# Large subgroups of the Hamming cube

Let's consider the abelian group $\mathbb{Z}^N_2$ equipped with the Hamming metric (the hypercube).

Suppose I have a subgroup of this hypercube (not necessarily a subcube) which is generated by a set of elements all of which are at least distance $d$ from the identity, i.e. it has a set of generators all of which are large.

How many elements of this subgroup can be within $R$ of the identity?

-
Question seems a little vague to me. With $N$ generators of the form $(1,1,\dots,1,0,1,\dots,1,1)$ you get the whole group (if $N$ is even). –  Gerry Myerson Feb 28 '11 at 22:54
Do you want a lower bound on $R$ or an upper bound on $R$. If you want to find subgroups with $R$ small, this is basically the theory of binary linear error-correcting codes, which has been extensively studied (for this case, the constraint that the group is generated by a set of large Hamming weight elements is completely superfluous). If you want to find subgroups with $R$ big, you should look at the MacWilliams identities (also from the theory of error correcting codes) and see what those tell you about $R$. –  Peter Shor Feb 28 '11 at 23:09
You can get half of the elements small. Let $e_k$ be the element $(0,0,\ldots,0,1,0, \ldots 0)$ with a single $1$ in the $k$th position. Let $v$ be the element $(1,1,1,1,1,\ldots,1)$. Now, consider the $t+1$ generators $v$ and $v+e_k$, $k = 1 \ldots t$. The subgroup generated contains the subgroup with $1$'s in any subset of the first $t$ positions, all of whose elements have Hamming weight at most $t$. The subgroup has order $2^{t+1}$, and there are $2^t$ small elements.
Unless $R$ is fairly big, you can't do better, because if you add an element with large Hamming weight to one with small Hamming weight, the result has large Hamming weight.