= Solution
A based loop is a continuous map $\alpha:[0,1]\to X$ with $\alpha(0)=\alpha(1)=x_0$. Two loops are equivalent when they are joined by a homotopy that keeps both endpoints at $x_0$. The <fundamental group> $\pi_1(X,x_0)$ is the set of these based-homotopy classes.
For loops $\alpha$ and $\beta$, define concatenation by
$$
(\alpha*\beta)(t)=
\begin{cases}
\alpha(2t),&0\leq t\leq\tfrac12,\\
\beta(2t-1),&\tfrac12\leq t\leq1.
\end{cases}
$$
If $H$ and $K$ are based homotopies from $\alpha$ to $\alpha'$ and from $\beta$ to $\beta'$, concatenating $H(s,-)$ and $K(s,-)$ gives a based homotopy from $\alpha*\beta$ to $\alpha'*\beta'$. Thus multiplication $[\alpha][\beta]=[\alpha*\beta]$ is well-defined.
Concatenating three paths with different breakpoints changes only the speed of traversal. Linear interpolation between the two increasing piecewise-linear parametrizations gives a based homotopy, proving associativity on classes. The constant loop $c(t)=x_0$ is an identity, since deleting its stationary half is another endpoint-fixing reparametrization. The inverse of $[\alpha]$ is represented by $\bar\alpha(t)=\alpha(1-t)$. Indeed, $\alpha*\bar\alpha$ contracts through
$$
H(s,t)=
\begin{cases}
\alpha(2t(1-s)),&0\leq t\leq\tfrac12,\\
\alpha(2(1-t)(1-s)),&\tfrac12\leq t\leq1,
\end{cases}
$$
and the analogous contraction handles $\bar\alpha*\alpha$. Hence the operation satisfies all group axioms.
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