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I've been struggling for a while so I ask your help. I sort of have the ''tools'' but I don't know how to apply them. Aside of that I'm not sure if I'm understanding the exercise correctly.

Find the numbers of solutions of the equation $x^k = 1$ and the number of elements of order $k$ in the group $C_{180}$ with:

$a)$ $k = 7$.

$b)$ $k = 36$.

As far as I understand the solutions of $x^k = 1$ would be all the generators in $C_{180}$.

THE FIRST PART

Find the numbers of solutions of the equation $x^k = 1$ in the cyclic group $C_{180}$

An element $x^k$ is a generator of $C_{180}$ iff $\mathrm{gcd}(n,k) = 1$. So the total number of solutions to this equation for $n = 180$ is

\begin{align} \phi(180) &= \phi(3^22^25)\\ &=\phi(3^2)\phi(2^2)\phi(5)\\ &= (3^2-3)(2^2-2)(5-1)\\ &=72 \end{align}

SECOND PART

Find the number of elements of order $k$ in the group $C_{180}$.

Honestly I don't know how to proceed. Anyway, below I write some thoughts:

$(\ast)$ For $k = 7$.

Here I'm confused. In one hand we have that $\gcd(180,7) = 1$, so $x^7$ is a generator in $C_{180}$, and $|x^7| = \frac{180}{\gcd(180,7)} = 180$. What does this mean?

$(\ast)$ For $k = 36$.

I took a look at this post. According to the most voted answer, since $36|180$ we have that there are $\phi(36) = 12$ elements, but I'm not able to understand why.

Thanks for any help.

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asd
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  • In your attempt for the first part, you found the number of generators. But what you want to find is the number of solutions to $x^k = 1$ for some fixed $k$ which is given in parts (a) and (b). So for example, in part (a) you want to find how many elements satisfy $x^7 = 1$ in $C_{180}$. – Pratyush Sarkar Mar 15 '17 at 22:10
  • @PratyushSarkar So for $k = 7$, we have that $\mathrm{gcd}(180,7) = 1 \implies x^7$ is a generator in $C_{180}$ with order $|x^7| = \frac{180}{\mathrm{gcd}(180,7) } = 180$. That implies that $x^7 = x^0 = 1$. Does it mean that there is just one element of order $k = 7$ in $C_{180}$? – asd Mar 15 '17 at 22:21
  • @PratyushSarkar Mmm maybe the number of solutions to the equation with $k = 7$ is $\phi(7) = 6$. I don't understand the difference between number of solutions and number of element for some fixed $k$. – asd Mar 15 '17 at 22:27
  • Here, $x$ is not any specific element in the group. In order to avoid confusion, lets say $C_{180} = \langle a \rangle$ where $a$ is some generator. Then you want to find which elements $a^j$ satisfy the equation $x^7 = 1$, i.e., for which $j$ is it true that $(a^j)^k = 1$. – Pratyush Sarkar Mar 15 '17 at 22:32

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Hint (Or try to search about this theorem)

If $G$ is a cyclic group of order $n$ and $d$ is a divisor of n, then the number of elements of $G$ of order $d$ is $$\varphi(d)$$ where $\varphi$ denotes the Euler's totient function.

So for the first part, $x^k=1$ iff $|x|$ divides $k$. So you have to find the number of elements with order dividing k.

For the second part, you have to find the number of elements with order equal to k.

The theorem stated can indeed be applied on both parts of question.

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