I have a group $G$ defined by $G = \langle x,y,z|x^2 = y^3 = z^3 = xyz \rangle$ and we know that $a$ $=$ $xyz$ belongs to the centre of $G$. But im struggling to show that $\frac{G}{\langle a\rangle} \cong A_4$.
Seeing as its a quotient, I thought the best way to show that it's isomorphic to $A_4$ would be to use the first isomorphism theorem, where I find a homomorphic such that its image is $A_4$ and it's kernel $\langle a\rangle$ but I'm having trouble finding such a homomorphism, if it exists.
I then have to go onto show that $|G| = 24$. Brute force seems a little strenuous for such a big group, but i worked out that $a^2 = 1$ but im not sure how i could use such a fact to show that the order of $G$ is $24$.
Let $a=(12)(34)$, $b=(123)$, and $c=(234)$ in $A_4$. These elements generate $A_4$. Notice that $a^2=1$, $b^3=1$, and $c^3=1$. We also have that $abc=1$. Thus $a^2=b^3=c^3=abc$, so if we map $x\mapsto a$, $y\mapsto b$, and $z\mapsto c$ we get a surjective homomorphism from $G$ to $A_4$. Call this homomorphism $\phi$. $A_4$ has the presentation $\langle a,b|a^2,b^3,(ab)^3\rangle$, and $c=(ab)^2$. Thus $\phi(xyz)=\phi(xy)(ab)^2=(ab)^3=1$, so $xyz$ is contained in the kernel. Comparing the presentations we see that the kernel must in fact be $\langle xyz\rangle$, which gives you the isomorphism you desire.
To see that $G$ has $24$ elements, note that $A_4$ has $12$ elements and $A_4$ is the quotient of $G$ by a subgroup of order $2$. The order of a quotient group is the number of cosets of the subgroup we are taking the quotient by, so since each coset has $2$ elements and they partition $G$, $G$ must have $24$ elements.