Subset of Linear Code with Even Weight Codewords
Theorem
Let $C$ be a linear code.
Let $C^+$ be the subset of $C$ consisting of all the codewords of $C$ which have even weight.
Then $C^+$ is a subgroup of $C$ such that either $C^+ = C$ or such that $\order {C^+} = \dfrac {\order C} 2$.
Proof
Note that the zero codeword is in $C^+$ as it has a weight of $0$ which is even.
Let $c$ and $d$ be of even weight, where $c$ and $d$ agree in $k$ ordinates.
Let $\map w c$ denote the weight of $c$.
Then:
\(\ds \map w {c + d}\) | \(=\) | \(\ds \map w c - k + \map w d - k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map w c + \map w d - 2 k\) |
which is even.
Since the negative of a vector $\mathbf v$ in $\Z_2$ equals $\mathbf v$, it follows that the inverse of $c \in C$ is also in $C$.
It follows from the Two-Step Subgroup Test that $C^+$ is a subgroup of $C$.
Let $C \ne C^+$.
Then $C$ contains a codeword $c$ of odd weight.
Let $C^-$ denote the subset of $C$ consisting of all the codewords of $C$ which have odd weight.
Adding $c$ to each codeword of $C^+$ gives distinct codewords of odd weight, so:
- $\order {C^-} \ge \order {C^+}$
Similarly, adding $c$ to each codeword of $C^-$ gives distinct codewords of even weight, so:
- $\order {C^-} \le \order {C^+}$
As $C = C^+ \cup C^-$ it follows that:
- $\order C = 2 \order {C^+}$
Hence the result.
$\blacksquare$
Sources
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $6$: Error-correcting codes: Exercise $3$