The Crusade of Answers

According to the average order of permutations by Richard Strong, the sum of the orders of all elements of $S_n$ has the following asymptotic: $$n!e^{C\sqrt{\frac{n}{\log(n)}} + O(\frac{\sqrt{n}log(log(n))}{log(n)})}$$ where $C \approx 2.99047$ is a constant.

$$[F_n : V_{\{x^2\}}(F_n)] = 3^{m + C_m^2 + C_m^3}$$ It was proved by Bartel van der Waerden and Friedrich Wilhelm Levi in "Über eine besonderen Klasse von Gruppen" in 1933.

As @LordSharktheUnknown describes in the comments . . . No, it's not an error. The inverses of the operation $\oplus$ can be delineated with a few examples, like so: $$\begin{align} 1\oplus (n-1)&=1+(n-1)-n \\ &=0, \\ 2\oplus (n-2)&=2+(n-2)-n\\ &=0, \\ &\vdots \end{align}$$

This is not an answer per se but a few places to look, according to "Generators and Relations for Discrete Groups (Reprint of Fourth Edition)," by H. S. M. Coxeter and W. O. J. Moser, page 90, are Nielsen, J. (1924b), "Die Isomorphismengruppen der friend Gruppen," Math. Ann. 91, 169-209. This h...

Yes, it's correct. The sentences need punctuation though.

You're correct . . . ish. The $\alpha$ is not necessarily countable, so your use of "$\Bbb Z\oplus\Bbb Z\oplus \Bbb Z\dots$" is not strictly appropriate (not least because it does not include the $x_{\alpha}$). This is what it really means: "the direct sum over all $\alpha<\beta$ of $\Bbb{Z}x_{...

As @DerekHolt states in the comments . . . Any diagonal matrix with entries from $\{-1,1\}$ and determinant $1$, excluding the identity, generates a subgroup of $\operatorname{SO}(6)$ isomorphic to $\Bbb Z_2$.

Your proof is correct. Well done. The phrasing at the end, though, is a little off; try a new sentence reading "Hence $\phi(G_1)$ is cyclic." It's just a minor suggestion.

Since $$M_2^n= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & n \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$ by induction on $n$ for $n\in\Bbb Z$, we have that the subgroup $\langle M_2\rangle $ generated by $M_2$ is isomorphic to the free group $\Bbb Z$.