A question from GRE math sub 9367, problem 59 Two subgroups H and K of a group G have orders 12 and 30, respectively. Which of the following could NOT be the order of the subgroup of G generated by H and K?
A. 30
B. 60
C. 120
D. 360
E. Countable infinity
A is the answer because H, with order 12 that doesn't divide 30, can't be a subgroup of K.
But anybody can help me construct a concrete example of E, the subgroup generated by H and K with order of countable infinity?
 A: Consider the free product (group coproduct) $G* H$. It is defined as follows:
If we start by considering $G$ and $H$ as disjoint sets, $G*H$ is the set of finite words whose letters are the elements of $G\cup H$ other than their identity elements. Group multiplication is given by simply writing one word after another, and the empty word is the identity in this group. What we've constructed so far is a free group.
Now, to make $G$ and $H$ subgroups of this group, we mod out by the relations given by multiplication in $G$ and multiplication in $H$. In other words, if a word has a sequence of letters that are all elements of $G$, we can write them as a single letter in $G$, the letter we get by computing the product (in $G$) of that sequence of letters. We do the same for $H$.
This group is countable since the set of all finite words on a finite set of letters is countable. It is infinite because we have words like $ghghghgh$ of arbitrary length (here $g\in G$ and $h\in H$). It contains copies of $G$--those words that consist only of letters in $G$ (and can thus be expressed with a single element of $G$), and a copy of $H$ likewise.
A: For any integers $m\ge3$ and $n\ge2$, any countable group $F$ can be embedded in a $2$-generator group $G=\langle a,b\rangle$ where $a$ has order $m$ and $b$ has order $n$ [F. Levin, Factor groups of the modular group, J. London Math. Soc. 43 (1968), 195-203; Theorem 2.1].
Let $F$ be any countably infinite group. If you apply Levin's construction with $m=12$ and $n=30$, then $G$ is a countably infinite group, $H=\langle a\rangle$ is a subgroup of order $12$, $K=\langle b\rangle$ is a subgroup of order $30$, and $E=\langle H,K\rangle=\langle a,b\rangle=G$.
The point of posting yet another answer is that the example $E$ you are asking for not only exists (as has already been shown), it can be an "arbitrarily large" countable group, in the sense that it can be made to contain any countable group you like.
A: With  $\alpha=\frac{2\pi}{12}$ and $\beta=\frac{2\pi}{30}$, let $H$ be the subgroup of $SO(3,\mathbb R)$ generated by $$\begin{pmatrix}\cos\alpha&\sin\alpha&0\\
-\sin\alpha&\cos\alpha&0\\
0&0&1\end{pmatrix} $$
and $K$ by 
$$\begin{pmatrix}1&0&0\\
0&\cos\beta&\sin\beta\\
0&-\sin\beta&\cos\beta\end{pmatrix}. $$
Together they generate a group that is not among the few well-known finite subgroups of $SO(3,\mathbb R)$. On the other hand, as a finitely generated group, it cannot be bigger than countably infinite.
