I don't quite understand about pushout does anyone help me in this matter?
Given $f: \mathbb{Z} \to \mathbb{Q}$ be the inclusion and $g: \mathbb{Z} \xrightarrow{n} \mathbb{Z}$ be multiplication by $n$. How can I show that the pushout of $f$ and $g$ is isomorphic in relation to $\mathbb{Q}/\mathbb{Z} \oplus \mathbb{Z}/n\mathbb{Z}$ ?
A common way to show an isomorphism between a pushout and a given group is to show that the given group satisfies the universal property for the desired pushout. That is, in your case, you would want to find homomorphisms $a : \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \oplus \mathbb{Z}_n$ and $b : \mathbb{Z} \to \mathbb{Q}/\mathbb{Z} \oplus \mathbb{Z}_n$ such that $a \circ f = b \circ g$, and such that for any group $G$ and homomorphism $a' : \mathbb{Q} \to G$ and $b' : \mathbb{Z} \to G$ satisfying $a' \circ f = b' \circ g$, there exists a unique homomorphism $h : \mathbb{Q}/\mathbb{Z} \oplus \mathbb{Z}_n \to G$ such that $a' = h \circ a$ and $b' = h \circ b$.
However, in this case, I do not think there can be any such isomorphism. My argument: if we let $g' : \mathbb{Q} \to \mathbb{Q}$ be multiplication by $n$, then $g' \circ f = f \circ g$. Therefore, by the universal property of the pushout, there would be a unique homomorphism $h : \mathbb{Q}/\mathbb{Z} \oplus \mathbb{Z}_n \to \mathbb{Q} \to \mathbb{Q}$ such that $g' = h \circ i_1$ and $f = h \circ i_2$. But now, $\mathbb{Q}/\mathbb{Z} \oplus \mathbb{Z}_n$ is a torsion group, whereas $\mathbb{Q}$ is a torsion-free group, which would force $h$ to be the zero homomorphism. Then that would imply $g' = h \circ i_1 = 0$ and $f = h \circ i_2 = 0$, both of which are clearly false.
(By a refinement of the argument in the previous paragraph, you can show that if $P$ is the pushout with corresponding homomorphisms $i_1 : \mathbb{Q} \to P$ and $i_2 : \mathbb{Z} \to P$, then $i_1$ and $i_2$ must both be injective -- which again would make it impossible to have $P \simeq \mathbb{Q}/\mathbb{Z} \oplus \mathbb{Z}_n$.)
If you had not come up with the previous argument, then here is one that's possibly more straightforward to come up with: we know that the pushout is isomorphic to the quotient of $\mathbb{Q} \oplus \mathbb{Z}$ by the subgroup $\{ (m, -nm) \mid m \in \mathbb{Z} \} = \langle (1, -n) \rangle$. However, again in the quotient group $(\mathbb{Q} \oplus \mathbb{Z}) / \langle (1, -n) \rangle$, no integer multiple of $\overline{(1, 0)}$ is equal to zero, so the pushout has at least one nonzero torsion-free element. Yet again, that makes it impossible for the pushout to be isomorphic to a torsion group such as $(\mathbb{Q} / \mathbb{Z}) \oplus \mathbb{Z}_n$.