Necessary Conditions for Two Groups to Be Isomorphic

Necessary Conditions for Two Groups to Be Isomorphic

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 bijection 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 function a function $f$ exists, then $f$ is said to be an Isomorphism between the groups $(G_1, *_1)$ and $(G_2, *_2)$.

In general, proving that two groups are isomorphic is rather difficult. However, there are some necessary conditions that must be met between groups in order for them to be isomorphic to each other. If a necessary condition does not hold, then the groups cannot be isomorphic. We will look at some of these necessary conditions in the following lemmas noting that these conditions are NOT sufficient to guarantee the existence of an isomorphism.

Lemma 1: If $(G_1, *_1)$ and $(G_2, *_2)$ are two finite groups and $G_1 \cong G_2$ then $\mid G_1 \mid = \mid G_2 \mid$.
  • Proof: Suppose that $m = \mid G_1 \mid \neq \mid G_2 \mid = n$. Then either $m < n$ or $m > n$.
  • Let $f : G_1 \to G_2$ be an isomorphism of $G_1$ and $G_2$. If $m < n$ then $f$ cannot be surjective and hence $f$ is not a bijection which is a contradiction. If $m > n$ then $f$ cannot be injective and hence $f$ is not a bijection which is also a contradiction. Hence $m = n$, i.e.,
(1)
\begin{align} \quad \mid G_1 \mid = \mid G_2 \mid \quad \blacksquare \end{align}
Corollary 1: If $(G_1, *_1)$ if a finite group and $(G_2, *_2)$ is and infinite group then $G_1 \not \cong G_2$.
  • Proof: If $(G_1, *_1)$ is a finite group and $(G_2, *_2)$ is an infinite group then $\mid G_1 \mid = m < \infty = \mid G_2$ for all positive integers $n$, so by Theorem 1 we have that $G_1 \not \cong G_2$. $\blacksquare$
Lemma 2: If $(G_1, *_1)$ is an abelian group and $(G_2, *_2)$ is a non-abelian group then $G_1 \not \cong G_2$.
  • Proof: Let $(G_1, *_1)$ be an abelian group and $(G_2, *_2)$ be a non-abelian group. Suppose that $G_1 \cong G_2$. Then there exists a bijection $f : G_1 \to G_2$ such that for all $x, y \in G_1$ we have that:
(2)
\begin{align} \quad f(x *_1 y) = f(x) *_2 f(y) \end{align}
  • Since $(G_2, *_2)$ is a non-abelian group, let $c, d \in G_2$ be elements that do not commute. That is, $c *_2 d \neq d *_2 c$. Since $f$ is bijective, we have that $f$ is surjective, and so there exists elements $a, b \in G_1$ such that $f(a) = c$ and $f(b) = d$. Now since $(G_1, *_1)$ is an abelian group, we must have that $f(a *_1 b) = f(b *_1 a)$. However:
(3)
\begin{align} \quad f(a *_1 b) = f(a) *_2 f(b) = c *_2 d \neq d *_2 c = f(b) *_2 f(a) = f(b *_1 a) \end{align}
  • So a contradiction arises. Hence our assumption that $G_1 \cong G_2$ was false. Therefore $G_1 \not \cong G_2$. $\blacksquare$

Once again, it should be emphasized that the lemmas/corollaries are NOT sufficient to prove whether two groups are isomorphic.

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