Subgroup of order 12 
Question: 
  Suppose that $H$ is a normal subgroup of a group $G$. If $\left | H \right |=4$ and $gH$ has order $3$ in $G/H$, find a subgroup of order $12$ in $G$.

By the property of cosets:
$\left | gH \right | = \left | H \right |=3$
Note also that $\left | gH \right | =3$ implies that $\left ( gH \right )^{3}= g^{3}H=H$
Now, $\left | H \right | =3$ so $\forall\: h \in H: h^{\left | H \right |}=h^{3}=e$
By Lagrange's theorem, the order of a subgroup $H$ divides the order of a group $G$.
From this:
$\left | H \right |$ divides $\left | G \right |$ and $\left | g \right |$ divides $\left | G \right |$
Respectively, we have $\left | G \right |=\left | H \right |q$ for some $q \in \mathbb{Z^{+}}$
This is just $4q$
and
$\left | G \right |=\left | g \right |\bar{q}$ for some $\bar{q} \in \mathbb{Z^{+}}$.
This is just $3 \bar{q}$
So, $4q=3\bar{q}$
This is just $q=4$ and $\bar{q}=3$.
Thus, $\left | G \right | =12n \forall n \in \mathbb{Z^{+}}$
By Lagrange's theorem, there must exists a subgroup of order $12$.
Is my argument valid?
Thanks in advance.
 A: The set $S=H\cup gH\cup g^2H$ forms a subgroup of $G$ (why?). It has order $12$ (why? your proof needs to use that cosets have empty intersection). Done.
As to the proof in your question, you seem to be using the "fact" that if $n$ divides the order of a group $G$ then $G$ has a subgroup of order $|G|/n$. This is not true, and any non-abelian simple group gives a counter-example (as they all have even order*, so take $n=...$what?).
A second (but minor) issue is that you state $|g|=3$. This is incorrect. Instead, $|gH|=3$. Can you come up with a counter-example?
*This is highly non-trivial, but just take $A_5$ as an example. It has order $60$, so even.
A: Now you know how to proceed, but let me point out two glaring errors in your proof:
While it is true that $\lvert H \rvert = \lvert gH \rvert$, this quantity is not the order of $gH$ in $G/H$. For example, $H = 3\mathbb{Z}$ has infinite order in $G = \mathbb{Z}$, but the order of $1+H$ in $G/H$ is $3$.
Also, the "converse" of Lagrange's theorem that you use that the end isn't valid. However, look up Cauchy's theorem and the Sylow theorems, which are tools that you can use for future reference to find subgroups of certain orders.
