Use of Fundamental structure of finite abelian group Let $G$ be a finite abelian group generated by $x, y$ i.e $G = \langle x, y \rangle$. By the Fundamental Theorem of a finite abelian group, we have $G \cong \mathbb (Z_m, +_m) \times (\mathbb Z_n, +_n)$ for some $m, n \in  \mathbb N$. But I want to show that 


$G \cong \langle x \rangle \times \langle y^d \rangle $, where $d = |\langle x \rangle \cap \langle y \rangle|$.


We define a map $\psi : \langle x \rangle \times \langle y^d \rangle \rightarrow G$ by $\psi((x^i, y^{dj})) = x^iy^{dj}$ which is homomorphism. Since $d$ divides $o(b)$ so that g.c.d $(d, o(b)) = d$. Thus we have $|G| = |\langle x \rangle \times \langle y^d \rangle|$. We need only to show that  $\psi$ is one-one map or $\langle x \rangle \cap \langle y^d \rangle = \{ e \}$, where $e$ is the identity element in $G$ as. I unable to prove $\langle x \rangle \cap \langle y^d \rangle = \{ e \}$. I would be thankful for any kind of help.
 A: Define a map $\phi : \mathbb Z \times \mathbb Z : \rightarrow \mathbf G$ such that $\phi(m, n) = a^m b^n$ which is an onto homomorphism and ker$\phi = K$ is a subgroup of  $\mathbb Z \times \mathbb Z$.  I have seen this link, $K$ is one of the three types. In the first case $K = \langle (0, b) \rangle$ for some $b > 0$. Thus, by Gallian paper https://www.jstor.org/stable/24337849?seq=1#page_scan_tab_contents,  $\mathbf G \cong \frac{\mathbb Z \times \mathbb Z}{K} \cong \mathbb Z_b \times \mathbb Z$, which is not possible, because $G$ is finite but $ \mathbb Z_b \times \mathbb Z$ is infinite. Similarly, $K$ is not of the form $\mathbb Z b e_2$. Thus only the third case is left. In the third case, we have $K= \langle (a, b) \rangle \oplus \langle (0,c) \rangle$ for some $a, b, c > 0$. Let $A = \langle (a, b) \rangle, B = \langle (0,c) \rangle $, by second isomorphism theorem, for a subgroup $A, B$ of $G$ such that $A \leq N_G(B)$, we have $\frac{AB}{B}\cong \frac{A}{A \cap B}$ or $\frac{A + B}{B}\cong \frac{A}{A \cap B}$ . Thus $\frac{K}{\langle a, b \rangle }\cong \langle (0,c) \rangle$. Use third isomorphism theorem, $(G/H)\big/(K/H) \cong G/K$, where $H, K$ are normal subgroup of $G$ with $H \leq K$. Take $H  = \langle (a, b) \rangle$ and $K = $Ker$\phi, G = \mathbb Z \times \mathbb Z$, we have $$\frac{\mathbb Z \times \mathbb Z}{K} \cong \big(\mathbb Z \times \mathbb Z/\langle (a, b) \rangle\big)\big/\big(K/\langle (a, b) \rangle\big) \cong (\mathbb Z_d \times \mathbb Z )/ \langle (0,c) \rangle$$
Next, we define a map $\psi : \mathbb Z_d \times \mathbb Z / \langle (0,c) \rangle \rightarrow \mathbb Z_d \times Z_c$ by $\psi\overline{(\bar x, y)} = (\bar x, \bar y)$, where $\bar y \in \mathbb Z_c$, which is an isomorphism.
