# Finding the group of invertible elements in the ring $\Bbb Z_{15}$

I need to find the group of invertible elements of $$\mathbb{Z}_{15}$$. So the invertible elements here are $$U(15)= \left\{1, 2, 4, 7, 8, 11, 13, 14\right\}.$$

By chinese theorem $$U(15)=U(5)\times U(3)$$.

I have found the generators, $$a=2$$ and $$b=7$$ So, $$a^{2}=4$$, $$a^{3}=8$$, $$ab=14$$, $$a^{2}b=13$$, $$a^{3}b=11$$. So here, $$a$$ takes order 3, and $$b$$ order 1

This is enough to say, that $$U(15)$$ isomorphic to $$\mathbb{Z}_4\times\mathbb{Z}_2$$ or no? Is my reasoning is correct and sufficent?

• The order of $2$ is not three since $2^{3} \neq 2$ in $\mathbb{Z}$. Similarly the order of $7$ is not one (the only element of order one is $1$). In fact the order of $2$ is $4$, and the order of $7$ is $4$. You are correct that $2,7$ generate $\mathbb{Z}_{15}^{\times}$ (what you call $U(15)$). – Adam Higgins Jan 15 at 15:41

## 1 Answer

You are correct that $$\mathbb{Z}_{15}^{\times} \cong \mathbb{Z}_4 \times \mathbb{Z}_2$$, but your reasoning is wrong. Firstly the trivial way to solve this is to note that $$\mathbb{Z}_{15}^{\times}$$ is a non-cyclic Abelian group of order 8. Since the only Abelian groups of order 8 are $$\mathbb{Z}_2^{3}$$, and $$\mathbb{Z}_{4} \times \mathbb{Z}_2$$ we would be done.

However on a more basic level, as you noted, $$\mathbb{Z}_{15}^{\times}$$ is an Abelian group generated by $$a = 2, b = 7$$. However $$\operatorname{ord}(a) = 4, \operatorname{ord}(b) = 4$$. But note also that $$a^{2}b^{2} = 1$$. Hence $$\mathbb{Z}_{15}^{\times}$$ has the presentation

$$\mathbb{Z}_{15}^{\times} \cong \left< a, b \mid ab-ba, a^4, b^4, a^{2}b^{2} \right>$$

This might allow you to see that $$\mathbb{Z}_{15}^{\times}$$ is also generated by $$a = 2$$ and $$c = ab = 14$$, with $$\operatorname{ord}(c) = 2$$. Hence

$$\mathbb{Z}_{15}^{\times} \cong \left< a, b \mid ab-ba, a^4, b^4, a^{2}b^{2} \right> \cong \left< a, c \mid ac-ca, a^4, c^2\right> \cong \left \times \left< c \mid c^2 \right>,$$

and so we're done.

Essentially we have have noted here is that all of the units (invertible elements) in $$\mathbb{Z}_{15}$$ can be written as $$\pm 2^{k}$$ for $$k=0,1,2,3$$, and that every such element is a unit.