$H\le G$ and $|H|=10$, $a^6\in H$, what could $|a|$ be? Without full Lagrange's theorem This is exercise 43 in chapter 4 of Gallian's Contemporary Abstract Algebra, 9th edition:

Suppose that $H$ is a subgroup of a group $G$ and $|H|=10$.  If $a$ belongs to $G$ and $a^6$ belongs to $H$, what are the possibilities for $|a|$?

The answer in the back of the book is "all divisors of 60".  It's easy to reach this conclusion using Lagrange's theorem: since $a^6\in H$, $|a^6|$ is a divisor of $|H|=10$, so $|a|$ is a divisor of $6\cdot 10 = 60$.  The problem is that Lagrange's theorem isn't proven until chapter 7.  Chapter 4 contains the special case of Lagrange's theorem when the ambient group is cyclic (Corollary 1 on p79: "In a finite cyclic group, the order of an element divides the order of the group"), but I don't see how to deduce this result using only that case, since neither $G$ nor $H$ is assumed cyclic.
Is there a way to do this exercise using only the facts introduced in the book up until this point?
 A: For every $h\in H$, $h^{10}=1$ (you do not need to know the full Lagrange theorem for this). Therefore $a^{60}=1$. So the order of $a$ divides $60$. Every divisor $d$ can occur. Indeed, let $C_{60}=\langle c\rangle$ be the cyclic group of order $60$. Then the subgroup $H$ generated by $c^6$ is of order $10$.Every $a\in C_{60}$ satisfies $a^6\in H$. For every divisor $d$ of $60$ the element $a=c^{60/d}$ has order $d$. So indeed every divisor of $60$ can occur as the order of your $a$.
Update The fact that in a group of order $n$ every element has order $\mid n$ is proved in the book only in Chapter 7. But here we need it just for $n=10$. Here is the proof. The order of $h\in H$ cannot be bigger than $10$. If it is not $1,2,5,10$ it must be $3, 4, 6, 7, 8$ or $9$. If it is $6,8, 9$ then $h^2$ or $h^3$ has order $3$ or $4$. So we can assume that the order of $h$ is $3,4$ or $7$. These cases are almost identical. Suppose that the order is $4$. So the different powers of $h$ are $1,h,h^2, h^3$. There is an element $u\in H\ne 1,h,h^2,h^3$. Then the $8$ elements $1,h,h^2,h^3, u, uh, uh^2, uh^3$ are all different. Let $v\in H$ be an element not in this set. Then the eleven elements $1,h,h^2,h^3, u, uh, uh^2, uh^3, v, vh, vh^2$ are all different, and all in $H$, a contradiction.
