Preservation of the Identities and Inverses Under Group Isomorphisms

# Preservation of the Identities and Inverses Under Group Isomorphisms

Informally we say that the groups $(G, \cdot)$ and $(H, *)$ are isomorphic if they have the same structure, and the existence of a bijection $f : G \to H$ where for all $x, y \in G$ we have that $f(x \cdot y) = f(x) * f(y)$ preserves this structure.

As we will see in the following propositions, if $e_1 \in G$ and $e_2 \in H$ are the identities with respect to $\cdot$ and $*$ then $f(e_1) = e_2$; and if $f(x) = y$ then we will also have that $f(x^{-1}) = y^{-1}$. In other words, the existence of an isomorphism $f$ preserves the identities and inverses (and in fact, all other special properties of the groups).

 Proposition 1: Let $(G, \cdot)$ and $(H, *)$ be groups such that $G \cong H$. If $e_1 \in G$ is the identity with respect to $\cdot$ and $e_2 \in H$ is the identity with respect to $*$ and $f : G \to H$ is an isomorphism from $G$ to $H$ then $f(e_1) = e_2$.
• Proof: Let $f : G \to H$ be an isomorphism from $G$ to $H$. Then for all $x, y \in G$ we have that $f(x \cdot y) = f(x) * f(y)$. Set $x = y = e_1$. Then:
(1)
\begin{align} \quad f(e_1) = f(e_1 \cdot e_1) = f(e_1) * f(e_1) \\ \quad f(e_1) * [f(e_1)]^{-1} = [f(e_1) * f(e_1)] * [f(e_1)]^{-1} \\ \quad e_2 = f(e_1) * [f(e_1) * [f(e_1)]^{-1}] \\ \quad e_2 = f(e_1) * e_2 \\ \quad e_2 = f(e_1) \quad \blacksquare \end{align}
 Proposition 2: Let $(G, \cdot)$ and $(H, *)$ be groups such that $G \cong H$. If $f : G \to H$ is an isomorphism from $G$ to $H$ then for all $x, x^{-1} \in G$, $y, y^{-1} \in H$ we have that if $f(x) = y$ then $f(x^{-1}) = y^{-1}$.
• Let $f : G \to H$ be an isomorphism from $G$ to $H$, let $e_1 \in G$ and $e_2 \in H$ be the identity elements with respect to $\cdot$ and $*$ and suppose that $x \in G$ and $y \in H$ is such that $f(x) = y$. Since $e_1 = x^{-1} * x$, we have that:
(2)
\begin{align} \quad f(e_1) = f(x^{-1} \cdot x) = f(x^{-1}) * f(x) \\ \end{align}
• By Proposition 1 we are given $f(e_1) = e_2$, so:
(3)
\begin{align} \quad e_2 = f(x^{-1}) * f(x) \\ \quad e_2 = f(x^{-1}) * y \\ \quad y^{-1} = [f(x^{-1}) * y] * y^{-1} \\ \quad y^{-1} = f(x^{-1}) * [y * y^{-1}] \\ \quad y^{-1} = f(x^{-1}) * e_2 \\ \quad y^{-1} = f(x^{-1}) \quad \blacksquare \end{align}