= Solution
The <Lagrange theorem> states that $|H|$ divides $|G|$ for every <subgroup> $H$ of a finite <group> $G$, with $|G|=|G:H||H|$. The left cosets partition $G$, and multiplication by a representative bijects $H$ with each coset, proving the formula.
The intersection $H\cap K$ contains the identity and is closed under $xy^{-1}$, so it is a <subgroup>. Its order divides both $|H|$ and $|K|$; if these are coprime, $H\cap K=\{e\}$.
The order $o(x)$ is the least positive $m$ with $x^m=e$. Division $k=qm+r$, $0\leq r<m$, shows that $x^k=e$ exactly when $r=0$, so exactly when $o(x)\mid k$.
If $x,y$ commute and have coprime orders $m,n$, then $(xy)^{mn}=e$. Conversely $(xy)^r=e$ implies $x^r=y^{-r}$ lies in $\langle x\rangle\cap\langle y\rangle=\{e\}$, so $m,n\mid r$ and $mn\mid r$. Hence $o(xy)=mn$.
<Cauchy theorem for groups> says that every prime divisor $p$ of $|G|$ occurs as the order of an element of $G$. For $|G|=26$, the Sylow $13$-subgroup $P\cong C_{13}$ is unique and normal, and Cauchy's theorem supplies a complement $C_2$. Thus $G=C_{13}\rtimes C_2$. The homomorphism $C_2\to\operatorname{Aut}(C_{13})\cong C_{12}$ is either trivial or has the unique image of order two, inversion. These give exactly $C_{26}$ and the dihedral <group> of order $26$.
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