Group Product Identity therefore Inverses

Theorem

Let $\left({G, \circ}\right)$ be a Group.

Let $x, y \in \left({G, \circ}\right)$.

Then if either $x \circ y = e$ or $y \circ x = e$, it follows that $x = y^{-1}$ and $y = x^{-1}$.

Proof

From the Division Laws for Groups:

$x \circ y = e \implies x = e \circ y^{-1} = y^{-1}$

Also by the Division Laws for Groups:

$x \circ y = e \implies y = x^{-1} \circ e = x^{-1}$

• The same results are obtained by exchanging $x$ and $y$ in the above.

$\blacksquare$

Sources

• George McCarty: Topology: An Introduction with Application to Topological Groups (1967): Chapter $\text{II}$: Theorem $1 \ \text{(iii), (iv)}$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 28 \ (1)$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 35.3$