= Solution
A linear length-$n$ <cyclic code> is a subspace closed under cyclic coordinate shifts. Identify a word with its <polynomial> modulo $X^n-1$; multiplying by $X$ performs the shift. Thus a cyclic code is an <ideal> of $\mathbb F_q[X]/(X^n-1)$. Its preimage in the <polynomial> ring is an <ideal> containing $X^n-1$, so the <Euclidean algorithm> gives a unique monic generator $g$ dividing $X^n-1$. The <generator polynomial of a cyclic code> is $g$ and the <check polynomial of a cyclic code> is $h=(X^n-1)/g$; a word lies in the code exactly when $hc\equiv0\pmod{X^n-1}$.
For the binary length-seven code, the given $h$ factors with
$$
\boxed{g(X)=X^3+X+1,\qquad (X^3+X+1)(X^4+X^2+X+1)=X^7+1.}
$$
This cubic has no binary root and is irreducible. Let $\alpha$ be its root in $\mathbb F_8$. Since the nonzero multiplicative group has <prime> order seven and $\alpha\ne1$, its seven powers enumerate all nonzero elements. The condition $c(\alpha)=0$ gives a three-row binary parity-check <matrix> whose columns are precisely the seven nonzero vectors of $\mathbb F_2^3$. This is <Hamming's code> in a cyclic coordinate order. It has dimension four, and minimum distance three: no one or two distinct nonzero columns sum to zero, while $g$ itself has weight three. Its sixteen radius-one balls have total size $16(1+7)=128$, confirming the original perfect Hamming code.
It actually contains its <dual code> in this coordinate order. Each parity-check row has four ones, and any two distinct rows have two simultaneous ones, so $HH^T=0$ over $\mathbb F_2$. The row space $C^\perp$ therefore lies in $\ker H=C$. Equivalently the dual generator is the reciprocal check <polynomial>
$$
h^*(X)=X^4+X^3+X^2+1=(X+1)g(X),
$$
which is divisible by $g$. Hence \b[there is a three-dimensional subcode equal to the dual], and in any other coordinate order it is equivalent to that dual.
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