The square of a k-th root of unity is a k/2-th root of unity

Last updated on Nov 12, 2025

If we take the set of $k$ -th roots of unity (with $k$ even) and square each element, the resulting set will be a set of half the size. The new set will be the $\frac{k}{2}$ -th roots of unity.

For example, suppose $k = 6$ . The 6th roots of unity would be

{1, ω, ω^{2}, ω^{3}, ω^{4}, ω^{5}}

If we square each element, we get the following set. Some elements have exponents greater than or equal to $k$ , but we will handle that in the next step.

{1^{2}, ω^{2}, ω^{4}, ω^{6}, ω^{8}, ω^{10}}

We can then factor the exponents as follows:

{1^{2}, ω^{2}, ω^{4}, ω^{6}, (ω^{6}) ω^{2}, (ω^{6}) ω^{4}}

Since $ω$ is a 6-th root of unity, $ω^{6} \equiv 1$ so we have:

{1^{2}, ω^{2}, ω^{4}, 1, (1) ω^{2}, (1) (ω^{4})}

Removing multiplication by $1$ , we get

{1^{2}, ω^{2}, ω^{4}, 1, ω^{2}, ω^{4}}

Now replace all the duplicate terms with a single element:

{1, ω^{2}, ω^{4}}

The new set is half the size of the original, and each element is a 3-rd root of unity:

$1^{3} \equiv 1$
${(ω^{2})}^{3} = ω^{6} \equiv 1$
${(ω^{4})}^{3} = ω^{12} = ω^{6} ω^{6} \equiv 1 \cdot 1 = 1$

If we plot the 6-th roots of unity on a circle, we can see that squaring “removes” every other element. We started with ${1, ω, ω^{2}, ω^{3}, ω^{4}, ω^{5}}$ and ended with ${1, ω^{2}, ω^{4}}$

To reiterate, if we take the set of $k$ -th roots of unity, and $k$ is even, then square each element, we get a set of half the size with each element being the $\frac{k}{2}$ -th root of unity.

Some more examples:

If $k = 10$ and we square each of the 10-th roots of unity, we get a set of size five which are the fifth roots of unity.
If $k = 8$ and we square each of the 8-th roots of unity, we get a set of size four which is the fourth roots of unity.
If $k = 4$ and we square each of the 4-th roots of unity, we get a set of size two which is the 2-nd roots of unity.
If $k = 2$ and we square each of the 2-nd roots of unity, we get a set of size 1 which is just the element 1.

The last point is easily illustrated. The second roots of unity are square roots of 1, which are always ${1, - 1} \equiv {1, ω^{k / 2}}$ . Squaring 1 results in 1 and squaring -1 results in 1. Equivalently, ${(ω^{k / 2})}^{2} = ω^{k} \equiv 1$ .

Example of squaring the 8-th roots of unity

Consider the subgroup of 8th roots of unity $⟨ 9 ⟩ = {1, 9, 13, 15, 16, 8, 4, 2}$ in the finite field $𝔽_{17}$ . We square all elements of this subgroup as follows:

\begin{matrix} 1^{2} = 1 (\mod 17), \\ 9^{2} \equiv 13 (\mod 17), \\ 13^{2} \equiv 16 (\mod 17), \\ 15^{2} \equiv 4 (\mod 17), \\ 16^{2} \equiv 1 (\mod 17), \\ 8^{2} \equiv 13 (\mod 17), \\ 4^{2} \equiv 16 (\mod 17), \\ 2^{2} = 4 (\mod 17) . \end{matrix}

The set obtained after squaring is ${1,13,16,4}$ , which is precisely the subgroup of 4th roots of unity.

Here is a visualization of the roots of unity before and after squaring. We started with the set ${1, 9, 13, 15, 16, 8, 4, 2}$ and ended with the set ${1,13,16,4}$

k must be even

If $k$ is odd, then there is no such thing as “half of the group” as an odd-sized set cannot be divided into two. For the purposes of NTT, we only deal with even-sized $k$ , so we aren’t interested in the case where $k$ is odd.

Proof of the claim that the new set is half the size

Let $ω$ be a primitive $k$ -th root of unity with $k$ even. Let $⟨ ω ⟩$ be the subgroup generated by $ω$ of order $k$ . We claim that $| {ω^{2} | ω \in ⟨ ω ⟩} | = k / 2$ .

The proof is actually quite simple and intuitive.

We established in an earlier chapter that $ω^{i}$ and $ω^{i + k / 2}$ are additive inverses. Since $k$ is even, we can partition the group into two sets, the first one being $0... (k / 2 - 1)$ and the second being $k / 2... k - 1$ :

{ω^{0}, ω^{1}, ω^{2}, . . .} {ω^{k / 2}, ω^{k / 2 + 1}, ω^{k / 2 + 2}, . . .}

Those elements are congruent to the following representation:

{ω^{0}, ω^{1}, ω^{2}, . . .} {- ω^{0}, - ω^{1}, - ω^{2}, . . .}

If we apply $f (x) = x^{2}$ to both sets, we get two sets with identical content and size $k / 2$

{1, ω^{2}, ω^{4}, . . .} {1, ω^{2}, ω^{4}, . . .}

Since the two sets are identical, the union of the two sets will be the same size, which is $k / 2$ .

Proof that squaring a $k$ -th root of unity produces a $k / 2$ -th root of unity

Suppose $a$ is a $k$ -th root of unity. We aim to show that $a^{2}$ is a $\frac{k}{2}$ -th root of unity, that is:

{(a^{2})}^{\frac{k}{2}} \equiv 1

Let’s simplify ${(a^{2})}^{\frac{k}{2}}$ :

{(a^{2})}^{\frac{k}{2}} = a^{k}

Since $a^{k} \equiv 1$ (because $a$ is a $k$ -th root of unity), it follows that ${(a^{2})}^{\frac{k}{2}} \equiv 1$ .

Therefore, $a^{2}$ is indeed a $\frac{k}{2}$ -th root of unity.