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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?

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  • $\begingroup$ 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 $U(15)$). $\endgroup$ – Adam Higgins Jan 15 at 15:41
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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.

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