Can the possible groups by determined by hand? Suppose, $G$ is a group of order $12$ containing an element $a$ with order $4$. Can I show the following facts by hand ?


*

*The group is either cyclic or isomorphic to the group $C3:C4$

*$a^2$ is the only element in the group with order $2$


Cauchy's theorem tells us that there is at least an element with order $2$ 
and at least an element with order $3$. If $b$ has order $3$, is it correct
to claim $$G=\{e,a,a^2,a^3,b,ba,ba^2,ba^3,b^2,b^2a,b^2a^2,b^2a^3\},$$ and 
if yes, does this help to show the facts above ?
 A: We can show it with Sylow's theorems. Let $\langle b\rangle$ be a Sylow-$3$ subgroup. The number of Sylow-$3$ subgroups is $\equiv 1 \pmod 3$ and it divides $12$; it is $1$ or $4$. 
Suppose Sylow-$3$ subgroup is not normal (so there should be $4$ Sylow-$3$ subgroups); collect their non-identity elements, we cover $8$ elements of order $3$. 
What about remaining $4$ elements? There is already a subgroup of order $4$, and hence we should have unique subgroup of order $4$; so $\langle a\rangle$ is unique (so normal) subgroup of order $4$. This forces that $a^2$ is the unique element of order $2$ (wait! proof is incomplete), hence it is in the center of $G$. 
Take product of this central element of order $2$ with elements of order $3$; we get elements of order $6$. This leads a contradiction, since we have 
$$\mbox{($8$ elements of order $3$) +($4$ elements of order dividing $4$)+some elements of order $6$ }$$
this is exceeding $|G|$. 
Hence, the Sylow-$3$ subgroup is normal, hence unique. This proves first bullet in question depending on whether subgroup of order $4$ is unique or non-unique.
We already proved that $\langle b\rangle$ is unique subgroup of order $3$, so $\langle b\rangle\trianglelefteq G$. Then we have $a\langle b\rangle a^{-1}=\langle b\rangle$. But then what is $aba^{1-}$? It is member in $\{1,b,b^2\}$ and $aba^{-1}$ can not be $1$ (why?)
If $aba^{-1}=b$ then $ab=ba$, so $G$ becomes cyclic group of order $12$; it contains unique element of order $2$. 
If $aba^{-1}=b^2$ then $a^2ba^{-2}=(b^2)^2=b$; so $a^2$ commues with $b$ (and obviously with $a$). Hence $a^2$ is central element of order $2$. 
Suppose, there is another element $c$ of order $2$. Then, since $a^2$ is central, $a^2$ with $c$ will form a Klein-4 group of order $4$; it is Sylow-$2$ subgroup. But hypothesis says that the Sylow-$2$ subgroup is cyclic, a contradiction. Thus $a^2$ is unique element of order $2$. 

Another way to show uniqueness of element of order $2$.
$\langle a^2\rangle$ is a given subgroup of order $2$, and is contained in a cyclic subgroup $H=\langle a\rangle$ of order $4$. Suppose $c$ is another element of order $2$. Then $\langle c\rangle$, being a $2$-subgroup, must be contained in a Sylow-$2$ subgroup, say $K$. Since $K$ is conjugate to $H$, $K$ is cyclic. Then $H\cap K=1$, hence $|HK|=\frac{|H|.|K|}{|H\cap K|}=4.4/1>|G|$, a contradiction. Thus, 

If $|G|=12$ and if $G$ contains an element of order $4$, then $G$ contains unique element of order $2$. 

