Find all elements in a cyclic group of a specific order?

Question

I am looking for the number of elements of order 4 in a cyclic group with $4k$ elements. I have tried everything I could think of. I know that the number of order 4 elements is divisible by $\varphi(4) =2$ and I found something online that said the number of $x$ such that $x^4 =1$ is given by the possible values for $\operatorname{ord}(x^4) = \frac{\operatorname{ord}(x)}{\gcd(4,4k)}$ which gives $\gcd(4,4k) = 4$, but I'm lost.

Answer 1

Let $g$ be a generator of your group. Then every element is of the form $g^i$ for some integer $i$ satisfying $0 \leq i < 4k$. If $g^i$ has order $4$, then $(g^i)^4 = g^{4i} = g^0$, so $4k$ divides $4i$, i.e., $k$ divides $i$. Thus your element is in the subgroup generated by $g^k$; this is a cyclic subgroup of order $4$. The question thus reduces to finding how many elements have order $4$ in the cyclic subgroup of order $4$; this should be easier.

Answer 2

Let $C=\langle x \rangle$ be a cyclic group of order $4k$, where $k$ is a positive integer. Let $m$ be any positive integer such that $m \leq 4k$.

What is the order of $x^m$? It is $$\frac{4k}{\gcd(m,4k)}$$ from the formula you have observed yourself.

Now what you want is to find those values of $m$ such that $$\frac{4k}{\gcd(m,4k)} = 4 \iff k = \gcd(m,4k) \iff 1 = \gcd(m/k,4).$$

Since $m$ must be a multiple of $k$, the only possible values for $m$ are $\{k,2k,3k,4k\}$ (remember that $m \leq 4k$). Accordingly, $m/k \in \{1,2,3,4\}$.

Since $1 = \gcd(m/k,4)$, however, we find that $m/k \in \{1,3\}$.

Thus the only two elements of order $4$ in $C$ are $x^k$ and $x^{3k}$.

Answer 3

Perhaps it is easier to start with this simpler question:

Which elements $x$ satisfy $x^4=1$?

The answer is clearly the powers of $g^k$: $1, g^k, g^{2k}, g^{3k}$.

Now check the orders of these elements:

• $1$ has order $1$

• $g^k$ has order $4$

• $g^{2k}$ has order $2$

• $g^{3k}$ has order $4$

Therefore, the answer to the original question is $g^k$ and $g^{3k}$.

Answer 4

Every cyclic group of $n$ elements is isomorphic to $n$ points evenly spaced in the circle $|z| = 1$ in the complex plane, with identity corresponding to $z = 1$.

All order 4 elements correspond to solutions of $z^4 = 1$. However, the converse does not hold, i.e. not all solutions are of order 4. So we have to look at the solutions: we have $1$ (order $1$), $-1$ (order $2$) and $\pm i$ (order $4$).

If $n$ is not divisible by $4$, then $\pm i$ will not correspond to elements of the group. If $n$ does divide $4$, then you would have to do some math to figure out which elements correspond, but you're not asked which elements correspond, just how many there are, so the answer is $2$.