Let $G$ a group of order $6$. Prove that $G \cong \Bbb Z /6 \Bbb Z$ or $G \cong S_3$. Let $G$ a group of order $6$.  Prove that:
i) $G$ contains 1 or 3 elements of order 2.
ii) $G \cong \Bbb Z /6 \Bbb Z$ or $G \cong S_3$.

I haven´t covered Sylow groups and normal groups. This is an exercise from the chapter about group actions. I have covered Lagrange and cosets.
Surely given $n\in \mathbb N, n\geq 1$, there exists a cyclic group of order $n$: In our case, there exists a cyclic group $G\cong \mathbb Z /\mathbb 6Z \cong \mathbb Z_6$.


So that covers one option: Now, $\mathbb Z_6$ has exactly how many elements of order $2$?   Use Lagrange, and determine the possible orders of subgroups. Any element of order $2$ will generate a subgroup of order $2$. $Z_6$ has only one subgroup of order $2$: What subgroup of $\mathbb Z$ has order $2$ ,and which element necessarily has order $2$?
For any group of even order, there can exist only an odd number of elements of order $2$. Why?
We can rule out $5$ such elements for a group of order $6\;\ldots\quad$WHY?
Hence, that leaves us with $G$ having $1$ or $3$ elements of order $2$.  $\mathbb Z_6$ covers one possibility. If a group $G$ of order $6$ has $3$ elements of order $2$, how does, how does this fully determine the corresponding group $G$? (For any $G\not\cong \mathbb Z_6$, all it's elements must of order $1$,$2$, or $3$.)


Now, use these facts to justify that the only groups of order $6$ must be isomorphic to $\mathbb Z_6$ or to $S_3$.
If $G$ isn't $\mathbb Z/6\mathbb Z$, then it isn't cyclic. Hence, the maximal order is 3 (see Lagrange). It easily follows that an element of order 2 exists as well. From here to show that it's actually $S_3$ should be trivial. This solves the second part.

The first part follows directly from the second.