# Necessary Conditions for Two Groups to be Isomorphic

Recall from the Group Isomorphisms page that the groups $(G, \cdot)$ and $(H, *)$ are said to be isomorphic denoted $G \cong H$ if there exists a bijection function $f : G \to H$ such that for all $x, y \in G$ we have that $f(x \cdot y) = f(x) * f(y)$. If function a function $f$ exists, then $f$ is said to be an **Isomorphism** between the groups $(G, \cdot)$ and $(H, *)$.

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 $(G, \cdot)$ and $(H, *)$ are two finite groups and $G \cong H$ then $\mid G \mid = \mid H \mid$. |

**Proof:**Suppose that $m = \mid G \mid \neq \mid H \mid = n$. Then either $m < n$ or $m > n$.

- Let $f : G \to H$ be an isomorphism of $G$ and $H$. 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.,

Corollary 2: If $(G, \cdot)$ if a finite group and $(H, *)$ is and infinite group then $G \not \cong H$. |

**Proof:**If $(G, \cdot)$ is a finite group and $(H, *)$ is an infinite group then $\mid G \mid = m < \infty = \mid H \mid$ for all positive integers $n$, so by Lemma 1 we have that $G \not \cong H$. $\blacksquare$

Lemma 3: If $(G, \cdot)$ is an abelian group and $(H, *)$ is a non-abelian group then $G \not \cong H$. |

**Proof:**Let $(G, \cdot)$ be an abelian group and $(H, *)$ be a non-abelian group. Suppose that $G \cong H$. Then there exists a bijection $f : G \to H$ such that for all $x, y \in G$ we have that:

- Since $(H, *)$ is a non-abelian group, let $c, d \in H$ be elements that do not commute. That is, $c * d \neq d * c$. Since $f$ is bijective, we have that $f$ is surjective, and so there exists elements $a, b \in G$ such that $f(a) = c$ and $f(b) = d$. Now since $(G, \cdot)$ is an abelian group, we must have that $f(a \cdot b) = f(b \cdot a)$. However:

- So a contradiction arises. Hence our assumption that $G \cong H$ was false. Therefore $G \not \cong H$. $\blacksquare$

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