For any given $k$ what is a non-abelian group $G$ of the smallest order that satisfies $\forall g\in G, g^k=e$?

Question

For any given $k \geq 1$ what is a non-abelian finite group $G$ of the smallest order that satisfies $\forall g\in G, g^k=e$?

An exercise in an abstract algebra textbook asks the reader to construct such $G$ for $k=3,4$.

These are the answers given by the book:

$G_1=\left\{\begin{pmatrix}1&a&c\\&1&b\\&&1\end{pmatrix}:a,b,c\in GF(3)\right\}$ for $k=3$,

$G_2=\left\{\begin{pmatrix}1&a&d&f\\&1&b&e\\&&1&c\\&&&1\end{pmatrix}:a,b,c,d,e,f\in GF(2)\right\}$ for $k=4$.

It is clear that $G_1$ is the group of the smallest order for $k=3$. However, the dihedral group $D_4$ (i.e., the isometry group of a square) is also a solution for $k=4$ but of a smaller order than $G_2$.

I wonder that for any given $k$, what is $G$ of the smallest order?

Answer 1

This is my fourth attempt!

Case 1. $k$ is a power of a prime $p$ (where we must have $k>2$). Then choose $G = E_p$, a nonabelian group of order $p^3$ and of exponent $p$ when $p$ is odd.

Case 2. There exist primes $p,q$ dividing $k$ with $p|(q-1)$, and either $p=2$ with $4 \not\vert k$ or $pq<r^3$, where $r$ is the smallest prime dividing $k$. Choose such a $p,q$ with $pq$ minimal, and let $G$ be the nonabelian group of order $pq$.

Case 3. Otherwise, choose $G = E_r$, where $r$ is the smallest prime dividing $k$.

Answer 2

Edit: The following answer reaches the same conclusion as that of Derek Holt, who expressed things more succinctly (by combining all the cases where $O(k) = pq$ together).

Answer: The question is void for $k = 1$ and $k = 2$. For $k \ge 3$, define $O(k)$ to be the smallest order of a non-abelian group $G$ satisfying $g^k = e$ for all $g \in G$. Then $O(k)$ goes as follows:

1. If $k$ is even:
   • If $6 | k$, then $O(k) = 6$ (as examplified by the symmetry group $S_3$);
   • If $3 \not | k$ and $4 | k$, then $O(k) = 8$;
   • If $3 \not | k$ and $4 \not | k$, then $O(k) = 2m$ where $m$ is the smallest odd prime factor of $k$.
2. If $k$ is odd, then $O(k)$ is the smallest of the following two numbers:
   • $r^3$ where $r$ is the smallest prime factor of $k$;
   • the smallest product $pq$ where $p < q$ are prime factors of $k$ with $p | q-1$. (If there are no such $p, q$ for $k$, then the answer is $r^3$.)

Explanations:

1. The question makes sense only for $k \ge 3$. Indeed, for $k = 1$, the only group of exponent $1$ is $\left\{ e \right\}$, which is abelian. Similarly, for $k = 2$, any group $G$ of exponent $2$ is abelian.

2. Let $r \ge 2$ be a prime. Then the smallest order of a non-abelian $r$-group is $r^3$ and it is always realized; see e.g. this Wikipedia page. In fact, there exists a non-abelian $r$-group $G_r$ of order $r^3$ such that:

   • if $r=2$, then $G_r$ has exponent $4 = r^2$, i.e. $g^4 = e$ for all $g \in G_2$. An example is the dihedral group $D_4$ of isometric symmetries of the square (identified by the OP);
   • for $r \ge 3$, then $G_r$ has exponent $r$, i.e. $g^r = e$ for all $g \in G_r$. For instance, the Heisenberg group modulo $r$ does the trick.

3. There is no non-abelian group of order $\le 5$, nor of order $p$ or $p^2$ for some prime $p$. (This follows from the previous fact.) It follows that $O(k) \ge 6$ and that it is not of the form $p$ or $p^2$ for some prime.

4. Let $k \ge 3$ and consider $G$ such that $g^k = e$ for all $g \in G$. Then the order of $G$ is a product of (powers of) prime factors of $k$.

Proof. Otherwise, there would be a prime $p$ dividing $|G|$ but not $k$; by Cauchy's theorem, there would exist $g \in G$ whose order is $p$, hence $g^a = e$ iff $p | a$; but $g^k = e$, hence $p|k$, which is a contradiction. QED

It follows that $O(k)$ is a product of (powers of) prime divisors of $k$.

5. Let $2 \le p < q$ be two primes.

   • If $p | q-1$, then there exists a non-abelian group of order $pq$, see for instance this MSE post. This group has exponent $pq$.

   • If $p \not | q-1$, then any group of order $pq$ is cyclic, hence abelian.

6. Let $k \ge 3$ be even.

   • If $6|k$, then consider the non-abelian symmetric group $S_3$ of order $6$ and of exponent $6$.

   • If $3 \not | k$, then $O(k) \neq 6$ (by 4) and so $O(k) \ge 8$ (by 3).

     (i) If $4 | k$, then consider the group $Dih_4$ of order $8$ and of exponent $4$.

     (ii) If $4 \not | k$, then $O(k)$ is not a power of $2$, for any non-abelian group of order $2^a$ for some $a \ge 2$ necessarily has an element $g$ of order $2^b$ with $b \ge 2$, so that $g^k \neq e$. Since $O(k)$ is not prime, we have $O(k) \ge 2m$ where $m$ is the smallest odd prime factor of $k$. By 5, it follows that $O(k) = 2m$.

7. Let $k \ge 3$ be odd.

   • Let $r$ be the smallest prime factor of $k$. By 2, consider the group $G_r$ of order $r^3$ and exponent $r$, we get $O(k) \le r^3$.

   • If $O(k) < r^3$, then necessarily, $O(k) = pq$ where $p < q$ are prime factors of $k$ such that $p|q-1$. (Indeed, by 4, $O(k)$ is a product of prime factors of $k$. This product involves at least 2 prime factors of $k$ -- so to have a non-abelian group -- and of at most 2 prime factors of $k$ -- so to have a group of order $< r^3$. Hence $O(k) = pq$ with $p, q$ prime factors of $k$. By 3, $p < q$. By 5, $p | q-1$.)