What does Frobenius reciprocity say in the case of $G = S_3$ and $H = C_2$? Using character theory the statement of Frobenius reciprocity would be:
$$ \langle \text{Ind}_H^G \psi , \phi \rangle_G = \langle \psi , \text{Res}_H^G \phi\rangle_H $$
Let's try to evaluate this expression with $G = S_3$ (the permutation group of three objects) and $H = C_2$ is a 2-cycle.

*

*The character table of $S_3$ looks like this:
$$ \begin{array}{c|crr} 
 & (1) & (12) & (123) \\ \hline 
\phi_1 & 1 & 1 & 1 \\ 
\phi_2 & 1 & -1 & 1 \\ 
\phi_3 & 2 & 0 & -1 \\ 
\end{array} $$

*The character table of $C_2$ is even simpler:
$$ \begin{array}{c|cr}  & (1) & (12) \\ \hline
\psi_1 & 1 & 1\\ 
\psi_2 & 1 & -1 \end{array} $$
How do we compute the induced characters $\text{Ind}_{H}^{G} \psi$ (from $H = C_2$ to $G = S_3$) and in the other direction, the restricted characters $\text{Res}_H^G \phi$.
We also might need the Schur orthogonality relations to compute the inner product.  So what does the equation look like in the end?

Related:

*

*Find character table for symmetric group $S_3$


*Character theory questions on Chemistry.SE Why does $(2z^2-x^2-y^2,x^2 - y^2)$ have -1 trace in the character table of $T_d$ for $C_3$ rotation?


*https://physics.stackexchange.com/questions/306995/decoding-the-character-table
 A: Prof. Holt is right, in the character table of $S_3$ the last two columns under the line should be switched: I corrected this for you. Now let us do some calculations. First let us find out what $\psi_1^G=(1_H)^G$ must be (I am using a less involved notation for the characters). Using Frobenius' Reciprocity: $[(1_H)^G,1_G]=[1_H,1_H]=1$, hence $(1_H)^G=1_G + \delta$, where $\delta$ is a character with $[\delta,1_G]=0$, so $\delta$ is a certain sum of the non-principal characters of the table of $S_3$. Note that $3=(1_H)^G(1)=1_G(1) + \delta(1)=1+\delta(1)$, so $\delta(1)=2$. Observe from the character table that $(\phi_3)_H=\psi_1+\psi_2=1_H+\psi_2$, This implies that $\delta = \phi_3$ (again use Frobenius' Reciprocity to calculate $[(1_H)^G, \phi_3]$). We conclude that $\psi_1^G=\phi_1+\phi_3$. In a similar vein now try to calculate $\psi_2^G$. Your answer should be $\psi_2^G=\phi_2 + \phi_3$.
The example you mentioned is just a small group example. For larger groups calculations get more involved and one needs to apply some efficiency, taking into account how conjugacy classes of $G$ and $H$ "behave". See for example I.M. Isaacs, Character Theory of Finite Groups, p. 64, where the character table of $G=A_5$ is constructed from $H=A_4$.
A: As an alternative to Nicky Hekster's answer, if you wanted to compute the induced characters directly, rather than by using Frobenius Reciprocity, you could do it using the formula
$$\psi_H^G(g) = \frac{1}{|H|} \sum_{x \in G,\,x^{-1}gx \in H}\psi(x^{-1}gx).$$
So, in your example $G=S_3$, $H=C_2$, for $g = (1,2,3)$, since $x^{-1}gx \not\in H$ for all $x \in G$, we have $\psi_H^G(g)=0$.
For $g = (1)$, we get $\psi_H^G(g) = 6/2 = 3$ for $\psi=\psi_1$ or $\psi_2$.
Finally, for $g=(1,2)$, we have $x^{-1}gx \in H$ iff $x=(1)$ or $(1,2)$, and we find $(\psi_1)_H^G(g) = 1$ and $(\psi_2)_H^G(g) = -1$.
Now you can see directly that $(\psi_1)_H^G = \phi_1+\phi_3$ and $(\psi_2)_H^G = \phi_2+\phi_3$
