Prove that every group of order $35$ is cyclic.
Now, the subgroups of this are ones whose orders divide the order of this group(by lagrange), these are of prime orders $7$ and $5$.
and I guess $\Bbb Z_7\times \Bbb Z_5$ is of order $35$ and since these are both cyclic, so is $\Bbb Z_{35}$.
But that doesn't prove anything about every subgroup of order $35$.
How do I do this.
The number of 7 Sylow subgroups divides 5 and is congruent to 1 mod $7$, hence there's a unique one, and hence normal. Similarly, the number of $5$ Sylow subgroups divides 7 and is congruent to $1$ mod $5$, hence there's a unique one. It follows that $G \cong \mathbb{Z}/7\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}\cong \mathbb{Z}/35\mathbb{Z}$ by Chinese remainder theorem.