I am an undergrad Mathematics student and I've been reading some additional literature for my lectures and came upon a quite short and seemingly elegant proof of the Second Sylow Theorem. Though, I already know other proofs of this, I would like to understand this one as well, maybe use it for my final exam. So, here goes:

Theorem: (2nd Sylow Thm.). In a finite group G, the Sylow p-subgroups are conjugate.

Proof: Let $P_1$ and $P_2$ be two Sylow p-subgroups. Then $G=P_1P_2+P_1x_2P_2+...+P_1x_sP_2$ (*). Let there be $b_i$ cosets of $P_2$ in $P_1x_iP_2$. Here $b_i=[x_i^{-1}P_1x_i : x_i^{-1}P_1x_i\cap P_2]$ and is $1$ or a power of $p$ (**). But $b_1+...+b_s=[G:P_2]$ is not a multiple of $p$. Hence for some $i, b_i=1$ and $x_i^{-1}P_1x_i=P_2$.

(from: Marshall Hall, Jr., The Theory of Groups, The Macmillian Company, New York, 1959.)

What I have yet to understand is the use of the "$+$" in (*). What does it mean exactly? I also have problems to understand the index at (**) for $b_i$, I'd highly appreciate an explanation of this as well, if possible.

  • $\begingroup$ Can you provide the source for this proof? This may help provide context for the notation used in your concern (*). $\endgroup$ – MightyTyGuy Jul 2 '18 at 15:47
  • 1
    $\begingroup$ I think the $+$ denotes disjoint union. But this proof is essentially the same as what I think of as the standard proof. You consider the action of $P_1$ by left multiplication on the set of all left cosets $xP_2$ of $P_2$. Since $P_1$ is a $p$-group, orbits have lenth $p^k$ for some $k \ge 0$, and since the number of cosets is coprime to $p$, there exists an orbit $xP_2$ of length $p^0=1$. So $P_1xP_2 = xP_2$ and hence $x^{-1}P_1x=P_2$. $\endgroup$ – Derek Holt Jul 2 '18 at 16:02

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