# Why would the last bit of the total parity in a Hamming code(8,4) be set to zero?

I'm trying to understand the overall parity bit in a Hamming Code specifically at the end of this wikipedia article. (You can compare my interpretation with the anchor linked example if this seems wrong in case I misunderstood the matrix) $$\mathbf{G} = \left( \begin{array}{cccc|cccc} 1 & 0 & 0 & 0 & 0 & 1 & 1 & \color{red}1 \\ 0 & 1 & 0 & 0 & 1 & 0 & 1 & \color{red}1 \\ 0 & 0 & 1 & 0 & 1 & 1 & 0 & \color{red}1 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 & \color{red}0 \end{array} \right)_{4,8}$$

$$\mathbf{H} = \left( \begin{array}{cccc|cccc} 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 \\ \color{red}1 & \color{red}1 & \color{red}1 & \color{red}0 & 0 & 0 & 0 & 1 \end{array} \right)_{4,8}$$

As a refresher and if conventions in the article differ from the conventions one might be familiar with, you multiply G by a vector (in the article $\vec{a}=[1,0,1,1]$) to get the original data plus individual parity bits $\vec{a}$G and then multiply the result by H on the receiving end to verify each parity bit equals zero if no error occurred. The last example in wikipedia uses a backwards out of order example where the columns of G 5,6, and 7 are parity bits 4,2, and 1 respectively. The final column 8 (coloring mine) is the overall parity to detect double errors and indicate the parity of every bit. Why is the last bit 0 and not 1 to detect if the overall parity changed (odd errors) or not (even). Is this a typo/error? Research(1 2 3)

Bit coverage for reference: Parity bits which indicate which data bit was wrong as the binary representation of the wrong bit: The overall parity bit indicates if an even or odd number of errors occurred.

It's perfectly correct. The final column in $G$ is defined so as to make the sum of each row of $G$ to be zero modulo $2$, that is that all rows of $G$ have even parity.