Basic Theorems Regarding Group Isomorphisms

# Basic Theorems Regarding Group Isomorphisms

Recall from the Group Isomorphisms page that the groups $(G_1, *_1)$ and $(G_2, *_2)$ are said to be isomorphic, denoted $G_1 \cong G_2$ if there exists a bijective function $f : G_1 \to G_2$ such that for all $x, y \in G_1$ we have that $f(x *_1 y) = f(x) *_2 f(y)$. If such a bijective function $f$ exists, then $f$ is said to be an isomorphism between these groups.

We will now look at some basic theorems regarding isomorphisms.

 Theorem 1: Let $(G_1, *_1)$ and $(G_2, *_2)$ be isomorphic where $f : G_1 \to G_2$ is an isomorphism from $G_1$ to $G_2$. Then $f^{-1} : G_2 \to G_1$ is an isomorphism from $G_2$ to $G_1$.
• 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:
(1)
\begin{align} \quad f(x *_1 y) = f(x) *_2 f(y) \end{align}
• Since $f$ is an isomorphism we have that $f$ is a bijection, so $f^{-1}$ exists and $f^{-1} : G_2 \to G_1$ is a bijection from $G_2$ to $G_1$. We only need to show that $f^{-1} (c *_2 d) = f^{-1} (c) *_1 f^{-1} (d)$ for all $c, d \in G_2$. Let $f(x) = c, f(y) = d \in G_2$. Then:
(2)
\begin{align} \quad f^{-1}(c *_2 d) = f^{-1}(f(x) *_2 f(y)) = f^{-1}(f(x *_1 y)) = x *_1 y \end{align}
• We also have that:
(3)
\begin{align} \quad f^{-1}(c) *_1 f^{-1}(d) = f^{-1}(f(x)) *_1 f^{-1}(f(y)) = x *_1 y \end{align}
• Therefore for all $c, d \in G_2$ we have that $f^{-1}(c *_2 d) = f^{-1}(c) *_1 f(d)$, so $f^{-1}$ is an isomorphism. $\blacksquare$
 Theorem 2: Let $(G_1, *_1)$, $(G_2, *_2)$, and $(G_3, *_3)$ be groups such that $G_1 \cong G_2$ and $G_2 \cong G_3$. If $f : G_1 \to G_2$ is an isomorphism from $G_1$ to $G_2$ and $g : G_2 \to G_3$ is an isomorphism from $G_2$ to $G_3$ then $G_1 \cong G_3$ and $g \circ f : G_1 \to G_3$ is an isomorphism from $G_1$ to $G_3$.
• Proof: Let $G_1 \cong G_2$ and $G_2 \cong G_3$ where $f : G_1 \to G_2$ is an isomorphism from $G_1$ to $G_2$ and $g : G_2 \to G_3$ is an isomorphism from $G_2$ to $G_3$.
• Since $f$ and $g$ are both bijections by definition, we have tat $g \circ f : G_1 \to G_3$ is a bijection. Let $x, y \in G_1$. Then:
(4)
\begin{align} \quad (g \circ f)(x *_1 y) = g(f(x *_1 y)) = g(f(x) *_2 f(y)) = g(f(x)) *_3 g(f(y)) = (g \circ f)(x) *_3 (g \circ f)(y) \end{align}
• So for all $x, y \in G_1$ we have that $(g \circ f)(x *_1 y) = (g \circ f)(x) *_3 (g \circ f)(y)$, so $g \circ f : G_1 \to G_3$ is an isomorphism from $G_1$ to $G_3$ and $G_1 \cong G_3$. $\blacksquare$