Basic Properties of Groups

From the definition of a group, the following can be proven in general.

Uniqueness of Identity

If \(e\) and \(e'\) are both identities for a group \(G\), then:

\[ e = e'.\]

Consider a group \(G\) with two identities \(e\) and \(e'\). Then, from the definition of the identity:

\[e \ast e' = e \ \text{and} \ e \ast e' = e' \implies e = e'\]

so the identity is unique.

Uniqueness of Inverse

Given \(a \in G\), there exists a unique element denoted by \(a^{-1}\) which satisfies

\[ a \ast a^{-1} = a^{-1} \ast a = e\]

for the identity \(e\).

Let \(b\) and \(b'\) be inverses of \(a\) in a group. This implies that \(a \ast b = e\) and \(a \ast b' = e\) for the identity \(e\).

Therefore:

\[\begin{align*} a \ast b &= a \ast b' \\ (b \ast a) \ast b &= (b \ast a) \ast b' \\ e \ast b &= e \ast b' \\ b &= b \end{align*}\]

\((a^{-1})^{-1} = a\)

For any \(a \in G\), we have that:

\[ (a^{-1})^{-1} = a.\]

Since \(a^{-1}\) has an inverse of \((a^{-1})^{-1}\), it follows that \((a^{-1})^{-1} \ast a^{-1} = e\). Then by multiplying on the right by \(a\) we find that:

\[\begin{align*} (a^{-1})^{-1} \ast (a^{-1} \ast a) &= e \ast a \\ (a^{-1})^{-1} \ast e &= a \\ (a^{-1})^{-1} &= a \end{align*}\]

\((a \ast b)^{-1} = (b^{-1} \ast a^{-1})\)

For any \(a, b \in G\), we have that:

\[ (a \ast b)^{-1} = (b^{-1} \ast a^{-1}).\]

\[\begin{align*} (b^{-1} \ast a^{-1}) \ast (a \ast b) &= b^{-1} \ast (a^{-1} \ast a) \ast b \\ &= b^{-1} \ast e \ast b \\ &= b^{-1} \ast b \\ &= e \end{align*}\]

Cancellation

Given \(a, b, c \in G\):

\[ a \ast b = a \ast c \implies b = c.\]

Cancellation from both sides follows very easily from the existence of an inverse. Here is just the case on the left hand side:

\[\begin{align*} a \ast b &= a \ast c \\ (a^{-1} \ast a) \ast b &= (a^{-1} \ast a) \ast c \\ e \ast b &= e \ast c \\ b &= c \end{align*}\]