Preservation of Special Properties under Group Isomorphisms

# Preservation of Special Properties under Group Isomorphisms

We will now look at some more special properties of groups for which group isomorphisms preserve in the following theorems.

Theorem 1: Let $(G_1, *_1)$ and $(G_2, *_2)$ be groups such that $G_1 \cong G_2$. If $(G_1, *_1)$ is an abelian group then $(G_2, *_2)$ is also an abelian group. |

**Proof:**Let $G_1 \cong G_2$. Then there exists a bijective function $f : G_1 \to G_2$ such that for all $x, y \in G_1$ we have that $f(x *_1 y) = f(x) *_2 f(y)$. Let $x, y \in G_1$. Since $(G_1, *_1)$ is an abelian group we have that $x *_1 y = y *_1 x$ and so:

\begin{align} \quad f(x) *_2 f(y) = f(x *_1 y) = f(y *_1 x) = f(y) *_2 f(x) \end{align}

- This holds for all elements $f(x), f(y) \in G_2$. But since $f$ is a bijection, this means that for all $w, z \in G_2$ we have that $w * z = z * w$ so $(G_2, *_2)$ is an abelian group. $\blacksquare$

Theorem 2: Let $(G_1, *_1)$ and $(G_2, *_2)$ be groups such that $G_1 \cong G_2$. If $(G_1, *_1)$ is a cyclic group then $(G_2, *_2)$ is also a cyclic group. |

**Proof:**Let $G_1 \cong G_2$. Then there exists a bijective function $f : G_1 \to G_2$ such that for all $x, y \in G_1$ we have that $f(x *_1 y) = f(x) *_2 f(y)$. Sinc $(G_1, *_1)$ is a cyclic group there exists an element $g \in G_1$ such that $G_1 = <g>$. So every element $x \in G_1$ is such that there exists an $m \in \mathbb{N}$ such that $x = g^m$. So:

\begin{align} \quad f(x) = f(g^m) = f(\underbrace{g *_1 g *_1 ... *_1 g}_{m \: \mathrm{factors}}) = \underbrace{f(g) *_2 f(g) *_2 ... *_2 f(g)}_{m \: \mathrm{factors}} = [f(g)]^m \end{align}

- But $f$ is bijective, so for every element $w \in G_2$ there exists an $x \in G_1$ with $f(x) = w$, i.e., for every $w \in G_2$ we have that there exists an $m \in \mathbb{N}$ such that $w = [f(g)]^m$, so $G_2$ is a cyclic group and $G_2 = <f(g)>$. $\blacksquare$

Theorem 3: Let $(G_1, *_1)$ and $(G_2, *_2)$ be finite groups such that $G_1 \cong G_2$. If $(G_1, *_1)$ has a subgroup $(H_1, *_1)$ of order $n$ then $(G_2, *_2)$ has a subgroup $(H_2, *_2)$ of order $n$. |

**Proof:**Let $(H_1, *_1)$ be a subgroup of $(G_1, *_1)$ of order $n$. Say $H_1 = \{ h_1, h_2, ..., h_n \}$. We will show that if $H_2 = \{ f(h_1), f(h_2), ..., f(h_n) \}$ then $(H_2, *_2)$ is a subgroup of $(G_2, *_2)$ and has order $n$.

- First we show that $(H_2, *_2)$ is a subgroup of $(G_2, *_1)$. Clearly $H_2 \subseteq G_2$ and so all that we need to show is that $(H_2, *_2)$ is closed under $*_2$ and that for every element in $H_2$ its inverse is also in $H_2$.

- First, let $f(h_i), f(h_j) \in H_2$. Then:

\begin{align} \quad f(h_i) *_2 f(h_j) = f(h_i *_1 h_j) \end{align}

- But $h_i * h_j \in H_1$ since $(H_1, *_1)$ is a group and is closed under $*_1$. So $f(h_i *_1 h_j) \in H_2$, i.e., $f(h_i) *_2 f(h_j) \in H_2$, so $(H_2, *_2)$ is closed under $*_2$.

- Now let $f(h_i) \in H_2$. Since $(H_1, *_1)$ is a group we have that for $h_i \in H_1$ that $h_i^{-1} \in H_1$ and so if $e_1$ is the identity of $G_1$ and $e_2$ is the identity of $G_2$ then;

\begin{align} \quad e_2 = f(e_1) = f(h_i *_1 h_i^{-1}) = f(h_i) * f(h_i^{-1}) = f(h_i) * [f(h_i)]^{-1} \end{align}

- Therefore $[f(h_i)]^{-1} \in H_2$. Thus $(H_2, *_2)$ is a subgroup of $(G_2, *_2)$.

- Lastly, since $f : G_1 \to G_2$ is a bijection this means that $f : H_1 \to H_2$ is a bijection, so $n = \mid H_1 \mid = \mid H_2 \mid$, i.e., $(H_2, *_2)$ has order $n$. $\blacksquare$]