Consider the map $\phi\colon\mathbb Z\times\mathbb Z\to (\mathbb Z/20\mathbb Z)^*\colon (a,b)\mapsto \overline 7^a\overline{11}^b$. Determine the kernel and image of $\phi$. Then show that $(\mathbb Z/20\mathbb Z)^*$ is isomorphic to $\mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z$.

I’ve shown this is a group morphism. And if I’m correct, $\phi$ is surjective. Also, we have $\overline 7^4=\overline 1$ and $\overline {11}^2=\overline 1$. I calculated $7\cdot 11,7^2\cdot 11$ and $7^3\cdot 11$, and none of those are 1 modulo 20, so my conclusion is that $$ \operatorname{ker}(\phi)=\{(4m,2n):m,n\in\mathbb Z\}. $$ Now I don't know if I can use this kernel to prove the asked isomorphism. I know I can construct a map $\tau:\mathbb Z/2\mathbb Z\times\mathbb Z/4\mathbb Z\to(\mathbb Z/20\mathbb Z)^*$, such that this is a well-defined bijection, because $\operatorname{ord}(\overline 7)=4$ and $\operatorname{ord}(\overline{11})=2$. I just need to show then that this is a morphism.

However, without resorting to group tables, I’m not sure how to do this. Is it possible to determine the order of each element and its image, and if those correspond, can I then say that we indeed have an isomorphism? (note that I haven't shown yet that it is a morphism) I was also thinking of maybe using the fact that both groups are abelian, and for each order have the same number of elements, but we haven't shown that in class, so I'm not sure I can use that directly (otherwise, a proof would be much appreciated!)

  • $\begingroup$ $3$ is in the image though: Take $(a,b)=(-1,0)$ $\endgroup$ – Notone Mar 26 '18 at 12:06
  • $\begingroup$ O you're right, I messed up my calculations. I'll edit my question. $\endgroup$ – Sha Vuklia Mar 26 '18 at 12:09

This is a morphism because $$\phi((a,b)+(a',b')) = 7^{a+a'}11^{b+b'} = (7^a 11^b) (7^{a'} 11^{b'}) = \phi(a,b) \phi(a',b')$$

By the first isomorphism theorem, $(\Bbb Z \times \Bbb Z)/\ker \phi = \operatorname{im} \phi = (\Bbb Z/20\Bbb Z)^*$.

One then verifies that $(\Bbb Z \times \Bbb Z)/\ker \phi = (\Bbb Z \times \Bbb Z)/(4 \Bbb Z \times 2 \Bbb Z) = (\Bbb Z/4\Bbb Z) \times (\Bbb Z/2\Bbb Z)$.

Alternatively, $(\Bbb Z/20\Bbb Z)^* \cong ((\Bbb Z/4\Bbb Z) \times (\Bbb Z/5\Bbb Z))^* \cong (\Bbb Z/4\Bbb Z)^* \times (\Bbb Z/5\Bbb Z)^* \cong (\Bbb Z/2\Bbb Z) \times (\Bbb Z/4\Bbb Z)$.

  • $\begingroup$ Right, I didn't realise that $(\Bbb Z \times \Bbb Z)/(4 \Bbb Z \times 2 \Bbb Z) = (\Bbb Z/4\Bbb Z) \times (\Bbb Z/2\Bbb Z)$, but of course. Well, that makes it easy then. Thanks! $\endgroup$ – Sha Vuklia Apr 10 '18 at 19:31

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