Reference for an English translation of a proof of Frobenius on the origins of representation theory Background.
Let $G$ be a finite group, $c_1,\dots,c_s$ be the conjugacy classes of $G$, $\sum_{g\in c_i}g=C_i\in Z(\mathbb{C}G)$ be the class sums [a basis of $Z(\mathbb{C}G)$] and finally let $C_jC_k=\sum_i a_{ijk}C_i$.
The following is a quote from the paper Representations of
Finite Groups:
A Hundred Years, Part I by Lam

... we know that $Z(\mathbb{C}G)$ is (commutative and) semisimple. Frobenius was not
equipped with all this modern jargon, so in- stead he had to do a lot
of ad hoc calculations with the counting numbers $a_{ijk}$ to check what
we now know as the trace condition for semisimplicity.

On the same subject, a second quote, from the book Pioneers of representation theory  by Curtis

He proved ... by a further, and intricate, analysis of the constants $h_{ijk}$.

Question. Where can I find an English translation of the original ad hoc and intricate calculations of Frobenius?
 A: For anyone interested in the answer, I used google translate to read the original paper Über Gruppencharaktere by Frobenius. Here is my summary (from a more modern perspective).


*Throughout all indices $\alpha,\beta,i,j,\dots $ are in $1,\dots,s$ and represent conjugacy classes. For an index $i$, Frobenius denotes the inverse class $\{c^{-1}:c\in C_i\}$ by $C_{i'}$.


*Let $h_{ijk}$ denote the number of triplets $a\in C_i,b\in C_j,c\in C_k$ such that $abc=e$. Since $abc=e\iff bca=e$ we have $h_{ijk}=h_{jki}$. Since $abc=e\iff b(b^{-1}ab)c=e$, we have $h_{ijk}=h_{jik}$. So $h$ is symmetric in its variables. Also $abc=e\iff c^{-1}b^{-1}a^{-1}=e$ so $h_{ijk}=h_{i'j'k'}$.


*Let $E=Z(\mathbb{C}G)$ be the commutative algebra spanned by "the standard basis" $C_1,\dots,C_s$. For $i\in S$ let $A_\gamma$ be the $s\times s$ matrix representing multiplication by $C_\gamma$ in $E$ with respect to standard basis. Note that by definition $A_k A_j=A_jA_k=\sum_i a_{ijk}A_i$ and the entries are simply $(A_\gamma)_{\alpha,\beta}=a_{\alpha \beta\gamma}$.


*Let $h_i=|C_i|$ and note that $a_{ijk}=\#\{(b,c)\in C_j\times C_k : bc=a\}$ for any fixed $a\in C_i$, so that $a_{ijk} h_i = h_{i'jk}$.


*The trace condition for semisimplicity is $\det(\text{trace}(A_\alpha A_\beta)_{\alpha,\beta})\neq 0$. Frobenius then permutes the columns $\beta\leftrightarrow\beta'$ to show $\det P_{\alpha\beta}\neq 0$ where $$P_{\alpha\beta}=\text{trace}(A_\alpha A_{\beta'})=\sum_{\gamma,\delta}{a_{\gamma \delta\alpha}a_{\delta\gamma\beta'}}=\sum_{\gamma,\delta}\dfrac{h_{\gamma'\delta\alpha}h_{\delta'\gamma\beta'}}{h_\gamma h_\delta}=\sum_{\gamma,\delta}\dfrac{h_{\gamma\delta\alpha}h_{\gamma\delta\beta}}{h_\gamma h_\delta}$$


*Let $Q$ be the real $s\times s^2$ matrix $Q_{\alpha,(\gamma,\delta)}=\dfrac{h_{\gamma\delta\alpha}}{\sqrt{h_\gamma h_\delta}}$. Then $P=QQ^t$. To show $P$ is nonsingular, one has to show that $Q$ has rank $s$. [This follows from the Cauchy Binet formula, or by noting that $Px=0\implies x^tPx=(Q^tx)^t(Q^tx)=||Q^t x||^2=0\implies Q^tx=0$.] However, this is a simple task, because if we reduce to the $s\times s$ submatrix of $Q$ where $C_{\gamma}=\{e\}$ (this is written as $\gamma=0$), then our element $Q_{\alpha,(0,\delta)}$ is equal to $\sqrt{h_{\delta}}$ if $\alpha=\delta'$ and $0$ otherwise.
A: You mention Lam and Curtis, but what about Hawkins? He discussed the work of Frobenius in representation theory in a few papers in Arch. Hist. Exact Sci. in the 1970s. See Section 13.3 of his book "The Mathematics of Frobenius in Context"; his earlier papers are references 266, 267, and 268 in the bibliography.
