Finding/Recognising non-cyclic proper subgroups. $Q$ is a multiplicative group of order $12$. You are given that two elements of $Q$ are $a$ and $r$ and that $r$ has order $6$ and $a^2=r^3$ You are also given that $a$ has order $4$, $a^2$ has order $2$, $a^3$ has order $4$; as $r^6=e$, $r^2$ has order $3$. Where $e$ is the identity element.

The table below shows the number of elements of $Q$ with each possible order.
  \begin{array}{|l|c|c|c|c|c|}
\hline
\text{Order of element} & 1 & 2 & 3 & 4 & 6\\
\hline
\text{Number of elements} & 1 & 1 & 2 & 6 & 2\\
\hline
\end{array}
  $G$ and $H$ are non-cyclic groups of order $4$ and $6$ respectively.
Construct two tables, similar to the one above, to show the number of elements with each possible order for the groups $G$ and $H$. Hence explain why there are no non-cyclic proper subgroups of $Q$.

***END OF QUESTION***
I have had difficulty in constructing these tables, as for $H$, I am confused as to whether there are $3$ elements of order $2$; or $2$ elements of order $3$. So if you could talk me through your logic I would be most grateful.
Thanks in advance,
Regards, Blaze. 
 A: Since we are told $H$ is non-cyclic, it has NO elements of order $6$. So it only has elements of order $1,2$ or $3$. We have a unique element of order $1$, the identity. If $h \in H$ has order $3$, so does its inverse, and $h \neq h^{-1}$, so we either have $0,2$ or $4$ elements of order $3$ (they come in pairs).
We can rule out $0$ right away: if every non-identity element has order $2$, then for any two such, say $a,b$, we have that $\{e,a,b,ab\}$ is a subgroup of order $4$ of $H$, but $4 \not\mid 6$. So we have at least one element of order $3$ and one element of order $2$, call them $x$ and $y$.
It can be shown that $\{e,x,x^2,y,xy,x^2y\}$ are all distinct, so these must be the elements of $H$. Since $yx \in H$ as well, it must equal one of the non-identity elements. Again, it can be shown (you should do this!) that we have just two possibilities: $yx = xy$ or $yx = x^2y$.
In the first case, we have that: $(xy)^2 = x^2y^2 = x^2; (xy)^3 = x^3y^3 = y$ which means $xy$ must have order $6$. Since that would make $H$ cyclic, it must be that $yx = x^2y$.
This actually allows us to calculate the order of $xy$:
$(xy)^2 = (xy)(xy) = x(yx)y = x(x^2y)y = x^3y^2 = ee = e$, so $xy$ has order $2$. Since this means we have at least TWO elements of order $2$, we can have at MOST $2$ elements of order $3$, which means we have exactly two elements of order $3$ (since they come in pairs), and thus $3$ elements of order $2$.
A: $|H|=6$, so every element of $H$ has order 1, 2, 3, or 6.  The identity is the only element with order 1.  Since $H$ is not cyclic, there is no element of order 6.  Therefore all five nonidentity elements have order 2 or 3.  Powers of an element of order 3 produce two new (nonidentity) different elements of order 3.  Powers of an element of order 2 only produce the element.  Thus, $H$ has at least one and up to five elements of order 2.  If $H$ has more than one element of order 2, then it has more elements of order 2 than does $Q$ so cannot be a subgroup of $Q$.  So we consider that $H$ has only one element of order 2.  Then $H$ has two elements of order 3.
Are you having further trouble showing that such an $H$ cannot be a subset of $Q$?
