Preservation of the Identities and Inverses Under Group Isomorphisms

Preservation of the Identities and Inverses Under Group Isomorphisms

Informally we say that the groups $(G_1, *_1)$ and $(G_2, *_2)$ are isomorphic if they have the same structure, and the existence of a bijection $f : G_1 \to G_2$ where for all $x, y \in G_1$ we have that $f(x *_1 y) = f(x) *_2 f(y)$ preserves this structure.

As we will see in the following theorems, if $e_1 \in G_1$ and $e_2 \in G_2$ are the identities with respect to $*_1$ and $*_2$ 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).

Theorem 1: Let $(G_1, *_1)$ and $(G_2, *_2)$ be groups such that $G_1 \cong G_2$. If $e_1 \in G_1$ is the identity with respect to $*_1$ and $e_2 \in G_2$ is the identity with respect to $*_2$ and $f : G_1 \to G_2$ is an isomorphism from $G_1$ to $G_2$ then $f(e_1) = e_2$.
  • Proof: Let $f : G_1 \to G_2$ be an isomorphism from $G_1$ to $G_2$. Then for all $x, y \in G_1$ we have that $f(x *_1 y) = f(x) *_2 f(y)$. Set $x = y = e_1$. Then:
(1)
\begin{align} \quad f(e_1) = f(e_1 *_1 e_1) = f(e_1) *_2 f(e_1) \\ \quad f(e_1) *_2 [f(e_1)]^{-1} = [f(e_1) *_2 f(e_1)] *_2 [f(e_1)]^{-1} \\ \quad e_2 = f(e_1) *_2 [f(e_1) *_2 [f(e_1)]^{-1}] \\ \quad e_2 = f(e_1) *_2 e_2 \\ \quad e_2 = f(e_1) \quad \blacksquare \end{align}
Theorem 2: Let $(G_1, *_1)$ and $(G_2, *_2)$ be groups such that $G_1 \cong G_2$. If $f : G_1 \to G_2$ is an isomorphism from $G_1$ to $G_2$ then for all $x, x^{-1} \in G_1$, $y, y^{-1} \in G_2$ we have that if $f(x) = y$ then $f(x^{-1}) = y^{-1}$.
  • Let $f : G_1 \to G_2$ be an isomorphism from $G_1$ to $G_2$, let $e_1 \in G_1$ and $e_2 \in G_2$ be the identity elements with respect to $*_1$ and $*_2$ and suppose that $x \in G_1$ and $y \in G_2$ 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} *_1 x) = f(x^{-1}) *_2 f(x) \\ \end{align}
  • By Theorem 1 we are given $f(e_1) = e_2$, so:
(3)
\begin{align} \quad e_2 = f(x^{-1}) *_2 f(x) \\ \quad e_2 = f(x^{-1}) *_2 y \\ \quad y^{-1} = [f(x^{-1}) *_2 y] *_2 y^{-1} \\ \quad y^{-1} = f(x^{-1}) *_2 [y *_2 y^{-1}] \\ \quad y^{-1} = f(x^{-1}) *_2 e_2 \\ \quad y^{-1} = f(x^{-1}) \quad \blacksquare \end{align}
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