Criteria for a Group

Criterion 1: The Definition of a Group

Let $G$ be a nonempty set, and suppose a binary operation “$\cdot$” is defined on $G$. If this binary operation satisfies the following conditions:

(1) Closure: $ \forall x,y \in G,\ x\cdot y\in G $

(2) Associativity: $ \forall x,y,z\in G,\ (x\cdot y)\cdot z=x\cdot (y\cdot z) $

(3) Identity element: $ \exists e\in G,\ \forall a\in G,\ e\cdot a=a\cdot e=a $

(4) Inverse element: $ \forall a\in G,\ \exists b\in G,\ a\cdot b=b\cdot a=e $

then $G$ is called a group with respect to “$\cdot$”.

Criterion 2: The One-Sided Definition of a Group

Let $G$ be a nonempty set, and suppose a binary operation “$\cdot$” is defined on $G$. If this binary operation satisfies the following conditions:

(1) Closure: $ \forall x,y \in G,\ x\cdot y\in G $

(2) Associativity: $ \forall x,y,z\in G,\ (x\cdot y)\cdot z=x\cdot (y\cdot z) $

(3) Left identity element: $ \exists e\in G,\ \forall a\in G,\ e\cdot a=a $

(4) Left inverse element: $ \forall a\in G,\ \exists b\in G,\ b\cdot a=e $

then $G$ is called a group with respect to “$\cdot$”.

Note: If “left” in (3) and (4) is replaced everywhere by “right”, the conclusion still holds.

Proof : First, we prove that $ \forall a\in G $, $ a $ has a right inverse.

Note that:
$$
ab=a(eb)=a(ba)b=(ab)(ab)
$$

Also, by (4),
$$
\exists c\in G,\ c(ab)=e
$$

Therefore,
$$
e=c(ab)=c(ab)(ab)=e(ab)=ab
$$

Hence $ a $ has a right inverse, that is, $ a $ has an inverse.

Next, we prove that $ G $ has a right identity element.
$$
\forall a\in G,\ ae=a(ba)=(ab)a=ea=a
$$

Therefore, $ G $ has a right identity element, that is, $ e $ is the identity element of $ G $.

Thus, $ G $ is a group.

Criterion 3: The Division Definition of a Group

Let $G$ be a semigroup. If $ \forall a,b\in G $, the equations $ ax=b $ and $ ya=b $ both have solutions, then $G$ is a group.

Proof : By the condition, $ ax=a $ has a solution $ e_a $, that is,
$$
ae_a=a
$$
We now verify that $ e_a $ is a right identity element in $G$.

For $ \forall b\in G $, since $ ya=b $ has a solution, there exists $ c\in G $ such that
$$
b=ca
$$

Therefore,
$$
b e_a=(ca)e_a=c(a e_a)=ca=b
$$

Hence $ e_a $ is indeed a right identity element in $G$.

Next, we prove that every element in $G$ has a right inverse.

For $ \forall b\in G $, since $ bx=e_a $ has a solution, we know that $ b $ has a right inverse. Therefore, by Criterion 2, $G$ is a group.

Criterion 4

A finite semigroup that satisfies the left and right cancellation laws must be a group.

Proof : Let $G$ be a finite semigroup of order $n$, and let
$$
|G|=\{a_1,a_2,\cdots,a_n\}
$$

Consider the equation
$$
a_ix=a_j,\ \forall i,j,\ 1\le i,j \le n
$$

By the left cancellation law,
$$
a_ia_1,a_ia_2,\cdots,a_ia_n
$$
are pairwise distinct, and hence one of them must be $ a_j $. That is, $ a_ix=a_j $ has a solution for all $ i,j,\ 1\le i,j \le n $. Similarly, $ xa_i=a_j $ also has a solution for all $ i,j,\ 1\le i,j \le n $.

By Criterion 3 for determining groups, $G$ is a group.

Criterion 5

If a semigroup has a right identity element and satisfies the left cancellation law, then it must be a group.

Proof : Let $G$ be a finite semigroup of order $n$, and let
$$
|G|=\{a_1,a_2,\cdots,a_n\}
$$

Consider the equation
$$
a_ix=a_j,\ \forall i,j,\ 1\le i,j \le n
$$

By the left cancellation law,
$$
a_ia_1,a_ia_2,\cdots,a_ia_n
$$
are pairwise distinct, and hence one of them must be $ a_j $. That is, $ a_ix=a_j $ has a solution, and therefore $ a_ix=e $ has a solution. Hence $ \forall a\in G $, $ a $ has a right inverse. Therefore, by Criterion 2, $G$ is a group.

The cover image of this article was taken at Lake Baikal in Russia.