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