\(\newcommand{\ind}[1]{{\uparrow^#1}}\newcommand{\res}[1]{{\downarrow_#1}}\newcommand{\gen}[1]{\langle #1\rangle}\)The restriction map sends \(kG\)-modules to \(kH\)-modules. There is a map going in the other direction, sending \(kH\)-modules to \(kG\)-modules, called induction.

The clean, formal way to define it is to state that if \(V\) is a \(kH\)-module then the induced module, \(V\ind G\), is defined to be \(V\otimes_{kH} kG\). Unfortunately, we have only defined tensor products over a field, not over a group algebra, so we have to explicitly construct this module.

4.1.1 INDUCED CHARACTERS

We will start with a simplified situation to help motivate the construction we are about to do. Suppose that \(N\) is a normal subgroup of a finite group \(G\), and that there is a set \(T\) of coset representatives (a transversal) for \(N\) in \(G\) that form a subgroup (so \(G\) is a split extension or semidirect product). Let \(W\) be a \(kN\)-module, and let \(V\) be a direct sum of \(|G:H|\) copies of \(W\), labelled \(W_t\) for \(t\in T\). We want to turn \(V\) into a \(kG\)-module. If \(w\in W\) then write \(w_t\) for the corresponding element of \(W_t\) for \(t\in T\).

Let \(H\) be a subgroup of \(G\) of index \(n\), and let \(W\) be a \(kH\)-module. Let \(t_1,\dots,t_n\) denote a right transversal to \(H\) in \(G\). If \(g\in G\), then there exists a function \(\sigma:\{1,\dots,n\}\to\{1,\dots,n\}\) and elements \(h_i\in H\) for \(1\leq i\leq n\) such that\[ t_ig=h_it_{i\sigma},\]since the \(t_i\) form a right transversal. Note that \(h_i=t_igt_{i\sigma}^{-1}\).

Exercise 4.1 asks you to prove that \(\sigma\) is a permutation of \(\{1,\dots,n\}\), and that the map \(g\mapsto \sigma_g\) is a homomorphism.

We construct a module \(V\) of dimension \(n\cdot \dim(W)\) as follows:

1. (i) Let \(W_i\) for \(i=1,\dots,n\) be a collection of \(kH\)-modules, each isomorphic to \(W\) under fixed isomorphisms. Let \(V=W_1\oplus \cdots \oplus W_n\).
2. (ii) Fix some \(w\in W\), and let \(w_i\in W_i\) be the image of \(w\) in \(W_i\) under the fixed isomorphisms above.
3. (iii) Define \(w_ig\) to be \(w_{i\sigma}(t_igt_{i\sigma}^{-1})\), which is well defined as \(t_igt_{i\sigma}^{-1}\) is an element of \(H\) and \(W_{i\sigma}\) is a \(kH\)-module.

We should check that this is a \(kG\)-module. Let \(g,g'\in G\), and let \(\sigma,\sigma'\) be the corresponding permutations of \(\{1,\dots,n\}\). We have that
\begin{align*} (w_ig)g'&=\left(w_{i\sigma}(t_igt_{i\sigma}^{-1})\right)g'
\\ &=w_{i\sigma\sigma'}(t_igt_{i\sigma}^{-1})(t_{i\sigma}g't_{i\sigma\sigma'}^{-1})
\\ &=w_{i\sigma\sigma'}(t_igg't_{i\sigma\sigma'}^{-1})
\\ &=w_i(gg').\end{align*}
Hence \(V=W\ind G\) is a \(kG\)-module. It has dimension \(|G:H|\dim(W)\).

If \(\chi\) is a character for \(H\) with corresponding module \(W\), we write \(\chi\ind G\) for the character of the induced module \(W\ind G\).

Proposition 4.1. Let \(H\) be a subgroup of \(G\) and let \(\chi\) be a character of \(H\). Define \(\hat{\chi}:G\to \mathbb{C}\) (not a class function) by\[ \hat\chi(g)=\begin{cases}\chi(g)&g\in H,\\0&\text{otherwise}.\end{cases}\]The character of \(\chi\ind G\) is given by
\[\chi\ind G(g)=\frac{1}{|H|} \sum_{x\in G} \hat\chi(g^x).\]Equivalently, if \(T\) denotes a right transversal to \(H\) in \(G\), then
\[ \chi\ind G(g)=\sum_{t\in T} \hat\chi(tgt^{-1}).\]
Proof: Let \(g\in G\), let \(T=\{t_1,\dots,t_n\}\) denote a right transversal to \(H\) in \(G\), and let \(\sigma:\{1,\dots, n\}\to\{1,\dots,n\}\) denote the function defined by \(t_i g=h_it_{i\sigma}\), where \(h_i\in H\). In the construction of \(W\ind G\), we compute the contribution to the trace of \(g\) on \(W\ind G\) from each \(W_i\).

Let \(\{w_1^i,\dots,w_r^i\}\) denote a basis of \(W_i\). We have that \(w_j^i g=w_j^{i\sigma}(t_igt_{i\sigma}^{-1})\). Since we are looking for the trace, the image of \(w_j^i\) needs to be in \(W_i\) itself (or else the diagonal entry cannot possibly be non-zero) and so \(i\sigma=i\).

We claim that \(i\sigma=i\) if and only if \(g^{t_i}\in H\). We see that \(i\sigma=i\) if and only if \(t_ig=h_it_i\), and in other words, \(t_igt_i^{-1}=h_i\).

This then means that \(w_j^ig\in W_i\) for all \(1\leq j\leq r\), and in fact the action of \(g\) on \(W_i\) is the same as the action of \(h_i\) on \(W\). Hence the trace of \(g\) on \(W_i\) is the trace of \(h_i\) on \(W\), so \(\chi(t_igt_i^{-1})\).

We have therefore seen that the value of \(\chi\ind G\) at \(g\) is the sum over all elements \(t\) in a transversal \(T\) to \(H\) in \(G\) such that \(tgt^{-1}\in H\), of the value \(\chi(tgt^{-1})\). Extending \(\chi\) to \(\hat\chi\) by setting \(\hat\chi(g)=0\) outside of \(H\), we can restate this as
\[ \chi\ind G(g)=\sum_{t\in T} \hat\chi(tgt^{-1}).\]This is the second of our two equations.

To see the equivalence with the first equation, let \(g\in G\) and \(t\in T\) such that \(tgt^{-1}\in H\), and let \(h\in H\). We see that \((ht)g(ht)^{-1}=(tgt^{-1})^{h^{-1}}\in H\), and hence \(\chi(tgt^{-1})=\chi((ht)g(ht)^{-1})\). Thus \(\hat\chi(tgt^{-1})=\hat\chi(g^{(ht)^{-1}})\) if \(tgt^{-1}\in H\). If \(tgt^{-1}\not\in H\) then \(h(tgt^{-1})h^{-1}\not\in H\) either, since conjugation by \(h^{-1}\) is a bijection on \(H\), so must be a bijection on \(G\setminus H\). Hence in all cases\[ \hat\chi(tgt^{-1})=\hat\chi(htgt^{-1}h^{-1}).\]Thus we can multiply by all \(h\in H\) to get\[ |H|\chi\ind G(g)=\sum_{t\in T}\sum_{h\in H} \hat\chi(htgt^{-1}h^{-1})=\sum_{x\in G}\hat\chi(g^x),\]by writing \(x=(ht)^{-1}\). Dividing by \(|H|\) yields the first equation. □

4.1.2 FROBENIUS RECIPROCITY

Frobenius reciprocity is easy to state: if \(\chi\) is a character of \(G\) and \(\psi\) is a character of \(H\), then\[\gen{\psi,\chi\res H}=\gen{\psi\ind G,\chi}.\]It offers the moral justification for writing induction and restriction as similar but opposite symbols. It also allows us to understand the expression of induced characters in terms of restrictions, which we can suppose that we understand.

Although we state this only for \(k=\mathbb{C}\), because we use the inner product and we will prove it with character theory rather than module theory, there is a version that exists for any field \(k\), called the Nakayama relations. We won't prove that here, as the version over \(\mathbb{C}\) is enough for our needs. We will use it later in this chapter to prove a theorem from finite group theory, one of the main applications of the basic representation theory of finite groups to straight group theory.

Theorem 4.2 (Frobenius reciprocity). Let \(H\) be a subgroup of \(G\), let \(\chi\) be a character of \(G\) and let \(\psi\) be a character of \(H\). Then\[ \gen{\psi,\chi\res H}=\gen{\psi\ind G,\chi}.\]
Proof: The left-hand side is given by\[ \frac{1}{|H|}\sum_{h\in H} \psi(h)\overline{\chi(h)}.\]The right-hand side is given by\[ \frac{1}{|G|}\sum_{g\in G} \frac{1}{|H|}\sum_{x\in G} \hat{\psi}(g^x)\overline{\chi(g)}.\]These look completely different, so we will have to do some manipulation to get them to look the same. We note that

1. (i) for fixed \(x\in G\), \(g^x\) runs over all elements of the group as \(g\) does,
2. (ii) \(\chi(g^x)=\chi(g)\) for all \(g,x\in G\), and
3. (iii) \(\hat\psi(g)=0\) for \(g\not\in H\).

Using these facts one at a time, we may write
\begin{align*} \gen{\psi\ind G,\chi}&=\frac{1}{|G|\cdot |H|}\sum_{g\in G}\sum_{x\in G} \hat{\psi}(g^x){\chi(g^{-1})}
\\&=\frac{1}{|G|\cdot |H|}\sum_{g\in G}\sum_{x\in G} \hat{\psi}(g^x){\chi((g^{-1})^x)}
\\&=\frac{1}{|G|\cdot |H|}\sum_{g\in G}\sum_{x\in G} \hat{\psi}(g)\chi(g^{-1})
\\&=\frac{1}{|H|}\sum_{g\in G} \hat{\psi}(g)\chi(g^{-1})
\\&=\frac{1}{|H|}\sum_{h\in H} \psi(h)\chi(h^{-1})
\\&=\frac{1}{|H|}\sum_{h\in H} \psi(h)\overline{\chi(h)}=\gen{\psi,\chi\res H}.
\end{align*}
This completes the proof. □

Example 4.3. We have the character table of \(\mathrm{PSL}_2(7)\) from Example 3.13. It is below.

	\(1\)	\(x_2\)	\(x_3\)	\(x_4\)	\(x_5\)	\(x_6\)
\(\chi_1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)
\(\chi_2\)	\(3\)	\(-1\)	\(0\)	\(1\)	\(\alpha\)	\(\bar \alpha\)
\(\chi_3\)	\(3\)	\(-1\)	\(0\)	\(1\)	\(\bar \alpha\)	\(\alpha\)
\(\chi_4\)	\(6\)	\(2\)	\(0\)	\(0\)	\(-1\)	\(-1\)
\(\chi_5\)	\(7\)	\(-1\)	\(1\)	\(-1\)	\(0\)	\(0\)
\(\chi_6\)	\(8\)	\(0\)	\(-1\)	\(0\)	\(1\)	\(1\)

The group \(G=\mathrm{PGL}_2(7)\) has \(H=\mathrm{PSL}_2(7)\) as a subgroup of index \(2\). We can induce the characters from \(H\) up to \(G\). There are nine classes: five inside \(H\) (as \(g_5\) and \(g_6\) are conjugate in the overgroup) and four classes \(h_1,h_2,h_3,h_4\) outside \(H\), with elements of orders \(2\), \(6\), \(8\) and \(8\), and class sizes 28, 56, 42 and 42. For subgroups of index \(2\) where there is a complement we saw at the start of this chapter that induction is much nicer, as the transversal can be chosen to be \(\{1,h_1\}\).

Frobenius reciprocity and our description of restriction to a subgroup of index \(2\) (Proposition 2.29 and Theorem 2.31) shows that \(\chi_2\) and \(\chi_3\) induce up to the same, irreducible character, and all other characters induce up to the sum of an irreducible character and the product with \(\psi_2\), the non-trivial linear character of \(G\). Also, the induction of a character from \(H\) to \(G\) is \(0\) outside of \(H\), by the formula. This already gives us the following partial table.

	\(1\)	\(x_2\)	\(x_3\)	\(x_4\)	\(x_5\)	\(h_1\)	\(h_2\)	\(h_3\)	\(h_4\)
\(\psi_1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)
\(\psi_2\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(-1\)	\(-1\)	\(-1\)	\(-1\)
\(\psi_3\)	\(6\)	\(-2\)	\(0\)	\(2\)	\(-1\)	\(0\)	\(0\)	\(0\)	\(0\)
\(\psi_4\)	\(6\)	\(2\)	\(0\)	\(0\)	\(-1\)
\(\psi_5\)	\(6\)	\(2\)	\(0\)	\(0\)	\(-1\)
\(\psi_6\)	\(7\)	\(-1\)	\(1\)	\(-1\)	\(0\)
\(\psi_7\)	\(7\)	\(-1\)	\(1\)	\(-1\)	\(0\)
\(\psi_8\)	\(8\)	\(0\)	\(-1\)	\(0\)	\(1\)
\(\psi_9\)	\(8\)	\(0\)	\(-1\)	\(0\)	\(1\)

Computing the rest of the table is not so simple. Column orthogonality between \(x_j\) and \(h_i\) will not work because \(\psi_4(h_i)=-\psi_5(h_i)\). But we do know that \(\mathrm{PGL}_2(7)\) still acts on the projective line, so \(\psi_6\) (without loss of generality) is the deleted permutation representation.

We know that \(x_5\) acts as two \(4\)-cycles so \(h_4\), which squares to \(x_5\), acts as an \(8\)-cycle. This is an odd permutation, so all \(h_i\) are odd permutations. We know that \(x_3\) acts as two \(3\)-cycles, so \(h_2\), which squares to \(x_3\), must have cycle type \(6,1^2\). The element \(h_1\) is either a transposition or a triple transposition. This yields some more entries.

	\(1\)	\(x_2\)	\(x_3\)	\(x_4\)	\(x_5\)	\(h_1\)	\(h_2\)	\(h_3\)	\(h_4\)
\(\psi_1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)
\(\psi_2\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(-1\)	\(-1\)	\(-1\)	\(-1\)
\(\psi_3\)	\(6\)	\(-2\)	\(0\)	\(2\)	\(-1\)	\(0\)	\(0\)	\(0\)	\(0\)
\(\psi_4\)	\(6\)	\(2\)	\(0\)	\(0\)	\(-1\)
\(\psi_5\)	\(6\)	\(2\)	\(0\)	\(0\)	\(-1\)
\(\psi_6\)	\(7\)	\(-1\)	\(1\)	\(-1\)	\(0\)	\(a\)	\(1\)	\(-1\)	\(-1\)
\(\psi_7\)	\(7\)	\(-1\)	\(1\)	\(-1\)	\(0\)	\(-a\)	\(-1\)	\(1\)	\(1\)
\(\psi_8\)	\(8\)	\(0\)	\(-1\)	\(0\)	\(1\)
\(\psi_9\)	\(8\)	\(0\)	\(-1\)	\(0\)	\(1\)

Here \(a=-1,5\). We can just take \(\gen{\psi_1,\psi_6}\) to determine which, and we get \(a=-1\). This gives us a complete row.

We can take the exterior square of \(\psi_3\), using the formula \((\psi(x)^2-\psi(x^2))/2\), and the fact we know the power map. It has character values \(15\), \(-1\), \(0\), \(3\), \(1\), \(-3\), \(0\), \(-1\), \(-1\), and has inner product \(1\) with \(\psi_2\) and \(\psi_3\), while having norm \(3\). The remaining character must be (up to sign) \(\psi_8\), and we obtain another two characters, \(\psi_8\) and \(\psi_9\).

	\(1\)	\(x_2\)	\(x_3\)	\(x_4\)	\(x_5\)	\(h_1\)	\(h_2\)	\(h_3\)	\(h_4\)
\(\psi_1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)
\(\psi_2\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(-1\)	\(-1\)	\(-1\)	\(-1\)
\(\psi_3\)	\(6\)	\(-2\)	\(0\)	\(2\)	\(-1\)	\(0\)	\(0\)	\(0\)	\(0\)
\(\psi_4\)	\(6\)	\(2\)	\(0\)	\(0\)	\(-1\)	\(a\)	\(b\)	\(c\)	\(d\)
\(\psi_5\)	\(6\)	\(2\)	\(0\)	\(0\)	\(-1\)	\(-a\)	\(-b\)	\(-c\)	\(-d\)
\(\psi_6\)	\(7\)	\(-1\)	\(1\)	\(-1\)	\(0\)	\(-1\)	\(1\)	\(-1\)	\(-1\)
\(\psi_7\)	\(7\)	\(-1\)	\(1\)	\(-1\)	\(0\)	\(1\)	\(-1\)	\(1\)	\(1\)
\(\psi_8\)	\(8\)	\(0\)	\(-1\)	\(0\)	\(1\)	\(-2\)	\(1\)	\(0\)	\(0\)
\(\psi_9\)	\(8\)	\(0\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(-1\)	\(0\)	\(0\)

Note that the Frobenius-Schur indicator of \(\psi_4\) is \(+1\): we easily see that it is positive since \(\psi_4(g^2)\in H\) so we may compute it, and \(\psi_4(g^2)\) is negative only if \(g^2\) has order \(7\). This yields a negative contribution of only \(|x_5^G|=48\), and this is more than balanced out by even just that of \(h_1\) to the sum. Thus \(\psi_4\) and \(\psi_5\) are real, and so \(a,b,c,d\) are real.

Now take column orthogonality of \(h_1\) with itself to get another entry, since \(|C_G(h_1)|=12\). We see \(2a^2+12=12\), so \(a=0\). Do the same with \(h_2\), as \(|C_G(h_2)|=6\). We see \(2b^2+6=6\), so \(b=0\) as well. We have \(|C_g(h_3)|=8\), so \(2c^2+4=8\). Thus \(c=\pm\sqrt2\). The same equation holds for \(h_4\), so \(d=\pm\sqrt2\). If \(c=d\) then columns 8 and 9 would be the same, so they must be \(\sqrt2\) and \(-\sqrt2\), and we complete the table.

	\(1\)	\(x_2\)	\(x_3\)	\(x_4\)	\(x_5\)	\(h_1\)	\(h_2\)	\(h_3\)	\(h_4\)
\(\psi_1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)
\(\psi_2\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(-1\)	\(-1\)	\(-1\)	\(-1\)
\(\psi_3\)	\(6\)	\(-2\)	\(0\)	\(2\)	\(-1\)	\(0\)	\(0\)	\(0\)	\(0\)
\(\psi_4\)	\(6\)	\(2\)	\(0\)	\(0\)	\(-1\)	\(0\)	\(0\)	\(\sqrt2\)	\(-\sqrt2\)
\(\psi_5\)	\(6\)	\(2\)	\(0\)	\(0\)	\(-1\)	\(0\)	\(0\)	\(-\sqrt2\)	\(\sqrt2\)
\(\psi_6\)	\(7\)	\(-1\)	\(1\)	\(-1\)	\(0\)	\(-1\)	\(1\)	\(-1\)	\(-1\)
\(\psi_7\)	\(7\)	\(-1\)	\(1\)	\(-1\)	\(0\)	\(1\)	\(-1\)	\(1\)	\(1\)
\(\psi_8\)	\(8\)	\(0\)	\(-1\)	\(0\)	\(1\)	\(-2\)	\(1\)	\(0\)	\(0\)
\(\psi_9\)	\(8\)	\(0\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(-1\)	\(0\)	\(0\)