Prove that $(\Bbb{Z}/8)^∗\cong \Bbb{Z}/2×\Bbb{Z}/2$ and $(\Bbb{Z}/9)^∗\cong \Bbb{Z}/6$

Prove that $$(\Bbb{Z}/8)^∗\cong \Bbb{Z}/2×\Bbb{Z}/2$$ and $$(\Bbb{Z}/9)^∗\cong \Bbb{Z}/6$$

Is there any way to do this using the Automorphism group or otherwise?

• For any odd $n$, $8|n^{2}-1$. Also, we can check that $2$ is a primitive root mod 9. – Seewoo Lee Nov 8 '17 at 9:51

Otherwise:

• $(\mathbf Z/8\mathbf Z)^\times$ has order $4$ and is not cyclic since all its elements have order $2$ (except $1$).
• $(\mathbf Z/9\mathbf Z)^\times$ has order $6$ and is cyclic: $2$ has order $6$ (and also $2^5=5$).

The group $(\mathbb{Z}/8)^{\ast}$ has $\phi(8)=4$ elements. Since we know that it is not cyclic, e.g.

Why multiplicative group $\mathbb{Z}_n^*$ is not cyclic for $n = 2^k$ and $k \ge 3$

it must be the other group of order $4$, namely $\mathbb{Z}/2\times \mathbb{Z}/2$. The second group has $\phi(9)=6$ elements and is abelian, hence it must be $\mathbb{Z}/6$. Actually, we know exactly when these abelian groups are cyclic:

When is the group of units in $\mathbb{Z}_n$ cyclic?

If you have to work for specific small number like $8$ and $9$ one can enumerate the numbers coprime to them.

IN the case $n=8$, the units are $\{1,3,5,7\}$ which is a group for multiplication modulo $8$ with every element having $1$ as square mod 8. ANd the product of two elements from 3,5,7 is the third. Then it is clear that it is the Klein's 4-group.

You can work out the case $n=9$ and verify the units form a cyclic group of order $6$. If you want a general theory you can consult, for example Ireland and Rosen's Invitation to Number Theory.

Since we are dealing with "small" groups, we can solve this problem with a bit of brute force.

Let's start making a multiplication table for $\Bbb Z_8^\times$: $$\begin{array}{|c|c|c|c|c|} \hline &1&3&5&7\\ \hline 1&1&3&5&7\\ \hline 3&3&1&7&5\\ \hline 5&5&7&1&3\\ \hline 7&7&5&3&1\\ \hline \end{array}$$

Note that there are no elements of order $4$. You can define $$f:(\Bbb Z_8^\times,\cdot)\to(\Bbb Z_2\times\Bbb Z_2,+)$$ in the following way (it is not unique): $f(1)=(0,0)$, $f(3)=(1,0)$, $f(5)=(0,1)$, $f(7)=(1,1)$.

It is easy to see that this function is an isomorphism.

Can you try it for the other pair of groups?

• I have googled for thick lines in tables in MathJax, with no results. – ajotatxe Nov 8 '17 at 10:11