# is U14 isomorphic to U18?

• August 21st 2010, 07:10 AM
is U14 isomorphic to U18?
Let U14={1,3,5,9,11,13} , U18={1,5,7,11,13,17}

(Ui is the group of all invert elements for multiplication in Zi)

Is U14 isomorphic to U18?

After a LONG check, I found that all the orders of the elements match orders of elements in the second group, just got down after somebody told me it's not enough for isomorphism. Jeez, what IS enough then?!

Thanks :)
• August 21st 2010, 10:52 AM
PaulRS
We have $\mathbb{Z}^{\times}_{a\cdot b} \cong \mathbb{Z}^{\times}_{a}\times \mathbb{Z}^{\times}_{b}$ whenever $\text{gcd}(a,b)=1$

Thus:
$\mathbb{Z}^{\times}_{14} \cong \mathbb{Z}^{\times}_{2}\times \mathbb{Z}^{\times}_{7}$ and $\mathbb{Z}^{\times}_{18} \cong \mathbb{Z}^{\times}_{2}\times \mathbb{Z}^{\times}_{9}$

So it's enough to check that $\mathbb{Z}^{\times}_{7} \cong \mathbb{Z}^{\times}_{9}$ but this is true since both are cyclic groups of the same order. $(*)$

$(*)$ $\mathbb{Z}^{\times}_{n}$ is cyclic if and only if $n=2,4,p^k,2p^k$ where p is an odd prime.
In your particular case you can check that $<3>=\mathbb{Z}^{\times}_{7}$ and $<2>=\mathbb{Z}^{\times}_{9}$ ...

The reason why you were told that what you did was not enough is because you have to consider the "multiplication" between distinct elements... - if you found a function $f$ and showed that it is a bijection and $f(ab)=f(a)f(b)$ you'd be done...