# Math Help - Isomorphism between U_(7) and Z_(7)

1. ## Isomorphism between U_(7) and Z_(7)

Problem:
There is an isomorphism of $U_{7}$ with $\mathbb{Z}_{7}$ in which $\zeta = e^{i(2pi/7)} \leftrightarrow 4$. Find the element in $\mathbb{Z}_{7}$ to which $\zeta^{m}$ must correspond for $m = 0, 2, 3, 4, 5, 6$.

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After arriving at an answer, I noticed that I said there was no answer for all of the even m values, so I'm not sure if I am correct.

To figure out the isomorphism, I used the example in the problem,

$\zeta^{1} \leftrightarrow 4$.

I reasoned that $4 + 4 \in \mathbb{Z}_{7} = 1$, in which that 1 corresponds to the exponent of zeta.

I used this to look for the other values asked in the problem.

For m=0, $3.5 + 3.5 \in \mathbb{Z}_{7} = 0$, but $3.5 \notin Z$, so I said that for m=0, there is no corresponding zeta function.

For m=3 (skipping ahead), $5 + 5 \in \mathbb{Z}_{7} = 3$, $3 \in \mathbb{Z}_{7}$, so $\zeta^{3} \leftrightarrow 5$.

I think I'm doing something incorrectly. Because since this is an isomorphism, and therefore a one-to-one correspondence, shouldn't all m=0,1,2,3,4,5,6 in $U_{7}$ map to a member of $\mathbb{Z}_{7}$? My method is producing answers for only the odd numbers.

Any help is extremely appreciated.

2. ## Re: Isomorphism between U_(7) and Z_(7)

Originally Posted by tangibleLime
Problem:
There is an isomorphism of $U_{7}$ with $\mathbb{Z}_{7}$ in which $\zeta = e^{i(2pi/7)} \leftrightarrow 4$. Find the element in $\mathbb{Z}_{7}$ to which $\zeta^{m}$ must correspond for $m = 0, 2, 3, 4, 5, 6$.

------------------------

After arriving at an answer, I noticed that I said there was no answer for all of the even m values, so I'm not sure if I am correct.

To figure out the isomorphism, I used the example in the problem,

$\zeta^{1} \leftrightarrow 4$.

I reasoned that $4 + 4 \in \mathbb{Z}_{7} = 1$, in which that 1 corresponds to the exponent of zeta.

I used this to look for the other values asked in the problem.

For m=0, $3.5 + 3.5 \in \mathbb{Z}_{7} = 0$, but $3.5 \notin Z$, so I said that for m=0, there is no corresponding zeta function.

For m=3 (skipping ahead), $5 + 5 \in \mathbb{Z}_{7} = 3$, $3 \in \mathbb{Z}_{7}$, so $\zeta^{3} \leftrightarrow 5$.

I think I'm doing something incorrectly. Because since this is an isomorphism, and therefore a one-to-one correspondence, shouldn't all m=0,1,2,3,4,5,6 in $U_{7}$ map to a member of $\mathbb{Z}_{7}$? My method is producing answers for only the odd numbers.

Any help is extremely appreciated.
Note that if $f:U_7\to\mathbb{Z}_7$ is a morphism then $f(\zeta^m)=mf(\zeta)=4m$. So, for $m=0$ you should get that $f(\zeta^0)=0$. Make sense? I'm not really sure what you did there?

3. ## Re: Isomorphism between U_(7) and Z_(7)

Hm... could you please expand on why $f(\zeta^m)=mf(\zeta)=4m$?

4. ## Re: Isomorphism between U_(7) and Z_(7)

Originally Posted by tangibleLime
Hm... could you please expand on why f(\zeta^m)=mf(\zeta)=4m?
Because, by definition you know that $f(xy)=f(x)+f(y)$ (since $f$ is a homomorphism). So, in particular by induction one has that $f(x^m)=f(x\cdots x)=f(x)+\cdots+f(x)=mf(x)$. and so if $f(x)=y$ then $f(x^m)=mf(x)=my$.

5. ## Re: Isomorphism between U_(7) and Z_(7)

well the isomorphism in question isn't the "usual" (or "obvious") isomorphism. but just looking at Z7 as a group, it should be clear that 4 is a generator.

so $\zeta^2$ should correspond to 4+4, which as you pointed out, is 1 (in Z7).

what drexel28 is getting at, is that "powers" (repeated group operations) in the 7th roots of unity (which inherits its group operation from complex multiplication) correspond to "multiples" in Z7 (which has a group operation of addtion modulo 7).

it's not what we are used to, but in the group <Z7,+> " $4^2$" means 4+4, or as is sometimes written (2)4 (this turns out to work out to be the same as multiplication modulo 7, but a word of caution: (2)4 doesn't really mean "two times 4" it means "4+4" (4 added to itself).). the reason for this caveat, is that you can actually write something like (8)4, meaning "4+4+4+4+4+4+4+4" (mod 7), which makes sense in Z7, but 8 is not an element of Z7.