SubGroups Question

Mar 2010
51
3
Need some help with this question:

"Find four subgroups of <J,+>. Can you describe all subgroups of <J,+>
 

Drexel28

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Nov 2009
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Need some help with this question:

"Find four subgroups of <J,+>. Can you describe all subgroups of <J,+>
This is weird notation, but assuming that this is how Rudin uses \(\displaystyle J\) you mean \(\displaystyle J=\mathbb{Z}\)? If so how about \(\displaystyle J,2J,3J,4J\)? What do you think all of the subgroups look like?
 
Mar 2010
51
3
Yeah, in this particular text J holds the same value as \(\displaystyle \mathbb{Z}\).

I'm teaching myself abstract algebra and I'm not familiar with the notation \(\displaystyle J, 2J, 3J, ... , nJ\). Is this purely implying a scalar quantity? Such that \(\displaystyle nJ = ... -3n, -2n, -n, 0, n, 2n, 3n ...\)?
 

Drexel28

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Nov 2009
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Yeah, in this particular text J holds the same value as \(\displaystyle \mathbb{Z}\).

I'm teaching myself abstract algebra and I'm not familiar with the notation \(\displaystyle J, 2J, 3J, ... , nJ\). Is this purely implying a scalar quantity? Such that \(\displaystyle nJ = ... -3n, -2n, -n, 0, n, 2n, 3n ...\)?
Exactly! Why is this a subroup? What are the three (or possibly one) requirements they must satisfy?
 
Mar 2010
51
3
Well obviously \(\displaystyle nJ \subset J\) and given that \(\displaystyle a + b^{-1} \in J\) it logically follows that \(\displaystyle n(a + b^{-1}) \in nJ\), which quickly shows that \(\displaystyle nJ\) has all inverses and an identity which are also in \(\displaystyle J\).
 

Drexel28

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Well obviously \(\displaystyle nJ \subset J\)
So? Don't you mean that \(\displaystyle nz+Jn\subseteq Jn,\text{ }z\in\mathbb{Z}\), for closure?

and given that \(\displaystyle a + b^{-1} \in J\) it logically follows that \(\displaystyle n(a + b^{-1}) \in nJ\), which quickly shows that \(\displaystyle nJ\) has all inverses and an identity which are also in \(\displaystyle J\).
I don't quite follow, it is true that you can show a subgroup is a subgroup by showing that \(\displaystyle nz,nz'\in nJ\implies nz+(nz)^{-1}\in nJ\). Is that what you mean?
 
Mar 2010
51
3
I don't quite follow, it is true that you can show a subgroup is a subgroup by showing that \(\displaystyle nz,nz'\in nJ\implies nz+(nz)^{-1}\in nJ\). Is that what you mean?
That is exactly what I meant. I have a lot of the understanding behind it but the formal syntax is still a little new to me.
 

Drexel28

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That is exactly what I meant. I have a lot of the understanding behind it but the formal syntax is still a little new to me.
Right, ok then. So yeah that's correct since \(\displaystyle (nz')^{-1}=-nz'\) and \(\displaystyle nz+(-nz')=n(z-z')\in nJ\)
 
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Mar 2010
51
3
Thanks for the help! How would you go about tackling this extremely similar problem?:

Find four subgroups of \(\displaystyle <Q_{+} , \cdot>\).

Something in the form \(\displaystyle nQ_{+}\) wouldn't work the same way as with \(\displaystyle J\), right?
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
Thanks for the help! How would you go about tackling this extremely similar problem?:

Find four subgroups of \(\displaystyle <Q_{+} , \cdot>\).

Something in the form \(\displaystyle nQ_{+}\) wouldn't work the same way as with \(\displaystyle J\), right?
\(\displaystyle nJ\leqslant J\leqslant \mathbb{Q}\)!