2
$\begingroup$

I have to solve this problem: If $G =\text{Drp}(G_p)$ where $G_p$ is a $p$-group, $\text{Drp}(G_p)$ denotes the direct product of the $p$-primary components of $G_p$, and if $H < G$, prove that $H =\text{Drp}(H\cap G_p)$.

I have tried to do it in this way:

  • $H\cap G_p$ is normal in $H$;
  • $(H\cap G_p)\cap\langle H\cap G_q: q\neq p\rangle = \{1\}$;

Now, how i can prove the third property of the direct product?

$\endgroup$
2
  • 1
    $\begingroup$ Is $Drp$ an standard notation? What does it mean? $\endgroup$
    – Pp..
    Commented Feb 8, 2015 at 11:39
  • $\begingroup$ It is a way to indicate the direct product of the primary components of G $\endgroup$
    – elisa
    Commented Feb 8, 2015 at 11:50

1 Answer 1

0
$\begingroup$

Proof

Start with $G$ given as $G=G_{p_1}\times \cdots \times G_{p_m}$ with $p_i\neq p_j$ for $i\neq j$. You need to show that for every $h=(h_1,\ldots,h_m)$ every element $(0, \ldots, h_i, \ldots,0)$ belongs to $H$. This is sufficient to prove what you want but also necessary: otherwise $H$ would not be a direct product.

To show that, define $a_i=|G|/|G_{p_i}|$: this number is a multiple of every $|G_{p_j}|$ with $j\neq i$; it is coprime to $p_i$ by construction and, hence, it is also coprime to the order of $h_i$ inside $G_{p_i}$. Using this, we get $h^{a_i}=(0, \ldots, h_i^{a_i}, \ldots, 0)$ belongs to $H$. Moreover, because $\mathrm{gcd}(a_i, |h_i|)=1$, we get that $h_i$ and $h^{a_i}$ generate the same cyclic subgroup of $G_{p_i}$: it follows that $(0, \ldots, h_i, \ldots, 0)$ is also in $H$, which is what we wanted to show.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .