Group Example: x inv c y

Theorem

Let $\left({G, \circ}\right)$ be a group, and let $c \in G$.

We define a new product $*$ on $G$ as:

$\forall x, y \in G: x * y = x \circ c^{-1} \circ y$

Then $\left({G, *}\right)$ is a group.

Proof

G0: Closure

Let $x, y \in G$.

Then:

$\forall x * y = x \circ c^{-1} \circ y \in G$ as $c^{-1} \in G$

G1: Associativity

Let $x, y, z \in G$.

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x * \left({y * z}\right)\)	\(=\)	\(\displaystyle x \circ c^{-1} \circ \left({y \circ c^{-1} \circ z}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of $*$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({x \circ c^{-1} \circ y}\right) \circ c^{-1} \circ z\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Associativity of $\circ$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({x * y}\right) * z\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of $*$

G2: Identity

Let $x \in G$.

• $x * c = x \circ c^{-1} \circ c = x$
• $c * x = c \circ c^{-1} \circ x = x$

So $c$ serves as the identity.

G3: Inverses

Let $x \in G$.

We need to find $y \in G$ such that $x * y = c \Longrightarrow x \circ c^{-1} \circ y = c$.

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x * y\)	\(=\)	\(\displaystyle c\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x \circ c^{-1} \circ y\)	\(=\)	\(\displaystyle c\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle c^{-1} \circ y\)	\(=\)	\(\displaystyle x^{-1} \circ c\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle y\)	\(=\)	\(\displaystyle c \circ x^{-1} \circ c\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Thus the inverse of $x$ under the operation $*$ is $c \circ x^{-1} \circ c$ where $x^{-1}$ is the inverse of $x$ under $\circ$.

And thus we see that $\left({G, *}\right)$ is a group.

$\blacksquare$

Sources

• John F. Humphreys: A Course in Group Theory (1996): $\S 3$: Exercise $3$