Let $G=\langle x,y,z:x^2=y^3=z^3=xyz=1\rangle$, let $H=\langle z\rangle$. If $S=\{1,y,y^2,y^2z\}$, show the following:
(a) $HSy\subseteq HS$ and $HSz\subseteq HS$
Now how would I be able to show (a) without using direct computations? This is a very long computation (I tried it), so if there is a more efficient way of showing this please fill me in.
Use the definition: $$HS = \{ h \cdot s : h \in H, s \in S \}.$$ If you fix an $a$, $$HSa = \{h\cdot s\cdot a : h \in H, s \in S\}.$$ Try to write an arbitrary element from the second set as an element of the first. For example, an arbitrary element of $HSy$ looks like $hsy$ where $h \in H$, $s \in S$. Can you use the fact that $y \in S$ to rewrite this as $hs^\prime$ for some $s^{\prime} \in S$ (i.e. show that if $s \in S$ then $sy \in S$)?
**After your comment, I think the better way to do this is to notice the following: your last relation says that $yz=x$, and since $x = x^{-1}$, we also have $z^2 y^2 = x$. The elements of $HSy$ are all clearly in $HS$ with the possible exception of those of the form $h\cdot(y^{2}z)y$ for some $h \in H$, these look like $z^a y^{2} z y$ for some $a$. But these can always be written as $z^{a-2}(z^{2}y^{2})(zy)$. Can you finish from here?