Let $G$ a group of order $6$. Prove that $G \cong \Bbb Z /6 \Bbb Z$ or $G \cong S_3$. [closed]

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 $S_3$.

For (ii), in the case there are three elements of order 2, let $G$ act on the set of elements of $G$ of order 2 by conjugation. Let $G \to S_3$ be the corresponding homomorphism. What can be the kernel?

If there is just one element of order $2$, all elements of $G$ must commute with that element. Consider the orders of the other elements.