Let $\pi$ be a set of prime numbers. A finite group is said to be a $\pi$-group if every prime that divides its order lies in $\pi$. If $G$ is finite, show that $G$ has a unique largest normal $\pi$-subgroup (which may be trivial and may be all of $G$).

What I did: suppose $|G|=p_1^{k_1}p_2^{k_2}\ldots p_n^{k_n}$. If $\pi$ contains a prime outside of the $p_i$'s, then the only $\pi$-subgroup is the trivial group. Then consider when $\pi$ contains a subset of the $p_i$'s, say $p_1,p_2,\ldots,p_m$. I don't know what I can say about a subgroup whose order is divisible by $p_1,p_2,\ldots,p_m$.


There exists a maximal normal $\pi$-subgroup of $G$ since $G$ is finite. If $H$ and $K$ are two such subgroups what can you say about $HK$?

  • $\begingroup$ Let's see... $HK$ is a normal subgroup since $H$ and $K$ are normal subgroups. It is larger than $H$ and $K$. What is its order? We have $HK = \cup_{k\in K}Hk$, so it is the union of some cosets of $H$, and its order is divisible by $|H|$. But why are all its divisors in $\pi$? $\endgroup$ – Paul S. May 26 '13 at 2:14
  • $\begingroup$ Okay thanks Serkan, that finishes the problem then. Is that last result hard to prove? (Should I try it on my own, or should I look it up somewhere?) $\endgroup$ – Paul S. May 26 '13 at 2:19
  • $\begingroup$ It's relatively simple to prove. you should try it. $\endgroup$ – Ted May 26 '13 at 2:21

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