Consider the monoid End($\mathbb{Z}/_2 \oplus \mathbb{Z}/_2$) of endomorphisms of the abelian group $\mathbb{Z}/_2 \oplus \mathbb{Z}/_2$ under composition. Show that the group of units of End($\mathbb{Z}/_2 \oplus \mathbb{Z}/_2)$ is isomorphic to the symmetric group $S_3$.
So far, I have:
The group of units of End($\mathbb{Z}/_2 \oplus \mathbb{Z}/_2$) contains all of the endomorphisms that are invertible. These are the endomorphims that do not map components to 0; thus, there are 6 elements in the group of units: \begin{align*} \rho_1: & (x,y) \mapsto (x,y) & \rho_2: & (x,y) \mapsto (y,x)\\ \rho_3: & (x,y) \mapsto (x, x+y) & \rho_4: & (x,y) \mapsto (x+y, x) \\ \rho_5: & (x,y) \mapsto (y, x+y) & \rho_6: & (x,y) \mapsto (x+y, y) \end{align*} On the other hand, the symmetric group $S_3$ also has 6 elements: \begin{align*} (1)(2)(3) & & (12)(3) & & (13)(2) & & (23)(1) & & (123) & & (321). \end{align*}
To show the isomorphism, I just need to exhibit a bijection. What do I need to do to define a bijection between these two groups? I can map the identity to the identity, so $\rho_1 \mapsto (1)(2)(3)$, but then can I just arbitrarily pair off the rest of the elements?
You could approach this the other way round. An endomorphism is a unit if and only it is bijective; moreover $0$ is mapped to $0$.
So every unit in the endomorphism will certainly give you a bijective map from the nonzero elements of $Z/2 \oplus Z/2$ to itself, so a permutation of the nonzero elements, and there are 3. Conversely, this permutation of courses determines the endomorphism uniquely.
Thus you directly get an injective map from the unit endomorphisms into $S_3$. Since you have written down 6 unit elements in End thus suffices to show that the map is also surjective (in fact finding more than 3 would suffice).