Isomorphism and cyclic groups

[Papapetros Vaggelis]

Prove that

1) the quotient group \(\displaystyle{\left(\left(\mathbb{Z}\times \mathbb{Z}\right)/\langle{\left(2,2\right)}\rangle,+\right)}\) is isomorphic to the group \(\displaystyle{\left(\mathbb{Z}_{2}\times \mathbb{Z},+\right)}\), where \[\displaystyle{+:\left(\mathbb{Z}_{2}\times \mathbb{Z}\right)\times \left(\mathbb{Z}_{2}\times \mathbb{Z}\right)\longrightarrow \left(\mathbb{Z}_{2}\times \mathbb{Z}\right)}\] \[\displaystyle{\left(\left(\left[k_1\right]_{2},m\right),\left(\left[k_2\right]_{2},n\right)\right)\mapsto \left(\left[k_1+k_2\right]_{2},m+n\right)}\,.\]

2) if \(\displaystyle{\left(G,\cdot\right)}\) and \(\displaystyle{\left(H,\cdot\right)}\) are cyclic groups such that \(\displaystyle{\circ(G)<\infty\,,\circ(H)<\infty}\) , then the group \(\displaystyle{\left(G\times H,\cdot\right)}\) is cyclic if, and only if, \(\displaystyle{\gcd\,\left(\circ(G),\circ(H)\right)=1}\).

[tziaxri]

For the first part we have:
Let the function \[ f : { \displaystyle{\left(\mathbb{Z}\times \mathbb{Z}\right)}}\longrightarrow \left(\mathbb{Z}{_2}\times \mathbb{Z}\right) \] defined as: \[ f((m,n)) = (\left[m]_{2},n-m\right) \,.\] It is easy to see that f is a well defined function. Also f is a group homomorphism for : \[ \forall (m,n),(m',n') \in {\left(\mathbb{Z}\times \mathbb{Z}\right)} f((m,n)+(m',n'))=f(m+m',n+n')\] \[ =(\left[m]_{2},n-m\right)+(\left[m']_{2},n'-m'\right)=f(n,m)+f(n',m').\] Moreover, f is an epimorphism for:
\[\forall \displaystyle (\left[m]_{2},n\right) \in (\mathbb{Z}{_2}\times \mathbb{Z}), \exists (m,n+m) \in{\left(\mathbb{Z}\times \mathbb{Z}\right)}\] so that :\[ f(m,n+m)= (\left[m]_{2},n\right).\] with \[ ker{f}=\bigl\{{(m,n)\in\mathbb{Z}\times{\mathbb{Z}}:f((m,n))=(\left[0]_{2},0\right)}\bigr\}\] \[ = \bigl\{(m,n)\in\mathbb{Z}\times{\mathbb{Z}}:m=2k,k\in\mathbb{Z} ,m=n\bigl\} \] \[ = \bigl\{(2k,2k) \in\mathbb{Z}\times{\mathbb{Z} } \bigl\} = \langle{(2,2)}\rangle \] So, from the 1st Isomorphism Theorem, we have that \[ \displaystyle{\left(\left(\mathbb{Z}\times \mathbb{Z}\right)/\langle{\left(2,2\right)}\rangle,+\right)}\cong \left(\mathbb{Z}{_2}\times \mathbb{Z}\right)\].