Any group of order $85$ is cyclic.
My attempt:
Let $|G|=85=5\times17$
Let $H_1$ be the sylow-$5$ and $H_2$ be sylow-$17$ subgroup of $G.$
Then $H_1\cap H_2=\{e\}$
So $|H_1\times H_2|=85$
and hence $G\simeq H_1\times H_2$
Since $H_1\times H_2$ is cyclic so is $G.$
I'm not sure about my proof. Please tell me whether it's correct! Apart from voting please leave comment.
Best Answer
One more step is needed. Recall that one of a definition of direct product $G = H \times K$ is $H, K \trianglelefteq G$, $G = \langle H, K \rangle$, and $H \cap K = 1$. Hence, you have to show that $H_1, H_2 \trianglelefteq G$. (This can be easily shown by Sylow's theorem.)
Otherwise your proof fails. Such an example is $S_3$. Though Sylow subgroups $C_2$, $C_3$ of $S_3$ have trivial intersection $C_2 \cap C_3 = 1$ and $\lvert C_3C_2 \rvert = 6$ (which implies $S_3 = C_3C_2$), $S_3$ is not the direct product of the two. (Observe that $C_2 \not \trianglelefteq S_3$ in this case.)