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I would like to know if this statement (which i just met and suspiciously never realized before) and its proof are true:

Let $p$, $q$ be distinct primes and $G$ a group of order $n=p^{\alpha}q^{\beta}$. If $H$ is a $p$-Sylow subgroup of $G$ and $K$ a $q$-Sylow subgroup, then $G=HK$.

$\textit{Proof : }$ the order of the set $HK$ is $\frac{|H||K|}{|H\cap K|}=p^{\alpha}q^{\beta}=|G|$.

I'm a bit surprised by that.

Louis La Brocante
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  • Doesn't your computation of $|HK|$ require that either $H$ or $K$ be normal? – Robert Maschal Aug 01 '12 at 17:55
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    @RobertMaschal I don't think so. The formula gives the order of the set $HK$, even if it's not a group. In general the subgroup generated by $H$ and $K$ may have bigger order. – Louis La Brocante Aug 01 '12 at 17:58
  • @RobertMaschal Wouldn't they necessarily be normal in this group, because there would only be one Sylow $p$-subgroup, and one Sylow $q$-subgroup, so that the $H$ and $K$ complete contain their respective conjugacy classes? I may be off the mark a little here... – Emily Aug 01 '12 at 18:01
  • @EdGorcenski if $q>p$ then the $q$-sylow can't be normal. However the remark RobertMaschal only requires $1$ of the two Sylow subgroups to be normal. Now as i said in the last comment, I don't think we need that $HK$ is a subgroup for the order formula... Still this seems strange to me. – Louis La Brocante Aug 01 '12 at 18:04
  • Looking at it again, it seems fairly straightforward. Any $h \in H$ must be $|h| = p^m$, and for $k \in K$, $|k| = q^n$ by Lagrange. So then $H \bigcap K = \left{ \rm{id.} \right}$. I just fail to remember the conditions for $|HK| = |H||K|/|H \bigcap K|$. – Emily Aug 01 '12 at 18:10
  • ok, funny that i never figured out this before. ty Yuan. So we can then state from this that if $n=ab$ and $|H|=a$, $|K|=b$ and $H\cap K={e}$, then $G=HK$. – Louis La Brocante Aug 01 '12 at 18:24
  • @Rolando: this is true, and does suggest a certain amount of solvability (see Bender's proof of Burnside's theorem), but it only applies in somewhat limited circumstances:If $H\cap K ={e}$ only due to order considerations, then you have that $H$ is a Hall $\pi$-subgroup and $K$ is a $\pi$-complement. If a group actually has $p$-complements for all primes $p$ (much less for all sets of primes $\pi$), then it is solvable by Hall's theorem. – Jack Schmidt Aug 01 '12 at 18:29

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This is true. The computation of $|HK|$ only requires that $H$ and $K$ are subgroups. Explicitly, note that $$h_1 k_1 = h_2 k_2 \Leftrightarrow h_2^{-1} h_1 = k_2 k_1^{-1}$$

so this element must be an element of $H \cap K$, and conversely if $g \in H \cap K$ then $hk = hg^{-1} g k$. So each possible product in $HK$ occurs exactly $|H \cap K|$ times.

Qiaochu Yuan
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  • @Rolando, yes it is remarkable that the left side of the equation is a set and its order can be expressed in terms of subgroups – Nicky Hekster Aug 01 '12 at 21:59