Group Product Identity therefore Inverses/Part 2

Theorem

Let $g$ and $h$ be elements of a group $G$ whose identity element is $e$.

Then:

$h g = e \implies h = g^{-1}$ and $g = h^{-1}$

Proof 1

From the Division Laws for Groups:

$h g = e \implies g = e h^{-1} = h^{-1}$

Also by the Division Laws for Groups:

$h g = e \implies h = g^{-1} e = g^{-1}$

$\blacksquare$

Proof 2

Let $h g = e$.

Then:

\(\ds g\)

\(=\)

\(\ds e g\)

Group Axiom $\text G 2$: Existence of Identity Element

\(\ds \)

\(=\)

\(\ds \paren {h^{-1} h} g\)

Group Axiom $\text G 3$: Existence of Inverse Element

\(\ds \)

\(=\)

\(\ds h^{-1} \paren {h g}\)

Group Axiom $\text G 1$: Associativity

\(\ds \)

\(=\)

\(\ds h^{-1} e\)

by hypothesis

\(\ds \)

\(=\)

\(\ds h^{-1}\)

Group Axiom $\text G 2$: Existence of Identity Element

and:

\(\ds h\)

\(=\)

\(\ds h e\)

Group Axiom $\text G 2$: Existence of Identity Element

\(\ds \)

\(=\)

\(\ds h \paren {g g^{-1} }\)

Group Axiom $\text G 3$: Existence of Inverse Element

\(\ds \)

\(=\)

\(\ds \paren {h g} g^{-1}\)

Group Axiom $\text G 1$: Associativity

\(\ds \)

\(=\)

\(\ds e g^{-1}\)

by hypothesis

\(\ds \)

\(=\)

\(\ds g^{-1}\)

Group Axiom $\text G 2$: Existence of Identity Element

$\blacksquare$

Sources

• 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: The Group Property: Theorem $1 \ \text{(iv)}$