Nth roots of unity

**DanielThrice** · December 12th 2010, 09:17 AM

So here was my initial postulate:

Show that Q/Z is isomorphic to the multiplicative group U∗ consisting of all roots of unity in C. (That is, U∗ = {z ∈ C|zn= 1 for some n ∈ Z+}.)

This is what I understand:

I know I probably need to use the first isomorphism/homomorphism theorem, which states that if you have a homomorphism f from G to G', then there is an isomorphism from the quotient group G/H to the image f(G), where H = Ker f.

So the idea is to exhibit a homomorphism between Q and U* whose kernel is precisely the integers. To do this, we first figure out what the identity in U* is (because we need to show that our eventual homomorphism takes the integers to this identity in U*).

Any thoughts on this problem?

**Drexel28** · December 12th 2010, 10:46 PM

Originally Posted by DanielThrice

So here was my initial postulate:

Show that Q/Z is isomorphic to the multiplicative group U∗ consisting of all roots of unity in C. (That is, U∗ = {z ∈ C|zn= 1 for some n ∈ Z+}.)

This is what I understand:

I know I probably need to use the first isomorphism/homomorphism theorem, which states that if you have a homomorphism f from G to G', then there is an isomorphism from the quotient group G/H to the image f(G), where H = Ker f.

So the idea is to exhibit a homomorphism between Q and U* whose kernel is precisely the integers. To do this, we first figure out what the identity in U* is (because we need to show that our eventual homomorphism takes the integers to this identity in U*).

Any thoughts on this problem?

What if we considered $\displaystyle f:\mathbb{Q}\to\mathbb{U}^*:\frac{p}{q}\mapsto e^{\frac{2\pi i p}{q}}$ . This is evidently a homomorphism since $\displaystyle f\left(\frac{p}{q}+\frac{r}{s}\right)=e^{2\pi i\left(\frac{p}{q}+\frac{r}{s}\right)}=e^{\frac{2\ pi i p}{q}}\cdot e^{\frac{2\pi i r}{s}}=f\left(\frac{p}{q}\right)\cdot f\left(\frac{r}{s}\right)$ . Also, we recall that if $\zeta$ is an $n^{\text{th}}$ root of unity then $\zeta=e^{\frac{2\pi i p}{n}}$ for some $p\in[n]$ and so $\dislaystyle f\left(\frac{p}{n}\right)=\zeta$ . Moreover, it's fairly clear that $\ker f=\mathbb{Z}$ from where it follows from the FIT that $\mathbb{Q}/\mathbb{Z}\cong\mathbb{U}^*$ .

Remark: You might find it interesting that what you call $\mathbb{U}^*$ is also the torsion subgroup $\text{Tor}\left(\mathbb{U}\right)$ of the circle group $\mathbb{U}$ .

**DanielThrice** · December 13th 2010, 07:32 AM

This is great, but how did you come up with the e ^ 2pi i p/q?

**Drexel28** · December 13th 2010, 07:42 AM

Originally Posted by DanielThrice

This is great, but how did you come up with the e ^ 2pi i p/q?

How did I come up with it? It just seemed natural, no? We know that all $n^{\text{th}}$ roots of unity are of the above form and just did what seemed right.

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