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.
 A: 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 $S_3$.
A: 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.
A: I propose a different solution (although the problem is already solved in other answers) for i, that generalises it a little bit: A group $G$ with even order has an odd number of elements of order 2. From here, i follows directly.
The proof is really simple: for an element $a\in G$, we define:
$$U_a=\left\lbrace a,a^{-1}\right\rbrace$$
We have that every set $U_a$ has two elements unless $a^{-1}=a$, that only happens if $a=1$ or $a$ has order two.
Now if we collect all the $U_a$ for all $a\in G$, $G$ must be the disjoint union of all of them. So the sum of all $|U_a|$ must the order of the group. We're adding a $1$ for the element $1_G$, and two for every element with order different from two, that means that the number of elements of order $2$ must be odd so they sum up to an even number
A: $G$ must have subgroups of orders 2 and 3 by Cauchy's theorem.  These subgroups have prime order, so are cyclic, so say that $a$ and $b$ generate them and thus have orders 2 and 3, respectively.
Consider the set $\{1, a, b, ab, b^2, ab^2\}$. These must all be distinct, or else one of $a$ or $b$ will end up with the wrong order.  For example, $ab = b^2$ gives $a=b$, which is impossible. Since these six products exhaust the group, $G$ is generated by $a$ and $b$.
Now, what is $ba$?  It must be the same as one of the six elements above. $ba=1$ implies $a=b^{-1}$ which gives the wrong order for $a$ or $b$.  Similarly $ba=a$ implies that $b$ is the identity, and has order 1 instead of order 3, and we can rule out $ba=b$ and  $ba=b^2$ similarly. This leaves $ba=ab$ and $ba=ab^2$.
$ab=ba$ means that any product of $a$s and $b$s in any order can be rearranged into the form $a^ib^j$, and then to $a^{i\bmod 2}b^{j\bmod 3}$, and so is exactly the abelian group $Z_2\times Z_3 = Z_6$.
The final possibility is $ab = b^2a = b^{-1}a$.  This is exactly the defining relation of $D_6$, since a reflection followed by a rotation is the same as a reverse rotation followed by a reflection.  $D_6$ is isomorphic to $S_3$, with $a=(1 2)$ and $b = (1 2 3)$.
A: 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.
