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I'm having difficulty showing the given a map, say $\phi(z)=z^k$, is surjective. This question is from D & F section 1.6 - #19

Let $G$ =$\{z \in \mathbb C|z^n=1 \text{ for some } n \in \mathbb Z^+\}$. Prove that for any fixed integer $k>1$ the map from $G$ to itself defined by $z \to z^k$ is a surjective homomorphism, but is not an isomorphism.

To show $\phi(z)=z^k$ is a homomorphism let $z_a=e^\frac{2\pi ia}{n}$ and $z_b=e^\frac{2\pi ib}{n}$ where $a,b\in \mathbb Z^+$ and $a\neq b$ then $\phi(z_az_b)=(e^\frac{2\pi ia}{n}$$\cdot$$e^\frac{2\pi ib}{n}$)=$e^\frac{2\pi ik(a+b)}{n}$=$e^\frac{2\pi ika}{n}$$\cdot$$e^\frac{2\pi ikb}{n}$=$\phi(z_a)\phi(z_b)$.

I had what I thought was a proof for $\phi$ being surjective until I thought about this specific example. Here's the example: Suppose $z_a$ is a root of unity then so is a power $z^t_a$. My thoughts on $G$ are that it consists of all integer powers of the "basic" roots of unity for a partiuclar $n$. Let $n=3$ and $k=3$, then $\phi(e^\frac{2t\pi ik_i}{3})=(e^\frac{2t\pi ik_i}{3})^3 =e^{2t\pi ik_i}= 1$ for any integer $t\in \mathbb Z$ and integer $k_i$ such that $0\leq$$k_i$$\leq$ $2$. So we have $\phi[G]=\{1\}$, $\phi$ is not surjective in this instance.

What have I misunderstood about the question? Are there any glaring errors I'm making? Is $n$ supposed to be a particular value and not just any integer value?

Your help is appreciated.

Davide Giraudo
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user41442
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    I believe the question means for some $n$ (that is all roots of unity, of all orders)- not the set of root of a given order. In the latter case, your proof indeed shows that $\phi$ is not surjective. – kneidell Mar 12 '13 at 22:25
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    btw- use what you've already found out to prove that $\phi$ is never injective if $k>1$ – kneidell Mar 12 '13 at 22:25
  • I don't understand how you got that $\phi$ always returns $1$ (is that your claim?). It's not true. Yeah, the cube of a root of degree 3 is always 1, but $G$ contains many more. For example, the roots of degree 9, when cubed, give roots of degree 3. – Yoni Rozenshein Mar 12 '13 at 22:34
  • Echoing @kneidell, what's meant is $$G={{,z\in{\bf C}\mid\exists n\in{\bf Z}^+{\rm\ s.t.\ }z^n=1,}}$$ – Gerry Myerson Mar 12 '13 at 23:08
  • @ Yoni, you're correct. I was stuck in thinking that we had to consider only a specific value of n, not all powers of all roots of unity for all n. – user41442 Mar 13 '13 at 00:12
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    @ Kneidell, thanks for clearing that up. – user41442 Mar 13 '13 at 00:14
  • I realized I completely misread the question and deleted my answer. – Alexander Gruber Mar 13 '13 at 02:01

2 Answers2

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G = $\{e^{i2\pi\over n}: n\in \mathbb Z^{+}\}$ so for $g,h\in G, (gh)^{k} = g^{k}h^{k}$ so this is easily seen to be a homomorphism.

Let $g\in G$ be given by $e^{i2\pi \over m}$. Then $w = e^{i2\pi \over mk} \in G$ and $w^{k} = e^{i2\pi \over m}$ so g is surjective.

This map is not injective because $e^{i2\pi}$ and $e^{i2\pi \over k}$ both get mapped to 1.

John Douma
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  • I do not agree with your set description of G. You have G = { e^(i2Pi/n) : n is in positive integers}. What about (for example) an element in G like z = e^(i2Pi2/5)? This is an nth root of unity because z^5 = 1. Right? But according to what you have above then it would not be in G. I think it should read G = { e^(i2Pir/s) : r/s is a rational number in [0,1) and r,s are relatively prime } – Geoffrey Critzer Aug 24 '15 at 17:48
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Of course, as you have pointed out $\phi$ is not an injective homomorphism.
But, the understanding of the group $G$ is a bit misleading.
$G= \{z\in \mathbb{C}|z^n=1\text{ for some } n \in \mathbb{Z}^+\}$. Here the group $G$ encapsulates all such elements which can be represented as $e^{\frac{2\pi i}{n}}$
To be a surjective homomorphism you have to show that all elements in G have an inverse under $\phi$.
So, take any element in $G$, let's say $e^{\frac{2\pi i}{a}}$, and observe that under $\phi^{-1}$ we have the element $e^{\frac{2\pi i}{ak}}$ which belongs to $G$. It need not have the same denominator as the mapped element as G has all the elements of the form $e^{\frac{2\pi i}{n}}$.
So, it is an surjective homomorphism

Rohan Raj
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