<s>[INST] Which is the value of $t$ ? It is given $t\in \{0,1,2,3,4\}$, a group $G$ and an element $a$ of $G$ such that $a^8=a^t$ and $a^{88}=a^{74}$ in $G$. If $a^2\neq a^0$ which is $t$ ? 
For $a\neq 0$ we have the following: 
\begin{align*}a^{88}=a^{74}&\Rightarrow a^{14}=a^0\Rightarrow a^{2\cdot 8-2}=a^0\\ & \Rightarrow \left (a^8\right )^2=a^2\Rightarrow a^{2t}=a^2\\ & \Rightarrow \left (a^{2t}\right )^4=\left (a^2\right )^4\Rightarrow a^{8t}=a^8\\ & \Rightarrow \left (a^8\right )^t=a^8\Rightarrow \left (a^t\right )^t=a^t\\ & \Rightarrow a^{t^2}=a^t\Rightarrow a^{t^2-t}=a^0\end{align*} 
Is everything correct so far? 
How could we continue? 
How could we use the fact that $a^2\neq a^0$ ? 
 [/INST] We know that $a^{14} = 1$, so the order of $a$ divides $14$.
We know that $a^2 \ne 1$, so the order of $a$ does not divide $2$.
We know that $a^{8-t} = 1$, so the order of $a$ is at most $8-t$, i.e. at most $8$.
Therefore, the order of $a$ is $7$, so $a^8 = a$, so $t = 1$.
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