Why do Z/7 have no cubic root of 2? I was reading a textbook and came across the following line:

Now we prove there is no cube root of 2 in $Z/7$. By noting that $(Z/7)^\times$ is cyclic of order 6, it will have only two third powers, which are ±1 by observation. Therefore by exclusion 2 can't be a cube in $Z/7$

I have two questions about this passage.


*

*Why are we talking about a corresponding cyclic group when the point is to prove facts in Z/7?

*Why does order 6 in a cyclic group imply only a two third power? Is there a formula of number of nth power based on the order of a cyclic group?
 A: Notice that since $x^6\equiv 1\pmod{7}$ for all $x$ not divisible by $7$, for any nonzero cube $y$ we have that
$$y^2\equiv 1\pmod{7}$$
The only two solutions to this equation are $1$ and $-1$.
A: Answers:


*

*In order to understand the ring $\Bbb Z_7$, we are examining the groups that the units form under multiplication, which is given by $(\Bbb Z_7)^\times$.  This group is a cyclic group of order $6$.

*In any group $G$ of order $n$ (i.e of $n$ elements) with identity $e$, $g^n = e$ for any $g \in G$.
A: If $G$ is cyclic of order $n$ and $d\mid n$ then multiplication by $d$ is a group endomorphism with kernel of size $d$ and image of size $\frac nd$. Here, taking third power in the (multiplicative) cyclic group of nonzero elements of $\mathbb Z/7$ has image size $\frac 63=2$. By exhibiting two elements of this image we then finally know that no other element is in the image (where "is in the image" means "has a cube root")
A: *

*Since $7$ is prime, the cyclic group $(\mathbb{Z}/7\mathbb{Z})^\times$ consists of all elements of $\mathbb{Z}/7\mathbb{Z}$ except $0$, which is not a cube root of $2$.

*Suppose $a$ is a third power in $(\mathbb{Z}/7\mathbb{Z})^\times$. This is to say that $a = x^3$ for some $x$ in $(\mathbb{Z}/7\mathbb{Z})^\times$. Since the order of $(\mathbb{Z}/7\mathbb{Z})^\times$ is $6$, it follows that $x^6 = 1$ and therefore $a^2 = x^6 = 1$. One then checks by brute force (or a little number theory :)) that the only two elements of $(\mathbb{Z}/7\mathbb{Z})^\times$ satisfying $a^2 = 1$ are $a = \pm 1$.
A: for what it's worth, reasoning is nice, but you can also just cube all seven elements and observe that you never get $2$
