I want to show that in general the amalgamated product of finite cyclic groups is not abelian.
Therefore, I define $A := \langle a \rangle$, $H_1 := \langle x \mid x^2 \rangle$ and $H_2 := \langle y \mid y^4 \rangle$, and $$ \phi_1 \colon A \to H_1, \quad a \mapsto x \,, \qquad \phi_2 \colon A \to H_2, \quad a \mapsto y^3 \,. $$
So the amalgamated product of $H_1$ and $H_2$ with respect to $A$ is $\langle x, y \mid x^2, y^4, x = y^3 \rangle$. Using Tietze transformation I get $\langle y \mid y^6, y^4 \rangle$.
Does this work or do I miss something?
Thanks for your answers!
Your computation of the amalgamated product $H_1 *_A H_2$ is correct. But it should be clear to you that $H_1 *_A H_2$ is again cyclic, and therefore abelian. (More explicitly, $H_1 *_A H_2 ≅ ⟨y \mid y^4, y^6⟩ ≅ ⟨y \mid y^2⟩ ≅ ℤ/2$.)