Given the following system:
$x_1 + x_2 \ \qquad \qquad+ x_5 = 0$
$x_1 \ \qquad + x_3 \qquad +x_5 = 0$
$x_1 \ \qquad +x_3 + x_4 \qquad = 0$
$\ \qquad x_2 + x_3 + x_4 \qquad = 0$
The augmented matrix of the system is
$ \left[ \begin{array}{ccccc|c} 1&1&0&0&1&0\\ 1&0&1&0&1&0\\ 1&0&1&1&0&0\\ 0&1&1&1&0&0 \end{array} \right] $
Row reduction in mod 2 gives the following matrix.
$ \left[ \begin{array}{ccccc|c} 1&1&0&0&1&0\\ 1&0&1&0&1&0\\ 1&0&1&1&0&0\\ 0&1&1&1&0&0 \end{array} \right] $
so the general solution is $x_1 = x_3 ,x_2 = x_3, x_3 = x_3, x_4 = 0, x_5 = 0$ since the only free variable is $x_3$ which implies there are $2^1 = 2 $ solutions, which would be either when $x_3 = 1$ or $0$ since the operation is mod 2 addition.
So that means the solution set is $(0,0,0,0,0)$ and $(1,1,1,0,0)$. I want to prove the solution set is a subgroup of $\Bbb Z_2^5$ under coordinatewise mod 2 addition.
So far, I let $A = (0,0,0,0,0) , B = (1,1,1,0,0)$, and I want to show $H = \{A,B\} \subset \Bbb Z_2^5$
I know for a subgroup you must prove the operation is associative and closed, identity exists, inverse exists. This is what I have so far..
I know the operation is closed and associative because:
$A +_2 A = (0,0,0,0,0) = A$
$A +_2 B = (1,1,1,0,0) = B$
$B +_2 A = (1,1,1,0,0) = B$
$B +_2 B = (0,0,0,0,0) = A$
I'm having some trouble with the identity/inverse.
$A \in H$ then $e = A +_2 A^{-1} = (0,0,0,0,0) +_2 (1,1,1,1,1) = (1,1,1,1,1)$, So $e \in H$, but that implies $e = (1,1,1,1,1)$, which causes a problem with inverse:
$B \in H$ then $B^{-1} = e +_2 B^{-1} = (1,1,1,1,1) +_2 (0,0,0,1,1) = (1,1,1,0,0)$ which is equal to $B$ and not $B^{-1}$.
My intuition initially told me that the identity element should be $(0,0,0,0,0)$, but that isn't what I got with the math... if it were $(0,0,0,0,0)$ then the inverse would work as well and there would be no problem.