I am trying to create an isomorphism between two groups:

$\phi : \mathbb{Z}_2 \times \mathbb{Z}_{12} \to \mathbb{Z}_4 \times \mathbb{Z}_{6} $

Every attempt I have done, I end up with only elements of order 12. For example, (1,1) has order 12 on the first group and (3,5) has order 12 on the second group, but I do not see how that helps me in any way.

The only other thing I know is that $\mathbb{Z}_2 \times \mathbb{Z}_{12} \cong \mathbb{Z}_2 \times \mathbb{Z}_{2} \times \mathbb{Z}_2 \times \mathbb{Z}_{3} \cong \mathbb{Z}_4 \times \mathbb{Z}_{6}$.

Any help is appreciated.


Careful! We have that $\mathbb Z_{xy} \cong \mathbb Z_x \times \mathbb Z_y$ iff $\gcd(x, y) = 1$. So for example, $\mathbb Z_4 \not\cong \mathbb Z_2 \times \mathbb Z_2$. Thus, we know that: \begin{align*} \mathbb Z_2 \times \mathbb Z_{12} &\cong \mathbb Z_2 \times (\mathbb Z_{3} \times \mathbb Z_4) \\ &\cong \mathbb Z_4 \times (\mathbb Z_2 \times \mathbb Z_{3}) \\ &\cong \mathbb Z_4 \times \mathbb Z_6 \\ \end{align*}

Following this logic, we see that: \begin{align*} (a, b) &\mapsto (a, b \text{ mod } 3, b \text{ mod } 4) \\ &\mapsto (b \text{ mod } 4, a, b \text{ mod } 3) \\ &\mapsto (b \text{ mod } 4, (4(b \text{ mod } 3) - 3a) \text{ mod } 6) \\ \end{align*}

  • $\begingroup$ Just to make sure I understand this function correctly, (1,4) maps to (0,1)? $\endgroup$
    – El Spiffy
    Nov 10 '16 at 12:47
  • $\begingroup$ That's correct! $\endgroup$
    – Adriano
    Nov 10 '16 at 12:52

Note that $\mathbb{Z}_2 \times \mathbb{Z}_2 \cong \mathbb{Z}_4$ doesn't hold, so your reasoning is not quite correct. However, $$\mathbb{Z}_2 \times \mathbb{Z}_{12} \cong \mathbb{Z}_2 \times (\mathbb{Z}_3 \times \mathbb{Z}_4) \cong (\mathbb{Z}_2 \times \mathbb{Z}_3) \times \mathbb{Z}_4 \cong \mathbb{Z}_6 \times \mathbb{Z}_4$$ does hold. How you just need to check which isomorphism is applied in each step.

  • $\mathbb{Z}_{12} \cong \mathbb{Z}_3 \times \mathbb{Z}_4$ via $x \mapsto (x \pmod 3, x \pmod 4)$.
  • $\mathbb{Z}_6 \cong \mathbb{Z}_2 \times \mathbb{Z}_3$ via $x \mapsto (x \pmod 2, x \pmod 3)$, so the inverse mapping is given by $(x, y) \mapsto 4y - 3x \pmod 6$ (this follows from the CRT).

This leaves the following construction for our isomorphism from $\mathbb{Z}_2 \times \mathbb{Z}_{12}$ to $\mathbb{Z}_6 \times \mathbb{Z}_4$: $$(x, y) \mapsto (x, y \pmod 3, y \pmod 4) \mapsto ((4(y\pmod 3) - 3x) \pmod 6, y \pmod 4)$$


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