11.9. Let $G$ be a finite abelian group and let $m$ be the least common multiple of the orders of its elements. Prove that $G$ contains an element of order $m$.
11.10. Supply a finite non-abelian group for which the conclusion of the previous exercise fails.
11.11. If $H$ is a subgroup of a finite group $G$, and if $|G|=m|H|$, adapt the proof of Lagrange's theorem to show that $g^{m!}∈H$ for all $g∈G$.
13.5. Let $G$ be a group of order $4n+2$. Use Cauchy's theorem, Cayley's theorem, and Exercise 6.6 to show that $G$ contains a subgroup of order $2n+1$.
11.9. 对任意$a,b∈G$有$o(ab)=\operatorname{lcm}(o(a),o(b))$,所以$G$的所有元素之积的阶是$G$的所有元素的阶的最小公倍数.(又见这帖)
11.10. $G=S_3,o((12))=2,o((123))=3$,但是$G$中没有元素阶数为6.
11.11. Consider the set $X$ of all cosets of $H$. Then $G$ acts by multiplication on $X$ and so there is a homomorphism $G \to Sym(X)$. By Lagrange's theorem, $g^{m!}$ acts as the identity. In particular $g^{m!}H =H$ and so $g^{m!}\in H$. (这帖和这帖)
13.5.