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Let $G$ be a group of order $pn$ , where $p$ is a prime and $p>n$ , then is it true that any subgroup of order $p$ is normal in $G$ ? ( I know that any subgroup of index smallest prime dividing order of the group would be normal , but this thing is far away from it . Please help . Thanks in advance )

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  • $\begingroup$ Yes, this follows from Sylow's Theorem. The number of Sylow $p$-subgroups is congruent to $1$ mod $p$ and also has to divide $n$. So if $p>n$ then there is a unique Sylow $p$-subgroup which must therefore be normal in $G$. $\endgroup$ – Derek Holt Feb 2 '16 at 16:24
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Here is a very elementary proof, which uses Lagrange's Theorem, but not Sylow's Theorem.

It is enough to prove that any two subgroups $P,Q$ of order $p$ are equal, because then we must have $gPg^{-1}=P$ for all $g \in G$, so $P$ is normal.

So suppose that $P \ne Q$. Then $P \cap Q = \{1 \}$ by Lagrange. They are both cyclic so $P = \{x^i : 0 \le i < p \}$ and $Q = \{y^i : 0 \le i < p \}$ for some $x,y$.

Since $|G| = pn < p^2$, the elements $x^iy^j$ with $0 \le i,j < p$ cannot all be distinct, so there is an equality $x^iy^j=x^ky^l$ with $(i,j) \ne (k,l)$.

But then $x^{i-k} = y^{l-j}$, contradicting $P \cap Q = \{1 \}$.

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  • $\begingroup$ I was about to post a solution considering the product of such subgroups, but then I realized that the counting done here is essentially the same. $\endgroup$ – Tobias Kildetoft Feb 2 '16 at 16:46
  • $\begingroup$ +1, nicely done. You are doing, I believe, a simplified version of the fact that if $H, K \le G$, with $G$ finite, then $\lvert H K \rvert = \lvert H \rvert \cdot \lvert K \rvert / \lvert H \cap K \rvert$, where $H K = \{ h k : h \in H, k \in K \}$ is not necessarily a subgroup. $\endgroup$ – Andreas Caranti Feb 2 '16 at 17:53
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The first theorem of Sylow says that there is a subgroup of order $p$.

The second theorem of Sylow says that all subgroups of $G$ of order $p$ are conjugates.

And most important, the third theorem of Sylow says that the number of subgroups of order $p$ divides $n$ and is congruent to $1$ mod $p$.

Since $n<p$ there can be only one subgroup $H$ of order $p$ and the second theorem of Sylow implies that $g^{-1}Hg=H$ for all $g\in G$. Thus, $H$ is normal.

So, not only "every subgroup of order $p$ is normal", but there is exactly one subgroup of order $p$ and it is normal.

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