Isomorphic to \((\mathbb Z \times\mathbb Z )\,\big/\langle(1,2)\rangle\)

Groups, Rings, Domains, Modules, etc, Galois theory
Post Reply
User avatar
Grigorios Kostakos
Founder
Founder
Posts: 461
Joined: Mon Nov 09, 2015 1:36 am
Location: Ioannina, Greece

Isomorphic to \((\mathbb Z \times\mathbb Z )\,\big/\langle(1,2)\rangle\)

#1

Post by Grigorios Kostakos »

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

Tags:
User avatar
Grigorios Kostakos
Founder
Founder
Posts: 461
Joined: Mon Nov 09, 2015 1:36 am
Location: Ioannina, Greece

Re: Isomorphic to \((\mathbb Z \times\mathbb Z )\,\big/\langle(1,2)\rangle\)

#2

Post by Grigorios Kostakos »

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 \(({\mathbb{Z}}\times{\mathbb{Z}})\big/\bigl\langle{(1,2)}\bigr\rangle\cong\mathbb{Z}\,.\)
Grigorios Kostakos
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 11 guests