A group of order $1645$ is cyclic This is what I have so far, but I get stuck at one point.
$|G| = 1645 = 5\cdot 7\cdot 47$  
I found with the Sylow theorem that there is a unique Sylow $5$-Subgroup, a unique Sylow $7$-Subgroup and a unique Sylow $47$-subgroup. Let's name them $H, E, D,$ respectively.
Because they are unique $H,E,D$ are normal in $G$. 
So we can say that $\triangle$
$$G\cong H\times E \times D$$
Here i get stuck...
I know it should become that $$H \cong \mathbb Z_5, \;E\cong \mathbb Z_7,\; D\cong \mathbb Z_{47}$$
but I can't find why...
 A: Any two finite cyclic groups are isomorphic
First since $\gcd(7,5)=1$ so $M=H\times  K\cong \mathbb Z_{35}$
$M=H\times K\cong\mathbb Z_{35}$ and $D=\mathbb Z_{47}$ and since $\gcd(35,47)=1$  so $G=M\times D\cong \mathbb Z_{35*47}$
and hence cyclic
A: Since $5$, $7$, and $47$ are prime numbers, there is only one isomorphism class of group with each of those respective orders. (proof: since the order of a subgroup divides the order of the group, there is no non-trivial subgroup of a group with a prime order. Therefore the cyclic subgroup generated by any non-trivial element must actually be equal to the whole group. So the group is cyclic. But all cyclic groups of the same order are isomorphic -- just send any generator to any generator.)
A: A (sub)group of prime order $p$ is cyclic and any of its elements $g\neq e$  is a generator: indeed, the order of the subgroup $\langle\, g\,\rangle$ generated by $g$ is a divisor of $p$ and it is $>1$  since $g\neq e$. Hence
$$H\simeq \mathbf Z/5\mathbf Z, \quad E\simeq \mathbf Z/7\mathbf Z, \quad D\simeq \mathbf Z/47\mathbf Z.$$
