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I mostly need help with proving $G$ is closed but a verification of the other parts is appreciated.

Let $G = \{z \in \mathbb{C} \mid z^n=1$ for some $n\in \mathbb{Z^+}\}$ I want to start by proving $G$ is closed under multiplication. So $z_1 z_2 = z_{1 \cdot 2}$ and I need to show that $z_{1\cdot2}^n = 1$ I was thinking of breaking $z_1$ and $z_2$ into prime factors(?) to show that the property is retained through multiplication. Is this the correct place to start?

To prove multiplication is associative, Let $z_1 = a+bi$ and $z_2 = c+di$ and $z_3 = e + fi$ $$(z_1 z_2) z_3 = ((a+bi) (c+di)) (e + fi) = (ac+adi+cbi-bd)(e + fi)$$ $$(z_1 z_2) z_3 = ace+adei+cbei-bde + acfi - adf - cbf - bdfi$$ $$z_1 (z_2 z_3) = (a+bi) ((c+di) (e+fi)) = (a+bi) (ce+cfi+edi-df)$$ $$z_1 (z_2 z_3) = cbei - cbf - bde - bdfi + ace + acfi + adei - adf$$ it's messy but this shows they are equal.

identity element would be 1 since $a+bi \cdot 1 = a+bi$ and the inverse would be $z^{-1} = \frac{1}{a+bi}$ $z z^{-1} = \dfrac{a+bi}{a+bi} = 1$.

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    $\begingroup$ You don't need to prove that multiplication is associative because multiplication is associative in $\mathbb C$. $\endgroup$
    – lhf
    May 16, 2016 at 18:47
  • $\begingroup$ Is $n$ fixed? In your set-theoretic description of $G$, each $n$ seemingly depends on $z$. $\endgroup$ May 16, 2016 at 18:47
  • $\begingroup$ $n$ is not fixed $\endgroup$
    – Obliv
    May 16, 2016 at 18:58
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    $\begingroup$ This question got two down-votes. Can anyone explain why? $\qquad$ $\endgroup$ May 16, 2016 at 19:19

1 Answer 1

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Hint: if $x^n = 1$ and $y^m = 1$, what is $(xy)^{mn}$?

The only other thing you really need to verify (given that associativity holds in $\mathbb C$) is that the reciprocal of a root of unity is a root of unity.

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    $\begingroup$ Oh I see. $(xy)^{mn} = x^{mn} * y^{mn} = 1^{m} * 1^{n} = 1$ for all $m,n \in \mathbb{Z^+}$. Thank you very much. How should I prove the reciprocal of a root of unity is a root of unity? It satisfies $z^n = 1$ since it's $\frac{1}{z^n} = 1$ and if the root that it is the reciprocal of satisfies $z^n = 1$ then $\frac{1}{1} = 1$. Is that all I need to do? $\endgroup$
    – Obliv
    May 16, 2016 at 18:57
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    $\begingroup$ As a side question: Is this group also abelian? It seems like $(a+bi)(c+di) = (c+di)(a+bi)$ $\endgroup$
    – Obliv
    May 16, 2016 at 19:23
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    $\begingroup$ Yes. And yes, of course it is abelian. $\endgroup$ May 16, 2016 at 19:56

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