This was a previously asked question that only had answers referring to Sylow's theorem, which is not something I can use here.
Let $G$ be a group, $|G|=35.$ We know that $G$ must contain an element of order $5$ and an element of order $7$.
(1) Prove that $G$ must have a normal subgroup of order $5$ or of order $7$ or both.
(2) Deduce that $G$ is isomorphic to $C_{35}$.
(1) Attempt: It is shown earlier in the question that if $\Omega$ is the set of subgroups of $G$ isomorphic to a subgroup $H\leq G,$ and $G$ acts on $\Omega$ by conjugation, then $$\text{Stab}(H)=G \iff H\; \text{is a normal subgroup}.(*)$$
Assume for contradiction $G$ has no normal subgroups of order $5$ or $7$. Take $g_1,g_2$ as order $5$ and order $7$ respectively, then $\langle g_1\rangle$,$\langle g_2\rangle \leq G$ are not normal.
Now by $(*)$ we can find an element in $G$ such that conjugating $\langle g_1\rangle$ by it gives a subgroup of $G$, isomorphic to but distinct from $\langle g_1\rangle$. By Lagrange's theorem then, and noting the intersection of subgroups is a subgroup, these two order $5$ subgroups must have trivial intersection
Similarly we can find two order $7$ subgroups with trivial intersection. And it is clear too that all non-identity elements in all four of these subgroups, of which there are $20$, are distinct.
This leaves $14$ non-identity elements outside of these subgroups, but I am stuck here.
I think if $G$ has no order $35$ element then noting the only non-negative integer solution to $4p+6q=14$ is $p=2,q=1$ shows that we can distribute these elements into two new distinct order $5$ subgroups and one new order $7$ subgroup, but I can't see how this would help.
I'm sure I can complete this as a counting argument but I'm not sure where to go.