Let $v_q(n,r)=\sum_{i=0}^r \binom{n}i (q-1)^i$ denote a number of points in a ball of radius $r$ in the Hamming metric on the cube $\Sigma^n$, where $|\Sigma|=q$. What is the maximal number of points in $\Sigma^n$ with mutual distances at least $d$ (later: $d$-distant point sets)? Gilbert's bound says that we may find $M$ such points if $(M-1)\cdot v_q(n,d-1)< q^n$: add points one by one, while we have less than $M$ points, the $(d-1)$-balls centered at them do not cover $\Sigma^n$ and we may add another point. If $q$ is a power of prime, Varshamov's bound improves the previous result by claiming that we may find $q^k$ $d$-distant points whenever $q^{k-1}v_q(n,d-1)<q^n$. For the proof, we identify $\Sigma$ with a finite field and consider the maximal subspace $X$ of $\Sigma^n$ into which all points are $d$-distant. If $\dim X\geqslant k$, we are done, else there exist a point $v$ not covered by $(d-1)$-balls centered at elements of $X$, and the span of $X$ and $v$ is larger $d$-distant subspace.

My question is: is Varshamov's bound true if we do not require that $q$ is a prime power (say, for $q=6$)? This often happens and always does astonish me when arithmetical properties like being prime powers are crucial for purely combinatorial or topological claims, and I wonder what is the situation in this particular case.

Let $v_q(n,r)=\sum_{i=0}^r \binom{n}i (q-1)^i$ denote a number of points in a ball of radius $r$ in the Hamming metric on the cube $\Sigma^n$, where $|\Sigma|=q$. What is the maximal number of points in $\Sigma^n$ with mutual distances at least $d$ (later: $d$-distant point sets)? Gilbert's bound says that we may find $M$ such points if $(M-1)\cdot v_q(n,d-1)< q^n$: add points one by one, while we have less than $M$ points, the $(d-1)$-balls centered at them do not cover $\Sigma^n$ and we may add another point. If $q$ is a power of prime, Varshamov's bound improves the previous result by claiming that we may find $q^k$ $d$-distant points whenever $q^{k-1}v_q(n,d-1)<q^n$. For the proof, we identify $\Sigma$ with a finite field and consider the maximal subspace $X$ of $\Sigma^n$ into which all points are $d$-distant. If $\dim X\geqslant k$, we are done, else there exists a point $v$ not covered by $(d-1)$-balls centered at elements of $X$, and the span of $X$ and $v$ is larger $d$-distant subspace.

My question is: is Varshamov's bound true if we do not require that $q$ is a prime power (say, for $q=6$)? This often happens and always does astonish me when arithmetical properties like being prime powers are crucial for purely combinatorial or topological claims, and I wonder what is the situation in this particular case.