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Let $G$ be a group of order $54$. Prove that there exists a normal subgroup of order $27.$ Is this normal subgroup unique?

Thoughts. Since $27$ divides $54$, by Lagrange's theorem we can not exclude the existence of such a subgroup.

Thank you in advance!

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    $\begingroup$ Have you heard of the Sylow theorems? $\endgroup$ – Arthur Jun 10 '16 at 5:56
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    $\begingroup$ Do you know either Cauchy's theorem or the Sylow theorems? If not, are you familiar with the fact that if all elements are of order dividing $2$ then the group is abelian? $\endgroup$ – Tobias Kildetoft Jun 10 '16 at 5:56
  • $\begingroup$ Another possible way forward is to consider the Cayley representation of the group. It allows us to view $G$ as a subgroup of $S_{54}$. Cauchy's theorem (applied to $p=2$) implies that $G$ is not contained in $A_{54}$. (This argument has been used many times on our site, but surprisingly I was unable to find one with the search engine). $\endgroup$ – Jyrki Lahtonen Jun 10 '16 at 6:07
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First Lagrange's theorem says that if $G$ has a subgroup $H$ then order of $H$ divides order of $G$. But converse may not hold.

In your case $|G|=54=3^3\times2 $. Now by applying Sylow theorem, we see that there is a Sylow-$3$ subgroup of $G$ having order $27$, say $H$. Also, $n_p$, the number of such a subgroup satisfies the conditions, $n_p|o(G)$ and $n_p\equiv1\pmod3.$ In this case only possible value for $n_p$ is $1$. Hence $H$ is a unique Sylow-$3$ subgroup. So $G$ has a normal subgroup of order $27$. (Unique Sylow-$p$ subgroups are normal.)

(This answer assumes the knowledge of Sylow theorems.)

Also note that one can use the fact that a subgroup of index $2$ is normal. Which makes things more simpler.

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    $\begingroup$ You should also add that the only number that is 1 mod 3 $\textit{and divides 2}$, is 1. $\endgroup$ – SquirtleSquad Jun 10 '16 at 6:08
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    $\begingroup$ A lot of the things here can be done a lot simpler. Subgroups of index $2$ are normal. And the important thing here is not that if there is just one Sylow subgroup then it is normal (this is not really something special about Sylow subgroups), but that if a Sylow subgroup is normal then there is just the one. $\endgroup$ – Tobias Kildetoft Jun 10 '16 at 6:11
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suppose that G is a group with order 2n where n is odd we can prove that G has a normal subgroup with order n .we know that G is isomorphic with H that H is a subgroup of symmetric group with order 2n now we have index of intersection of H and A2n in H less than index of A2n in S2n that equals to 2 since G has a member with order 2 (for example g) then H has a odd permutation then index of intersection of H and An in H is 2 now according to third teorm's isomorphic G is a subgroup with index 2 that is normal in G

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