Group Isomorphisms
Table of Contents

Group Isomorphisms

We have looked at a ton of different groups so far, though, one question that might arise is whether some of the groups we have looked at so far has a fundamentally similar structure. We will be able to compare the structures of specific groups after we look at the following definition.

Definition: Let $(G_1, *_1)$ and $(G_2, *_2)$ be groups. An Isomorphism between $G_1$ and $G_2$ is 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 there exists an isomorphism between the groups $(G_1, *_1)$ and $(G_2, *_2)$ then these groups are said to be Isomorphic which is denoted $G_1 \cong G_2$.

For example, consider the group of integers modulo $2$ under the operation $*_1$ which we define for all $x, y \in \mathbb{Z}_2 = \{0, 1 \}$ by:

(1)
\begin{align} \quad x *_1 y = (x + y) \mod 2 \end{align}

The following table illustrates the structure of the group $(\mathbb{Z}_2, *_1)$:

Screen%20Shot%202015-09-10%20at%207.30.06%20AM-1.png

Now consider the group $(A, *_2)$ where $A = \{ a, b \}$ and $*_2$ is defined for all $x, y \in \{ a, b \}$ such that:

(2)
\begin{align} \quad x * y = \left\{\begin{matrix} a & \mathrm{if} \: x, y = a \: \mathrm{or} \: x, y = b\\ b & \mathrm{if} \: x = a : \mathrm{and} \: y =b \: \mathrm{or} \: x =b \: \mathrm{and} \: y = a \end{matrix}\right. \end{align}

You should verify that $(A, *_2)$ is indeed a group and that the following table illustrates the structure of $(A, *_2)$:

Screen%20Shot%202015-09-10%20at%207.31.15%20AM-1.png

Notice how similar the structures of $(\mathbb{Z}_2, *_1)$ and $(A, *_2)$ are. In fact, both of these groups are isomorphic to one another. To show this, we need to prove that a bijection function $f : \mathbb{Z}_2 \to A$ exists such that for all $x, y \in \mathbb{Z}_2$ we have that $f(x *_1 y) = f(x) *_2 f(y)$. Consider the function defined by $f(0) = a$ and $f(1) = b$. It is not hard to verify that $f$ is indeed a bijection.

Now we note that:

(3)
\begin{align} \quad f(0 *_1 0) = f(0) = a , \quad \mathrm{and}, \quad \: f(0) *_2 f(0) = a *_2 a = a \\ \quad f(0 *_1 1) = f(1) = b , \quad \mathrm{and}, \quad \: f(0) *_2 f(1) = a *_2 b = b \\ \quad f(1 *_1 0) = f(1) = b, \quad \mathrm{and}, \quad \: f(1) *_2 f(0) = b *_2 a = b \\ \quad f(1 *_1 1) = f(0) = a, \quad \mathrm{and}, \quad \: f(1) *_2 f(1) = b *_2 b = a \end{align}

Hence, for all $x, y \in \mathbb{Z}_2$ we have that $f(x *_1 y) = f(x) *_2 f(y)$, so $f$ is an isomorphism between $(\mathbb{Z}_2, *_1)$ and $(A, *_2)$ and $\mathbb{Z}_2 \cong A$.

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