# Let $G=\langle X\mid R\rangle$ and $H=\langle X\mid S\rangle$. If $R\subseteq S$, the $H$ is isomorphic to a quotient of $G$.

Assume that a group $G$ has a presentation $\langle X \mid R \rangle$ and a group $H$ has a presentation $\langle X \mid S \rangle$. If $R \subseteq S$, the $H$ is isomorphic to a quotient of $G$.

In fact, if $F_X$ is the free group and $\varphi: F_X \rightarrow G$ and $\psi:F_X \rightarrow H$ are the homomorphism, then $\ker \varphi = R^{F_X} \leq S^{F_X} = \ker \psi$. So, $H \cong F_X/\ker\psi$ is isomorphic to a quotient of $F_X/\ker\varphi \cong G$.

• There might be a typo in your group presentation of $G$. If not, then $G$ isn't generated by a single element. To see this note that it has torsion (since $(aba)^4 = 1$) and that $a$ is of infinte order in $G$. – Stefan Mesken Apr 8 '16 at 6:46
• Okay. What are your thoughts? Do you see that $aba = bab$ in $H/ < (aba)^4 >$? – Stefan Mesken Apr 8 '16 at 7:09
• I thought of showing that $ker \varphi / \ker \psi \cong <(aba)^4>$ (Third isomorphism theorem), so that the quotient of the normal closures is $<(aba)^4>$ – math91 Apr 8 '16 at 7:15
• I'm the \langle \rangle fairy, here to let you know that $\langle, \rangle$ plays nicer with TeX than <, > does :) – Patrick Stevens Apr 8 '16 at 8:17
• Also \mid ($\mid$) gives you better spacing as a separator than the pipe character | does. – Patrick Stevens Apr 8 '16 at 8:17