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 results.

Proposition 1: Let $(G, \cdot)$ and $(H, *)$ be groups such that $G \cong H$. If $(G, \cdot)$ is an abelian group then $(H, *)$ is also an abelian group.

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

(1)

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

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

Proposition 2: Let $(G, \cdot)$ and $(H, *)$ be groups such that $G \cong H$. If $(G, \cdot)$ is a cyclic group then $(H, *)$ is also a cyclic group.

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

(2)

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

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

Proposition 3: Let $(G, \cdot)$ and $(H, *)$ be finite groups such that $G \cong H$. If $(G, \cdot)$ has a subgroup $(G_1, \cdot)$ of order $n$ then $(H, *)$ has a subgroup $(H_1, *)$ of order $n$.

• Proof: Let $(G_1, \cdot)$ be a subgroup of $(G, \cdot)$ of order $n$. Say $G_1 = \{ G_1, H_1, ..., h_n \}$. We will show that if $H_1 = \{ f(G_1), f(H_1), ..., f(h_n) \}$ then $(H_1, *)$ is a subgroup of $(H, *)$ and has order $n$.

• First we show that $(H_1, *)$ is a subgroup of $(H, \cdot)$. Clearly $H_1 \subseteq H$ and so all that we need to show is that $(H_1, *)$ is closed under $*$ and that for every element in $H_1$ its inverse is also in $H_1$.

• First, let $f(h_i), f(h_j) \in H_1$. Then:

(3)

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

• But $h_i * h_j \in G_1$ since $(G_1, \cdot)$ is a group and is closed under $\cdot$. So $f(h_i \cdot h_j) \in H_1$, i.e., $f(h_i) * f(h_j) \in H_1$, so $(H_1, *)$ is closed under $*$.

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

(4)

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

• Therefore $[f(h_i)]^{-1} \in H_1$. Thus $(H_1, *)$ is a subgroup of $(H, *)$.

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