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Suppose $G$ is a group of order $n$ . Let $p$ and $q$ be distinct primes both of which divide $n$. Can we say that $G$ has a subgroup of order $p\cdot q$?

Yanko
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user566574
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1 Answers1

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No. For example $A_5$ is a group of order $60=2^2\times3\times5$ but if you know this group is simple then you can easily check that it can't have any subgroups of order $15$.

Mark
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  • but simple means that the normal subgroups are trivial. What happens if you take the group generated by a permutation of order $5$ and a permutation of order $3$? – Yanko Feb 09 '19 at 16:46
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    @Yanko It depends what permutations you take, but for sure it is not going to be a subgroup of order $15$. Let's suppose $A_5$ has a subgroup of order $15$ and let's call it $H$. We can define an action of $A_5$ on the left cosets (which is a set of size $4$) by $g.xH=gxH$. That gives us a homomorphism $\phi:A_5\to S_{A_5/H}$which is defined by $\phi(g)(xH)=gxH$. The group $A_5$ is simple so the homomorphism must be either trivial or injective. But it's easy to see that both options fail. So we have a contradiction. – Mark Feb 09 '19 at 18:33
  • Actually, from here we conclude that if you take a $3$-cycle and a $5$-cycle then the generated subgroup is $A_5$ itself. – Mark Feb 09 '19 at 18:40