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.

Any hint is appreciated. Thank you in advance for your help.

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.

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

If you want a direct proof, the proof of the classification theorem is not so hard anyways.

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 $p$.

The decomposition follows.


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