5

To show that show that $\mathbb{Z}^*_{15} \cong \mathbb{Z}^*_{16},$ that is, that $\mathbb{Z}^*_{15} $ and $ \mathbb{Z}^*_{16},$ are isomorphic, must I look at their respective Cayley tables and discover which element corresponds to which?

Or is there a smarter way to do it?

I tried the following method:

If $\alpha$ is an isomorphism from $\mathbb{Z}^*_{15}$ to $\mathbb{Z}^*_{16},$ then

\begin{align} \alpha(1) &= 1 \\ \alpha(4)^2 &= \alpha(4)\alpha(4) = \alpha(16) = \alpha(1) = 1 \\ \alpha(11)^2 &= 1 \\ \alpha(14)^2 &= 1 \\ \alpha(2)\alpha(8) &= 1 \\ \alpha(7)(13) &= 1 \end{align}

Then $\{\alpha(4), \alpha(11), \alpha(14)\} = \{7, 9, 15\},$ because $7^2 = 9^2 = 15^2 = 1$ in $\mathbb{Z}^*_{16}.$ Moreover, $\alpha(2) = 3, 11, 5,$ or $13.$ Since $\alpha(2)\alpha(2) = \alpha(4),$ and $3^2 = 11^2 = 5^2 = 13^2 = 9$ in $\mathbb{Z}^*_{16},$ $\alpha(4) = 9.$ However, I haven't been able to deduce any other values. I can't seem to eliminate possibilities via contradiction. Is there a smarter way to do it? Or should I continue via this method?

Shaun
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Rafael Vergnaud
  • 1,394

2 Answers2

6

Since $\vert (\Bbb Z / n\Bbb Z)^\times\vert = \phi(n)$, we see that both groups have order $8$. They are clearly abelian, so they must be isomorphic either to $$\Bbb Z/8\Bbb Z,$$ $$\Bbb Z /4\Bbb Z \times \Bbb Z / 2 \Bbb Z$$ or $$\Bbb Z / 2 \Bbb Z \times \Bbb Z / 2 \Bbb Z \times \Bbb Z / 2 \Bbb Z.$$

Computing the orders of the elements shows that for both groups there are none of order $8$ while not all are of order $2$, so we are done.

Shaun
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bsbb4
  • 3,751
4

There is a "smarter" method if you have some more results available. The group of units $U(pq)$ for two distinct odd primes $p$ and $q$ are isomorphic to $U(p)\times U(q)$, i.e., $$ U(15)\cong U(3)\times U(5)\cong C_2\times C_4. $$ And we have $$ U(2^n)\cong C_2\times C_{2^{n-2}} $$ in general, for all $n\ge 2$. Hence $U(16)\cong C_2\times C_4\cong U(15)$.

References:

Is my proof that $U_{pq}$ is not cyclic if $p$ and $q$ are distinct odd primes correct?

How to prove that $U_{2^n}$ is isomorphic as group to $\mathbb Z_2 \times \mathbb Z_{2^{n-2}}$ for $n \ge 3$?

Dietrich Burde
  • 140,055