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Math Help - Nth roots of unity

  1. #1
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    Nth roots of unity

    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?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by DanielThrice View Post
    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}.
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  3. #3
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    This is great, but how did you come up with the e ^ 2pi i p/q?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by DanielThrice View Post
    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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