# SubGroups Question

#### jameselmore91

Need some help with this question:

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

#### Drexel28

MHF Hall of Honor
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?

#### jameselmore91

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

MHF Hall of Honor
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?

#### jameselmore91

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

MHF Hall of Honor
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?

#### jameselmore91

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

MHF Hall of Honor
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 $$\displaystyle nz+(-nz')=n(z-z')\in nJ$$

• jameselmore91

#### jameselmore91

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
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?
$$\displaystyle nJ\leqslant J\leqslant \mathbb{Q}$$!