# Identifying the group $G = \langle x,y \space | \space xy=yx, x^4=y^2 \rangle$ from the given presentation [duplicate]

I'm trying to solve a problem in my textbook which asks me to identify the groups $$G_1 = \langle x,y \space | \space x^3y=y^2x^2=x^2y\rangle$$ and $$G_2 = \langle x,y \space | \space xy=yx, x^4=y^2 \rangle$$ from the given presentations.

For $$G_1$$, I'm pretty sure I can say that $$x^3y=x^2y \implies x^3y(x^2y)^{-1} = e \implies x = e$$ (where $$e$$ is the identity), and then $$x^3y=y^2x^2 \implies y = y^2$$ as $$x=e$$ to get $$G_1 = \langle x,y \space | \space x=y=e \rangle \cong \{e\}$$ (though I'd appreciate it if you could tell me if I have this wrong).

What I'm struggling with is trying to do the same sort of thing for $$G_2$$ - I can't see any way of getting this into a form where I can see the represented group.

## marked as duplicate by user1729 group-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 7 at 10:17

• $xy=yx$ in $G_2$, so it's a (finitely-generated) abelian group. What do these look like? – Jane Doé May 5 at 21:38
• For $G_1$, you are correct; $x^3y=x^2y$ means $x=e$, so you get $y=y^2$, hence $y=e$. For $G_2$, note that $xy=yx$ means that the group is abelian, so every element can be written as $x^ay^b$ with $a,b\in\mathbb{Z}$. Now use the other relation to further simplify. – Arturo Magidin May 5 at 21:39
• $G_2$ is also in this question – Arnaud D. May 7 at 9:04
• The question about $G_2$ can be viewed as "find the abelianisation of the group $\langle x, y\mid x^4=y^2\rangle$", and so the techniques in the answers to this question are applicable here. – user1729 May 7 at 10:13
• In the mean time, unless anyone has serious objections I'll close this question as a duplicate of @ArnaudD.'s link: the answer to the $G_2$ question can be found there, while the $G_1$ solution was covered (correctly) in the question. – user1729 May 7 at 10:16
\begin{align}G_2 &= \langle x, y | xy = yx, x^4 = y^2 \rangle \\ &= \frac{ \langle x, y | xy = yx \rangle }{\langle \langle x^4y^{-2} \rangle \rangle} \\ &= \frac{\langle x \rangle_{\infty} \times \langle y \rangle_{\infty} }{\langle x^4y^{-2} \rangle} \\ &\cong \frac{\langle x \rangle_{\infty} \times \langle y \rangle_{\infty} }{\langle x^2y^{-1} \rangle} \times C_2 \\ &\cong C_{\infty} \times C_2.\end{align}