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:
(1)
\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:
(2)
\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:
(3)
\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;
(4)
\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 $n$. $\blacksquare$]
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