If $\lvert\,G\,\rvert=p^{2}\cdot\,q$ where $q,p$ are primes, How can i show $G$ has a normal p-sylow subgroup ? I tried working with the sylow theorem but i cant reach any contradiction when $Np = q$ and $Nq = p,p^2$ Also tried searching the same question but couldn't find in Stack Exchange
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If the order of $G$ equals the product of two primes then we know that $G$ has a normal subgroup of prime order equal to the prime divisor because the group may be generated by that prime. – JohnColtraneisJC Apr 26 '18 at 12:52
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https://en.wikipedia.org/wiki/Sylow_theorems Theorem 3 – JohnColtraneisJC Apr 26 '18 at 12:53
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@JohnColtraneisJC okay, Sorry forgot to say, It need to be a p-syllow subgroup – Daniel Moraes Apr 26 '18 at 12:54
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@zkzkzkzkzk the number of q-sylow subgroups being denotated by Nq, from the 3rd theorem It has to divide $p^2$, so $ Nq= 1,p$ or $p^2$ – Daniel Moraes Apr 26 '18 at 13:06
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It should be $n_p=1,q,q^2$ and $n_q=1,p$. – lhf Apr 26 '18 at 13:06
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@lhf yes Sorry messed up notation haha – Daniel Moraes Apr 26 '18 at 13:07
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Is there any relation between $p$ and $q$? – Bernard Apr 26 '18 at 13:07
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@Bernard nop, they are primes, bezout theorem could be used as they are also relatively primes – Daniel Moraes Apr 26 '18 at 13:08
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I thought of something like $p<q<2p$, for instance. – Bernard Apr 26 '18 at 13:09
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@Bernard there is not something like this – Daniel Moraes Apr 26 '18 at 13:12
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This is false for $q=2, p=3$. The $3$-Sylow subgroups of $A_4$ are not normal: there are four copies of $A_3$ in $A_4$.
lhf
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2@user528821, that's not what your question says. Ask a separate question fixing this. – lhf Apr 26 '18 at 13:12
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