Prove that $S_7$ cannot be written as a union of fewer than $459$ cyclic subgroups.
I have no idea about this. I know that $S_7$ has $7!$ elements. However, this has not been helpful. We have not learned about Sylow yet. Any help would be appreciated.
$S_7$ has elements of order $12$, $10$, $7$, $6$, $5$, $4$, $3$, $2$, and $1$. We need not care for the only element of order $1$ as it will be covered by any cyclic subgroup. An element of order $2$ consists of one, two, or three two-cycles. In each case, it is a power of an element of higher order: $$ (1\,2)=\bigl((1\,2)(3\,4\,5)\bigr)^3,\quad (1\,2)(3\,4)=(1\,3\,2\,4)^2,\quad (1\,2)(3\,4)(5\,6)=(1\,3\,5\,2\,4\,6)^3.$$ Hence elements of order two will automatically be covered once we handle those higher orders. The same argument works for order $3$: $$(1\,2\,3)=\bigl((1\,3\,2)(4\,5)\bigr)^2, \quad (1\,2\,3)(4\,5\,6)=(1\,4\,2\,5\,3\,6)^2 $$ and partially for order $4$: $$(1\,2\,3\,4)=\big((1\,4\,3\,2)(5\,6\,7)\bigr)^3. $$ However $(1\,2\,3\,4)(5\,6)$ is not the power of an element of higher order. Hence we do need the cyclic groups of order $4$ generated by elements of this form. There are ${7\choose 4}\cdot 3!\cdot {3\choose 2}$ elements of this form, but each cyclic group of order $4$ has two possible generators. Hence we need $$ \frac12{7\choose 4}\cdot 3!\cdot {3\choose 2}=315$$ cyclic groups of order $4$.
Order $5$ can again be ignored: $$(1\,2\,3\,4\,5)=\bigl((1\,4\,2\,5\,3)(6\,7) \bigr)^2.$$
For order six we have $$(1\,2\,3)(4\,5)(6\,7)=\bigl((1\,3\,2)(4\,5\,5\,7)\bigr)^2,$$ but $(1\,2\,3)(4\,5)$ and $(1\,2\,3\,4\,5\,6)$ are not such powers. There are ${7\choose 3}2!{4\choose 2}$ elements of the first type and ${7\choose 6}5!$ of the second type. As each cyclic group of order $6$ has two possible generators, we need to use $$\frac12\left({7\choose 3}2!{4\choose 2}+{7\choose 6}5!\right)= 630$$ cyclic groups of order $6$.
Elements of orer $7$ are all of the form $(1\,2\,3\,4\,5\,6\,7)$ and are not powers of higher-order elements. There are $6!$ such elelemts and each cyclic group of oredr $7$ has $6$ generators, hence we need $$\frac{6!}6=120 $$ cyclic groups of order $7$.
Elements of order $10$ look like $(1\,2\,3\,4\,5)(6\,7)$, there are ${7\choose 5}4!$ such elements, each cyclic group of order $10$ has $\phi(10)=4$ generators, hence we need $$ \frac14{7\choose 5}4!=126$$ cyclic groups of order $10$.
Finally, elements of order $12$ look like $(1\,2\,3\,4)(5\,6\,7)$, there are ${7\choose 4}3!2!$ of them, and each group has $\phi(12)=4$ generators, hence we need $$ \frac14{7\choose 4}3!2!=105$$ cyclic subgroups of order $12$. In total, we need at least $$ 315+630+120+126+105=1296$$ cyclic subgroups to cover $S_7$. As $1296>459$, the claim follows.
For convenience, here's a short and simple argument showing the original weak claim directly:
The maximal order of an element of $S_7$ is $12$. Hence any cyclic subgroup of $S_7$ has at most $11$ non-neutral elements. Hence $458$ cyclic subgroups together cover at most $11\cdot 458=5038$ non-neutral elements - but $S_7$ has $7!-1=5039$ non-neutral elements!