Isomorphic to \((\mathbb Z \times\mathbb Z )\,\big/\langle(1,2)\rangle\)
- Grigorios Kostakos
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Isomorphic to \((\mathbb Z \times\mathbb Z )\,\big/\langle(1,2)\rangle\)
Which group is isomorphic to the quotient group \(({\mathbb{Z}}\times{\mathbb{Z}})\big/\langle{(1,2)}\rangle\) , where \(\langle{(1,2)}\rangle\) is the normal subgroup of \({\mathbb{Z}}\times{\mathbb{Z}}\) generated by \((1,2)\) ? Justify your answer.
Grigorios Kostakos
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- Grigorios Kostakos
- Founder
- Posts: 461
- Joined: Mon Nov 09, 2015 1:36 am
- Location: Ioannina, Greece
Re: Isomorphic to \((\mathbb Z \times\mathbb Z )\,\big/\langle(1,2)\rangle\)
We give a solution:
The function \(f:{\mathbb{Z}}\times{\mathbb{Z}}\longrightarrow\mathbb{Z}\) defined as \[f((m,n))=2m-n\,,\] it is a well defined epimorphism (surjective homomorphism) with \begin{align*}
\ker{f}&=\bigl\{{(m,n)\in\mathbb{Z}\times{\mathbb{Z}}:f((m,n))=0}\bigr\}\\
&=\bigl\{{(m,n)\in\mathbb{Z}\times{\mathbb{Z}}:n=2m}\bigr\}\\
&=\bigl\{{(m,2m):m\in\mathbb{Z}}\bigr\}\\
&=\bigl\langle{(1,2)}\bigr\rangle\,.
\end{align*}
So, by the 1st Isomorphism theorem we have that \(({\mathbb{Z}}\times{\mathbb{Z}})\big/\bigl\langle{(1,2)}\bigr\rangle\cong\mathbb{Z}\,.\)
The function \(f:{\mathbb{Z}}\times{\mathbb{Z}}\longrightarrow\mathbb{Z}\) defined as \[f((m,n))=2m-n\,,\] it is a well defined epimorphism (surjective homomorphism) with \begin{align*}
\ker{f}&=\bigl\{{(m,n)\in\mathbb{Z}\times{\mathbb{Z}}:f((m,n))=0}\bigr\}\\
&=\bigl\{{(m,n)\in\mathbb{Z}\times{\mathbb{Z}}:n=2m}\bigr\}\\
&=\bigl\{{(m,2m):m\in\mathbb{Z}}\bigr\}\\
&=\bigl\langle{(1,2)}\bigr\rangle\,.
\end{align*}
So, by the 1st Isomorphism theorem we have that \(({\mathbb{Z}}\times{\mathbb{Z}})\big/\bigl\langle{(1,2)}\bigr\rangle\cong\mathbb{Z}\,.\)
Grigorios Kostakos
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