The purpose of this question is to ask: for $H \leq G$ groups, when doing induction from the character table of $H$ to the character table of $G$, is there a faster method by hand than the one I outline below?
The method I present here is what I have gathered is standard from my lecture notes. It seems it would not be so bad if it were done entirely by a computer, but entirely by hand or with the aid of a computer it takes a long time and is error prone.
Say $\chi_{\mathbf{1}_n}$, $\chi_{\epsilon_{n}}$, $\chi_{s_{n}}$ are the characters of the trivial, sign, and standard representations of $S_n$ respectively. Write $\mathrm{ccl}_G(g)$ for the conjugacy class of $g$ in the group $G$. Write $C_G(g)$ for the centraliser.
The character table of $S_4$ is
| irred. character of $S_4$ | $e$ (size 1) | $(1 2)$ (size 2) | $(1 2)(3 4)$ (size 3) | $(1 2 3)$ (size 8) | $(1 2 3 4)$ (size 6) |
|---|---|---|---|---|---|
| $\chi_{\mathbf{1}_4}$ | $1$ | $1$ | $1$ | $1$ | $1$ |
| $\chi_{\epsilon_{4}}$ | $1$ | $-1$ | $1$ | $1$ | $-1$ |
| $\chi_{s_{4}}$ | $3$ | $1$ | $-1$ | $0$ | $-1$ |
| $\chi_{\epsilon_{4}} \otimes \chi_{s_{4}}$ | $3$ | $-1$ | $-1$ | $0$ | $1$ |
| $\chi_5$ | $2$ | $0$ | $2$ | $-1$ | $0$ |
for $\chi_5$ some other irreducible character.
For $\chi$ a character of $S_4$, $g \in S_5$, I can construct a character of $S_5$ called $r^{*}(\chi)$ via
$$r^{*}(\chi)(g) = \sum_{\mathrm{ccl}_{S_4}(h) \subseteq \mathrm{ccl}_{S_5}(g)} \frac{|C_{S_5}(g)|}{|C_{S_4}(h)|} \chi(h).$$
This then gives me some not-necessarily irreducible characters of $S_5$ in a table with $7$ columns, $5$ rows (for the values of the characters) (in fact they turn out to all be reducible)
We can then calculate $\langle \chi_{\mathbf{1}_5}, r^{*}(\psi_1) \rangle$, $\langle \chi_{\epsilon_5}, r^{*}(\psi_1) \rangle$, $\langle \chi_{s_5}, r^{*}(\psi_1) \rangle$, $\langle r^{*}(\psi_1), r^{*}(\psi_2) \rangle$ for $\psi_1, \psi_2 \in \{ \chi_{\mathbf{1}_4}, \chi_{\epsilon_{4}}, \chi_{s_{4}}, \chi_{\epsilon_{4}} \otimes \chi_{s_{4}}, \chi_5\}$ to work out orthogonality relations between each of these characters, and use that to figure out how they decompose into irreducible characters.
By my calculations, calculating $\langle - , - \rangle$ in this table is $7 * 3 + 7 + 1=29$ operations (three multiplications - multiplying by the size of the conjugacy class, and the values of each representation on the conjugacy class, adding all seven of these conjugacy classes together, and dividing by $|S_5| = 120$.) Of course for representations like the trivial representation this is slightly easier but in most cases this is what we have to do.
Then we wish to calculate $5 + 5 + 5 + 5 + \frac{5 \times 4}{2} = 30$ such inner products (all possible pairs of the known characters of $S_5$, ignoring ordering since $\langle a, b \rangle = \langle b, a \rangle$) giving a total of
$$30 \times 29 = 870$$
operations (or so) that need to be performed. This obviously takes a long time by hand and is very error-prone. Even if we were to have the aid of a computer in, say, Excel to enter all the data and to set up the formulas to make it perform the correct calculations takes a while.
After that is deduced we can find out how these characters decompose as irreducible characters - but from just this data there are multiple ways of doing it - we also need to combine various facts such as $\chi(e) \in \mathbb{Z}$ for $\chi$ a character and row/column orthogonality to rule out certain cases, which is again even more mundane calculation. Eventually if everything is done correctly, all cases except for the correct one will be ruled out and it will allow us to deduce the character table for $S_5$.
So, is this method I am doing the standard way to do this, or is there a better way?