I believe you are asking a detailed question where most of the time we only answer if it is easy to do.
Answer to the first question
Using group theorist's notation so I don't get lost:
$$[\theta,\phi] = \dfrac{1}{|G|} \sum_{g \in G} \theta(g) \overline{\phi(g)}$$
We have the nice relation: if there are non-negative integers $d_\chi$ so that
$$\theta = \sum_{\chi \in \operatorname{Irr}(G)} d_{\chi} \cdot \chi$$
Then we have the nice projection / use dot products to get coefficients:
$$[\theta,\chi] = d_\chi$$
and the standard sort of pythagorean thing
$$[\theta,\theta]
= \left[ \theta, \sum_{\chi \in \operatorname{Irr}(G)} d_{\chi} \cdot \chi \right]
= \sum_{\chi \in \operatorname{Irr}(G)} d_{\chi} \cdot [ \theta, \chi ]
= \sum_{\chi \in \operatorname{Irr}(G)} d_{\chi}^2$$
So, I think this means 1. No, there is no typo in the formula (or at least there is a similar formula with no typo, and it does have squares)
Answer to the second question
So how do you determine the $d_{\chi}$ from the sum of their squares? Well, this is pretty hard in general. It is like finding the angle inside a circle from its radius ... you can't in general. However, we have assumed the $d_{\chi}$ are non-negative integers, and as you have mentioned, often there are not many possibilities.
- $1 = 1$
- $2 = 1^2 + 1^2$
- $3 = 1^2 + 1^2 + 1^2$
- $4 = 1^2 + 1^2 + 1^2 + 1^2$ or $4 = 2^2$
- $5 = 1^2$ plus one of the fours
Unless the number is small, you don't have much hope. However, the number is often small, so this trick gets used a lot.
A better version not only computes $[\theta,\theta]$, but also $[\theta,\phi]$ and looks for all the possible $d_\chi$ for $\theta$ and $\phi$ that would create that matrix. GAP has a command for this, OrthogonalEmbeddings and it is super-fun when it works (as in, gives you only a single solution -- usually it gives you many solutions).
Sorry on the third
I'm not sure about your third question. You might find those numbers early or late in the process. I think of them as "decomposition numbers", but that might not be the right word in this case. Another word is "multiplicities of the restriction".
Sometimes that is a good idea. In a recent question I asked, I need 7 positive integers that add up to 7 and that include 7 copies of "1". So I guess that was easy? But the next part needs some positive integers that add up to 14. I don't know how many (maybe I know it is 1, 7, or 14?).
I think the case you are studying is called spin and often the questions you ask are very successfully answered. A more general version is clifford theory and is also very successful. The even more general version is just induced characters. I am not having much luck personally there, but I am hoping my question will be Clifford theory once I figure out the next step.
Vocabulary
compound character is "reducible character"
group embedding is fine, or you can say "subgroup" if the way the two groups are related is clear. "fusion" is the name of how the conjugacy classes between the two groups match up (the way you added Class 2 and Class 3 of A4 to get Class 2 of S4, and the way you got 0 for Class 2 and Class 5 of S4).
Unless I'm missing something here, the exercise is wrong as stated.
Let $k$ of characteristic $p>0$ and let $G=H=\Bbb G_m$. Let $\phi:G \to G$ be the map that raises everything to the $p$-th power. Then $\phi$ is injective (as $p$-th roots in char $p$ are unique), but on $X(G)=\Bbb Z$, $\phi$ induces the "multiplication by $p$"-map, which is not surjective.
So what is going on here?
I think this behaviour is a good example of why in finite characteristic, algebraic groups via classical algebraic geometry (even over an algebraically base closed field) is not the right approach. You just lose too much when you don't allow nilpotents.
Let me sketch you, even if you don't know schemes, why group schemes are "better" in this case. The thing is that the homomorphism I gave above is not "injective" as a morphism of group schemes: it has a kernel. Now describing this kernel might be tricky if you don't know schemes, but you can think of it as the "$p$-th roots of unity", let's call it $\mu_p$.
I think it should be clear that the kernel of raising to the $p$-th power has something to do with $p$-th roots of unity. If you want to think in terms of Hopf algebras, then we have a Hopf algebra $k[T]/(T^p-1)$ (with the comultiplication defined uniquely by $\Delta(T)=T \otimes T$). This is non-reduced, so it won't give rise to an algebraic group. It still gives rise to a perfectly fine group scheme, though. The classical approach can't distinguish this group scheme from the trivial group scheme, because there are no non-trivial $p$-th roots of unity in $k$, but $\mu_p$ is still nontrivial as a group scheme.
One way one can "see" that $\mu_p$ is non-trivial is to use the "functor of points" approach. A group scheme can be identified with (basically) a representable functor on $k$-algebras with values in groups. Now $\mu_p$ represents the functor $A \mapsto \{a \in A \mid a^p=1\}$. If we allow non-reduced $A$ this is clearly distinct from the terminal functor that sends everything to the one-point set.
Best Answer
Yes, please don't feel bad about this confusion. It is for semi-historical reasons as KReiser rightly pointed out above, and can be quite confusing.
Namely, let us fix a field $K$ and an (affine) algebraic group $G$ over $K$. Then, for any separable$\color{red}{^{(\ast)}}$ algebraic extension $L/K$ one can form the group of $L$-rational characters
$$X^\ast_L(G):=\mathrm{Hom}(G_L,\mathbb{G}_{m,L}),$$
where these are homomorphisms of group $L$-schemes. There is somewhat varying terminology in this regard, but often people call $X_{K^\mathrm{sep}}$ the group of characters of $G$, and denote it just $X^\ast(G)$ (or $X(G)$). They then might call the elements of $X_K^\ast(G)$ the rational characters of $G$. Although, again, this is author dependent, and so you have to sort of feel out the definition in each case.
To understand the relationship between the rational characters over differing fields, it's useful to introduce some extra structure. If $L/K$ is Galois, with Galois group $\Gamma_{L/K}$, then there is a natural action of $\Gamma_{L/K}$ on $X^\ast_L(G)$ given as follows:
$$\sigma\cdot \chi:= \sigma_G\circ \chi\circ\sigma_{\mathbb{G}_{m,L}}^{-1}.$$
Here for a $K$-scheme $X$ and $\sigma$ in $\Gamma_{L/K}$, I am denoting by $\sigma_X$ the natural morphism $X_L\to X_L$ which is the identity on the $X$-factor and the induced map on $\mathrm{Spec}(L)$ given by $\sigma^{-1}\colon L\to L$ (you can remove the inverse here if you are OK with it being a right, opposed to left, action). It is a good exercise to check that this is a well-defined action (e.g. that $\sigma\cdot\chi$ really can still be made sense of as an element of $X^\ast_L(G)$).
From Galois descent for affine schemes (e.g. see [Poonen,§4.4]) there is a natural identification
$$X^\ast_M(G)=X_L^\ast(G)^{\Gamma_{L/M}},$$
for any tower of extensions $L/M/K$ (with $L/K$ Galois). Here I am being somewhat imprecise and identifying $X^\ast_M(G)$ with a subset of $X_L^\ast(G)$ via the (injective) group map
$$X^\ast_M(G)\to X^\ast_L(G),\qquad \chi\mapsto \chi_L,$$
(the base change map).
In particular, one sees that the natural map
$$X_K^\ast(G)\to X^\ast_{K^\mathrm{sep}}(G)=X^\ast(G),$$
is a bijection if and only if $\Gamma_{K^\mathrm{sep}/K}$ acts trivially on $X^\ast(G)$. As you mentioned, if $G$ is a torus, then this is equivalent to $G\cong \mathbb{G}_{m,K}^n$, i.e. that it is split (and in fact, you can use closely related ideas to define the maximal split subtorus -- see [Springer, Proposition 13.2.4]).
I hope this clarifies things!
$\color{red}{(\ast):}$ This is a little too technical to mention above, but you can really have $L/K$ be any algebraic extension. But, if $L/K$ is purely inseparable then the natural map $X^\ast_K(G)\to X_L^\ast(G)$ is a bijection.
References:
[Poonen] Poonen, B., 2017. Rational points on varieties (Vol. 186). American Mathematical Soc..
[Springer] Springer, T.A. and Springer, T.A., 1998. Linear algebraic groups (Vol. 9). Boston: Birkhäuser.