Product with Inverse on Homomorphic Image is Group Homomorphism

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Theorem

Let $G$ be a group.

Let $H$ be an abelian group.

Let $\theta: G \to H$ be a (group) homomorphism.


Let $\phi: G \times G \to H$ be the mapping defined as:

$\forall \tuple {g_1, g_2} \in G \times G: \map \phi {g_1, g_2} = \map \theta {g_1} \map \theta {g_2}^{-1}$


Then $\phi$ is a homomorphism.


Proof

First note that from Group Homomorphism Preserves Inverses:

$\map \theta {g_2}^{-1} = \paren {\map \theta {g_2} }^{-1} = \map \theta { {g_2}^{-1} }$

and so $\map \theta {g_1} \map \theta {g_2}^{-1}$ is not ambiguous:

$\map \theta {g_1} \map \theta {g_2}^{-1} = \map \theta {g_1} \paren {\map \theta {g_2} }^{-1} = \map \theta {g_1} \map \theta { {g_2}^{-1} }$


From External Direct Product of Groups is Group, $G \times G$ is a group.


Let $a_1, a_2, b_1, b_2 \in G$.

We have:

\(\ds \map \phi {a_1 b_1, a_2 b_2}\) \(=\) \(\ds \map \theta {a_1 b_1} \map \theta {a_2 b_2}^{-1}\)
\(\ds \) \(=\) \(\ds \map \theta {a_1 b_1} \paren {\map \theta {a_2 b_2} }^{-1}\) Group Homomorphism Preserves Inverses
\(\ds \) \(=\) \(\ds \map \theta {a_1} \map \theta {b_1} \paren {\map \theta {a_2} \map \theta {b_2} }^{-1}\) Definition of Group Homomorphism
\(\ds \) \(=\) \(\ds \map \theta {a_1} \map \theta {b_1} \paren {\map \theta {b_2}^{-1} } \paren {\map \theta {a_2}^{-1} }\) Inverse of Group Product
\(\ds \) \(=\) \(\ds \map \theta {a_1} \paren {\map \theta {a_2}^{-1} } \map \theta {b_1} \paren {\map \theta {b_2}^{-1} }\) Definition of Abelian Group: $H$ is Abelian
\(\ds \) \(=\) \(\ds \map \phi {a_1, a_2} \map \phi {b_1, b_2}\) Definition of $\phi$

Hence the result by definition of homomorphism.

$\blacksquare$


Examples

Mapping from Dihedral Group $D_3$ to Parity Group

Let $D_3$ denote the symmetry group of the equilateral triangle:

\(\ds e\) \(:\) \(\ds (A) (B) (C)\) Identity mapping
\(\ds p\) \(:\) \(\ds (ABC)\) Rotation of $120 \degrees$ anticlockwise about center
\(\ds q\) \(:\) \(\ds (ACB)\) Rotation of $120 \degrees$ clockwise about center
\(\ds r\) \(:\) \(\ds (BC)\) Reflection in line $r$
\(\ds s\) \(:\) \(\ds (AC)\) Reflection in line $s$
\(\ds t\) \(:\) \(\ds (AB)\) Reflection in line $t$


SymmetryGroupEqTriangle.png


Let $G$ denote the parity group, defined as:

$\struct {\set {1, -1}, \times}$

where $\times$ denotes conventional multiplication.


Let $\theta: D_3 \to G$ be the homomorphism defined as:

$\forall x \in D_3: \map \theta x = \begin{cases} 1 & : \text{$x$ is a rotation} \\ -1 & : \text{$x$ is a reflection} \end{cases}$

Let $\phi: D_3 \times D_3 \to G$ be the mapping defined as:

$\forall \tuple {g_1, g_2} \in D_3 \times D_3: \map \phi {g_1, g_2} = \map \theta {g_1} \map \theta {g_2}^{-1}$


Then the kernel $\map \ker \phi$ is the set of all pairs $\tuple {g_1, g_2}$ of elements of $D_3$ such that:

$g_1$ and $g_2$ are both rotations
$g_1$ and $g_2$ are both reflections.


Sources

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