Show that $G=\langle g,h| hg^3, h^2g^5\rangle $ is equal to $\{e\}$ Show that:
$$G=\langle g,h| hg^3, h^2g^5\rangle $$
is equal to $\{e\}$.
Is
$$H=\langle g,h| hg^3, h^2g^4\rangle $$
equal to $\{ e \}$?
I feel like this question should be relatively easy however I think I am confused by the notation does the first group mean the group generated by elements of the form $hg^3$ and $h^2g^5$ or is it when these elements occur they are equal to the identity?
 A: $G$ is generated by $g$ and $h$ and we have $hg^3=1$ and $h^2g^5=1$ (and to be precise, is the "most general" group with these conditions).
That is, $G=F/N$ where $F$ is the free group on the two generators $g,h$ and $N$ is the smallest among all normal subgroups of $F$ containing $hg^3$ and $h^2g^5$.

To wit, we find
$$ g=1\cdot g=h^2g^5\cdot g=h\cdot hg^3\cdot g^3=hg^3=1$$
and hence also
$$ h=h\cdot 1^3=h\cdot g^3=1.$$
Therefore $G=\{1\}$.

On the other hand, for $H=\langle g,h\mid hg^3, h^2g^4\rangle$ one verifies that $g\mapsto 1+2\Bbb Z$, $h\mapsto 1+2\Bbb Z$ induces a homomorphism $\phi\colon H\to \Bbb Z/2\Bbb Z$. The verification of this consists in checking that all relations map to the neutral element. Indeed $\phi(hg^3)=\phi(h)+3\phi(g)=0+2\Bbb Z$ and $\phi(h^2g^4)=2\phi(h)+4\phi(g)=0+2\Bbb Z$. As $\phi$ is clearly onto, $H$ cannot be the trivial group.
How did I get the idea to define this $\phi$? By playing around with the relations similar to the case of $G$, one finds $hg=h\cdot hg^3\cdot g=h^2g^4=1$, so $h$ cane be eliminated using $h=g^{-1}$. Then the two relations both turn into $g^2=1$. This suggests that $H\cong \Bbb Z/2\Bbb Z$.
