# Direct product decomposition of the group of complex roots of unity

I'm studying $$p$$-adic numbers (Robert's "A course in $$p$$-adic analysis) and, at page 41, the author states that, for every prime $$p$$, the group $$\mu$$ of all complex roots of unity has a direct product decomposition in terms of $$\mu_{(p)}$$, the group of roots of unity of order prime to $$p$$, and $$\mu_{p^{\infty}}$$, the group of roots of unity having order a power of $$p$$; explicitly $$\mu=\mu_{(p)}\cdot\mu_{p^{\infty}}.$$

I don't understand how I can prove it: if $$z\in \mu$$ and $$m\ge2$$ (the case $$m=1$$ is obvious), maybe I can use the exponential form of complex numbers $$z=exp(2\pi i\frac{k}{m})$$, but here I stuck.

What I found in the web until now it sufficies to observe that $$\mu$$ is abelian torsion and apply the corresponding classification theorem, right?

Note: however, I would like a direct proof.

• If you know that classification theorem, then yes – Max Mar 21 at 20:35
• Actually I don't know it... I read about it in the web after I posted the question... – LBJFS Mar 21 at 20:37
• I was looking for a "direct" answer, whenever possible – LBJFS Mar 21 at 20:40

Indeed, clearly the two subgroups in question have trivial intersection, and they're both normal since $$\mu$$ is abelian. Moreover if $$x\in \mu$$ has order $$n=p^\alpha m$$ with $$m\land p = 1$$, then write $$um + vp^\alpha = 1$$ with Bezout's theorem, so that $$x= x^{um+vp^\alpha} = x^{um}x^{vp^\alpha}$$ and you can easily check that $$x^{um}$$ has order $$p^\alpha$$ and $$x^{vp^\alpha}$$ has order $$m$$, that is, order prime with $$p$$.