# Isomorphism between Additive Group Modulo 16 and Multiplicative Group Modulo 17

## Theorem

Let $\struct {\Z_{16}, +}$ denote the additive group of integers modulo $16$.

Let $\struct {\Z'_{17}, \times}$ denote the multiplicative group of reduced residues modulo $17$.

Let $\phi: \struct {\Z_{16}, +} \to \struct {\Z'_{17}, \times}$ be the mapping defined as:

$\forall \eqclass k {16} \in \struct {\Z_{16}, +}: \map \phi {\eqclass k {16} } = \eqclass {3^k} {17}$

Then $\phi$ is a group isomorphism.

## Proof

Let $\eqclass x {16}, \eqclass y {16} \in \struct {\Z_{16}, +}$.

Then:

 $\ds \map \phi {\eqclass x {16} } \times \map \phi {\eqclass y {16} }$ $=$ $\ds \map \phi {x + 16 m_1} \times \map \phi {y + 16 m_2}$ Definition of Residue Class: for some representative $m_1, m_2 \in \Z$ $\ds$ $=$ $\ds 3 \uparrow \paren {x + 16 m_1} \times 3 \uparrow \paren {y + 16 m_2}$ using Knuth uparrow notation $3 \uparrow k := 3^k$ $\ds$ $=$ $\ds 3 \uparrow \paren {x + 16 m_1 + y + 16 m_2}$ Product of Powers $\ds$ $=$ $\ds 3 \uparrow \paren {\paren {x + y} + 16 \paren {m_1 + m_2} }$ $\ds$ $=$ $\ds 3 \uparrow \paren {\eqclass {x + y} {16} }$ Definition of Residue Class and Definition of Modulo Addition $\ds$ $=$ $\ds \map \phi {\eqclass x {16} + \eqclass y {16} }$ Definition of $\phi$

Thus it is seen that $\phi$ is a group homomorphism.

$\Box$

It remains to be seen that $\phi$ is a bijection.

Because $17$ is prime: $\forall x \in \Z, 1 \le x < 17: x \perp 17$ where $\perp$ denotes coprimality.

Thus by definition of multiplicative group of reduced residues modulo $17$:

$\order {\struct {\Z'_{17}, \times} } = 16$

where $\order {\, \cdot \,}$ denotes the order of a group.

Similarly, by definition of additive group of integers modulo $16$:

$\order {\struct {\Z_{16}, +} } = 16$

So:

$\order {\struct {\Z'_{17}, \times} } = \order {\struct {\Z_{16}, +} }$

which is a necessary condition for group isomorphism.

$\Box$

Now we have:

 $\ds 16$ $\equiv$ $\ds 0$ $\ds \pmod {16}$ $\ds \leadsto \ \$ $\ds \map \phi {\eqclass {16} {16} }$ $=$ $\ds \map \phi {\eqclass 0 {16} }$ $\ds \leadsto \ \$ $\ds \map \phi {\eqclass {16} {16} }$ $=$ $\ds \eqclass 1 {17}$ Group Homomorphism Preserves Identity $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds 3^{16}$ $\equiv$ $\ds 1$ $\ds \pmod {17}$ Definition of $\phi$

Now let $\eqclass x {16}, \eqclass y {16} \in \Z_{16}$ such that $\map \phi {\eqclass x {16} } = \map \phi {\eqclass y {16} }$.

We have:

 $\ds \map \phi {\eqclass x {16} }$ $=$ $\ds \map \phi {\eqclass y {16} }$ $\ds \leadsto \ \$ $\ds \forall m_1, m_2 \in \Z: \,$ $\ds \map \phi {x + 16 m_1}$ $=$ $\ds \map \phi {y + 16 m_2}$ Definition of Residue Class $\ds \leadsto \ \$ $\ds 3 \uparrow \paren {x + 16 m_1}$ $=$ $\ds 3 \uparrow \paren {y + 16 m_2}$ Definition of $\phi$ $\ds \leadsto \ \$ $\ds 3^x \paren {3^{16} }^{m_1}$ $=$ $\ds 3^y \paren {3^{16} }^{m_2}$ Product of Powers, Power of Power $\ds \leadsto \ \$ $\ds 3^x \times 1^{m_1}$ $=$ $\ds 3^y \times 1^{m_2}$ as $3^{16} \equiv 1 \pmod {17}$ from $(1)$ $\ds \leadsto \ \$ $\ds 3^x$ $=$ $\ds 3^y$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds y$

Thus $\phi$ is an injection.

From Equivalence of Mappings between Finite Sets of Same Cardinality it follows that $\phi$ is a bijection.

$\Box$

Thus $\phi$ is a bijective group homomorphism.

Hence the result by definition of group isomorphism.

$\blacksquare$