Group of order $545$ is cyclic?

Question

I just wanted to check that the following is correct. I have a group of order $545 = 5\cdot 109$. Using Sylow's Third theorem I found that there are two normal subgroups $H$ and $K$ of order $5$ and $109$ respectively. Now $H\cap K = \{e\}$. So $G = H\times K\simeq \mathbb{Z}_{545}$.

Is this correct?

Answer 1

Yes, well done! Note that $5$ does not divide $108$