<s>[INST] Code is not cyclic for any q I have code $C$ over $F_p$ with generator matrix which looks like 
$G = 
\begin{pmatrix} 
0 &0& 0& 1& 0& 1& 1 &1\\
1& 0 &0& 0 &1 &0 &1& 1\\
1& 1& 0& 0& 0& 1& 0& 1\\
1 &1& 1& 0 &0 &0& 1 &0\end{pmatrix}$
I need to show that this code is not cyclic for any $p$.
I constructed vector which looks like 
$$(c_2+c_3+c_4, c_3+c_4, c_4, c_1, c_2, c_1+c_3, c_1+c_2+c_4, c_1+c_2+c_3)$$ and its shifts but it didn't help me a lot. 
Does somebody have any ideas how it can be proven?
 [/INST] $C$ must contain every cyclic shift of its codewords. Consider the cyclic shift of the last row of $G$, $x = \left( {01110001} \right)$. If $C$ is a cyclic code, $x$ must be a linear combination of the rows of $G$, i.e.
${x^T} = a\left( {\begin{array}{*{20}{c}}
0\\
0\\
0\\
1\\
0\\
1\\
1\\
1
\end{array}} \right) + b\left( {\begin{array}{*{20}{c}}
1\\
0\\
0\\
0\\
1\\
0\\
1\\
1
\end{array}} \right) + c\left( {\begin{array}{*{20}{c}}
1\\
1\\
0\\
0\\
0\\
1\\
0\\
1
\end{array}} \right) + d\left( {\begin{array}{*{20}{c}}
1\\
1\\
1\\
0\\
0\\
0\\
1\\
0
\end{array}} \right)$
for some $a,b,c,d \in {F_p}$.
Then, it is not hard to see that 
$b + c + d = 0\left( p \right)$ and $c + d = 1\left( p \right)$ from the sum of first and second components of the rows, respectively. From the fifth components, $b=0$. So, 
$c + d = 1\left( p \right)$ and $c + d = 0\left( p \right)$. This is not possible for all prime $p$.
</s>