider the first restriction $A\to K_s\to G_i$ by $x\mapsto x$. Then for $x\in A$, we have $\tilde{\rho}(a)=\rho(1)=\text{id}_V$ so

\(x\mapsto (\chi_i\otimes \tilde{\rho})(x) =\chi_i(x)\text{id}_V.\) Consider the second restriction $x\mapsto s^{-1}xs$. If $x\in A$ then $s^{-1}xs\in A$ so similarly, we have $\tilde{\rho}(s^{-1}xs)=\text{id}_V$ so

\[x \mapsto \chi_i(s^{-1}xs)\text{id}_V=(s\chi_i)(x)\text{id}_V\]

Now we prove (2):

The restriction of $\theta_{i,\rho}$ to $A$ only involves characters $\chi$ belonging to the orbit $H\chi_i$ of $\chi_i$. This shows that $\theta_{i,\rho}$ determines $i$.

As the statement said, we consider character $\chi^i$ of $\theta_{i,\rho}$ for $a\in A$. Recall the character for induced representation for $s\in G$, $H$ subgroup of $G$, $f:H\to \mathbf{C}^{\times}$, is

\(\text{Ind}_H^G(f)(s)=f'(s)=\frac{1}{|H|}\sum_{t\in G, t^{-1}st\in H}f(t^{-1}st)\) Hence,

\[\begin{aligned} \chi^i(a) & = \frac{1}{|G_i|}\sum_{t\in G, t^{-1}at\in G_i} \chi_{\chi_i\otimes \tilde{\rho}}(t^{-1}at), \\ & = \frac{1}{|G_i|}\sum_{t\in G} \chi_i(t^{-1}at)\chi_{\tilde{\rho}}(t^{-1}at), \; (A \text{ normal so } t^{-1}at\in A\subset G_i \; \forall t\in G), \\ & = \frac{\text{dim }V}{|G_i|}\sum_{t\in G}\chi_i(t^{-1}at), \; (t^{-1}at\in A \Rightarrow \tilde{\rho}(t^{-1}at)=\rho(1)=\text{id}_V), \\ & = \frac{\text{dim }V}{|G_i|} \sum_{a'\in A,h\in H}\chi_i((a'h)^{-1}a(a'h)),\\ & = \frac{\text{dim }V}{|G_i|} \sum_{a'\in A,h\in H} \chi_i(h^{-1}ah), \; (A \text{ abelian}) \\ & = \frac{|A|\text{dim }V}{|G_i|} \sum_{h\in H} (h\chi_i)(a). \end{aligned}\]

We would like to show that $\chi^i\ne \chi^{i’}$ for $i\ne i’$. Observe that $|G_i|$ divides $|G|$ and that $h\chi_i\in \text{Hom }(A,\mathbf{C}^*)$ is $1$-dimensional irreducible representation of $A$ so the character defined by $\tau^i(a)=\frac{|G|}{|G_i|}\sum_{h\in H}(h\chi_i)(a)$ corresponds to direct sum of representations from $H\chi_i$. Since with $i\ne i’$ then $H\chi_i$ and $H\chi_{i’}$ are disjoint so $\tau^i\ne \tau^{i’}$, implying $\chi^i\ne \chi^{i’}$.

Let $W$ be the representation space for $\theta_{i,\rho}$, let $W_i$ be the subspace of $W$ corresponding to $\chi_i$ (the set of $x\in W$ such that $\theta_{i,\rho}(a)x=\chi_i(a)x$ for all $a\in A$). The subspace $W_i$ is stable under $H_i$ and the representation of $H_i$ in $W_i$ is isomorphic to $\rho$; whence $\theta_{i,\rho}$ determines $\rho$.

Recall from $\S 7.1$ of Serre’s book, $W$ can be identified with $W=\mathbf{C}[G]\otimes_{\mathbf{C}[G_i]}V$ where $V$ is the representation space of $\chi_i\otimes \tilde{\rho}$. In particular, this is a $\mathbf{C}[G]$-module where $g$ acts on $g’\otimes v$ by $(gg’)\otimes v$. Also note that, $(gg_i)\otimes v$ and $g\otimes (\chi_i\otimes \tilde{\rho})(g_i)v$ are considered the same in $W$ for $g_i\in G_i$.

Now, we identify $W_i$. For $a\in A, g\otimes v\in W$, we have $\theta_{i,\rho}(a)(g\otimes v)= (ag)\otimes v=g\otimes(g^{-1}ag)v$. Since $g^{-1}ag\in A$ so $(\chi_i\otimes \tilde{\rho})(g^{-1}ag)v=\chi_i(g^{-1}ag)v$. Thus, $\theta_{i,\rho}(a)(g\otimes v)=\chi_i(g^{-1}ag)(g\otimes v$. Thus, in order for $\theta_{i,\rho}(a)(g\otimes v)=\chi_i(a)(g\otimes v)$ for all $a\in A$, we must have $\chi_i(a)=(g\chi_i)(a)$ for all $a\in A$. This follows $g\in H_i$. Thus, $W_i=\mathbf{C}[H_i]\otimes_{\mathbf{C}[G_i]} V$.

With this, it is obvious that $W_i$ is stable under $H_i$. Furthermore, observe $W_i$ is spanned by $1\otimes v_j$ where $v_j$’s basis of $V$, $1$ identity in $H_i$ (this holds since $h\otimes v=1\otimes \rho(h)v$ for all $h\in H_i,v\in V$). Hence, one can easily construct isomorphism representation of $H_i$ in $W$ to $\rho$, as desired.

To show (3), it suffices to show that

\[\sum_{i\in X/H}\sum_{\rho}(\text{dim } \theta_{i,\rho})^2=|G|\]

where $\rho$ is summed over all irreducible representations of $H_i$. We have

\[\begin{aligned} \text{dim }\theta_{i,\rho} &=\text{dim }\text{Ind}_{G_i}^G (\chi_i\otimes \tilde{\rho}),\\ & = \frac{|G|}{|G_i|}\text{dim } (\chi_i\otimes \tilde{\rho}), \\ & = \frac{|G|}{|G_i|}\text{dim }\rho \end{aligned}\]

so for fixed $i$ then

\[\sum_{\rho}(\text{dim }\theta_{i,\rho})^2 = \frac{|G|^2}{|G_i|^2}|H_i|= \frac{|H|^2}{|H_i|}.\]

Hence, it suffices to show that $\sum_{i\in X/H}\frac{|H|^2}{|H_i|}=|G|$ or $\sum_{i\in X/H}\frac{|H|}{|H_i|}=|A|$. Note that $X=\text{Hom }(A,\mathbf{C}^*)$ is isomorphic to $A$ (see Conrad’s note) so

\[|A|=|X|=\sum_{i\in X/H}|H\chi_i|= \sum_{i\in X/H} \frac{|H|}{|H_i|}.\]

We are done.