Proof of Group Homomorphisms Preserve the Identity Element and Inverses

Throughout, the operations of $G$, $H$ and $K$ are associative with neutral elements $e_{G}$, $e_{H}$ and $e_{K}$ (Groups and Abelian Groups (i), Monoids (i)).

For (i), since $f$ is a homomorphism and $e_{G}$ and $e_{H}$ are neutral,

$$f(e_{G})f(e_{G})=f(e_{G}e_{G})=f(e_{G})=f(e_{G})e_{H},$$

and cancelling $f(e_{G})$ on the left (Cancellation Laws and Basic Inverse Identities in a Group (i)) gives $f(e_{G})=e_{H}$.

For (ii), let $g\in G$. By (i),

$$f(g)f(g^{-1})=f(gg^{-1})=f(e_{G})=e_{H},$$

and likewise $f(g^{-1})f(g)=f(g^{-1}g)=e_{H}$. So $f(g^{-1})$ is an inverse of $f(g)$, and since $f(g)$ has only one inverse (Uniqueness of the Identity Element and of Inverses in a Group (ii)), $f(g^{-1})=f(g)^{-1}$.

For (iii), the composite $k\circ f$ is a function from $G$ to $K$ with $(k\circ f)(g)=k(f(g))$ for every $g\in G$ (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ii)). Since $f$ and $k$ are homomorphisms, for all $g,h\in G$

$$(k\circ f)(gh)=k\bigl(f(g)f(h)\bigr)=k(f(g))\,k(f(h))=(k\circ f)(g)\,(k\circ f)(h),$$

so $k\circ f$ is a group homomorphism (Group Homomorphisms and Isomorphisms (i)).

For (iv), suppose that $f$ is an isomorphism. Since $f$ is bijective, its inverse $f^{-1}$ is a bijective function from $H$ to $G$ with $f\circ f^{-1}=\mathrm{id}_{H}$ (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (vi)). For $x,y\in H$, since $f$ is a homomorphism,

$$f\bigl(f^{-1}(x)f^{-1}(y)\bigr)=f(f^{-1}(x))\,f(f^{-1}(y))=xy=f\bigl(f^{-1}(xy)\bigr),$$

and since $f$ is injective (Injective, Surjective and Bijective Functions between Classes (i)), $f^{-1}(xy)=f^{-1}(x)f^{-1}(y)$. Thus $f^{-1}$ is a bijective homomorphism from $H$ to $G$, that is, an isomorphism (Group Homomorphisms and Isomorphisms (ii)).