= Solution
Every <permutation> is a product of <transpositions>, since a cycle can be written
$$
(a_1\,a_2\,\ldots\,a_k)=(a_1\,a_k)\cdots(a_1\,a_2)
$$
with rightmost factors acting first. Define the <sign of a permutation> to be $(-1)^r$ for a product of $r$ <transpositions>. To prove that it is well defined without assuming parity, let $P_\sigma$ be the <permutation matrix> specified by $P_\sigma e_i=e_{\sigma(i)}$. Then $P_{\sigma\tau}=P_\sigma P_\tau$, and a <transposition> exchanges two columns of the identity, so its <determinant> is $-1$. Any decomposition into $r$ <transpositions> therefore has
$$
\det P_\sigma=(-1)^r.
$$
The left-hand side depends only on $\sigma$, proving well-definedness. Multiplicativity of the <determinant> now gives the <sign homomorphism>:
$$
\boxed{\operatorname{sgn}(\sigma\tau)=\operatorname{sgn}(\sigma)\operatorname{sgn}(\tau).}
$$
For the first requested <group embedding>, use the <faithful four-point action of GL2 over F2>: label the four <vectors> in $\mathbb F_2^2$ and let an invertible $2\times2$ <matrix> act on them by multiplication. Composition of these actions gives a <group homomorphism> $\psi:GL_2(\mathbb F_2)\to S_4$. If a matrix fixes every vector, it fixes both standard basis vectors and is the identity, so the action is faithful. The matrix
$$
B=\begin{pmatrix}0&1\\1&0\end{pmatrix}
$$
fixes $(0,0)$ and $(1,1)$ and exchanges $(1,0)$ with $(0,1)$. Its image is a single <transposition>, hence $\operatorname{sgn}(\psi(B))=-1$. \b[The composite sign map is nontrivial.]
For the second <group embedding>, simply take
$$
\boxed{\phi(\sigma)=P_\sigma\in GL_n(\mathbb R),\qquad\det\phi(\sigma)=\operatorname{sgn}(\sigma).}
$$
The preceding multiplication formula proves the <group homomorphism> property, and $P_\sigma=I$ forces $\sigma(i)=i$ for every $i$, proving injectivity.
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